Chapter 1. Variational Method

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1 Chapter 1 Variational Method Variational method is an important means in mathematical physics. It is used to study the extremes of functionals that are a kind of special variables. In this chapter, some essential knowledge and its applications are presented Functional and Its Extremal Problems The conception of functional First of all, some simplest examples are given in order to introduce the concept of functionals. Example 1. Suppose that two points x = and x = x 1 on the x axis are given, and y = y(x) is a function defined on the interval [,x 1 ] and has a continuous first derivative. Then the length of the curve y = y(x) is l[y(x)] = 1+y 2 dx. (1.1.1) As long as the functions y(x) is changed, say, to be another function y 1 (x), the length l[y(x)] will be accordingly changed to be l[y 1 (x)]. That is to say, the variable l defined by (1.1.1) depends on the whole function y(x). Definition 1. A function set having some common features is defined as a function class. 1

2 2 Mathematics for Physicists For instance, a set of functions that are continuous on the interval [,x 1 ] is denoted as C[,x 1 ]. A set of functions that have first continuous derivatives on the interval [,x 1 ] is called C 1 class functions on [,x 1 ], and is denoted as C 1 [,x 1 ]. In this analogy, a set of functions that have the n-th continuous derivatives on the interval [,x 1 ] is called C n class functions on [,x 1 ], and is denoted as C n [,x 1 ]. If a function y(x) hasthen-th continuous derivative in the interval [,x 1 ], it belongs to C n [,x 1 ], and is simply denoted as y(x) C n. This denotation is also used for multi-variable functions. For instance, z(x, y) C 2 (D) meansthat the function z(x, y) has the second continuous partial derivatives on domain D. Example 2. Suppose that D is a given domain in the xy plane, and a function z(x, y) C 1 (D). Then the area of the surface corresponding to the function is S[z(x, y)] = 1+z x 2 + z y 2 dxdy. (1.1.2) D Obviously, the variable S depends on the whole function z(x, y). Having given the examples above, we now introduce the concept of functional. Definition 2. Let R be a number domain, and Y be a given function class denoted as {y(x)}. For every function y(x) belonging to Y, there is a number J R. In this case, the variable J is defined as a functional of the function y(x), and denote as J = J[y(x)]. The function class is called the definition domain of the functional J[y(x)]], which sometimes is also called the admissible function of the functional. In short, a functional is a map from a function set Y to a number domain R. Each argument of the map is a function, and each function y(x) belonging to Y is called an admissible function. It is easy for readers to give similar definition of the functionals depending on multifunctions. According to this definition, the integral (1.1.1) and (1.1.2) are respectively the functionals at C 1 [,x 1 ]andc 1 (D).

3 Variational Method 3 Example 3. The Fourier transformation F [f(x)] = f(x)e ikx dx is a functional. It has a parameter k. Once the k value is set, for each function f(x), the value of the functional is determined. The function f(x) is the admissible one of the functional. Alternatively, this functional can be regarded as a function of argument k, andis denoted as F (k). However, the value of F (k) does not come from the definition of a function, but from the integral above which relies on the concrete form of f(x) The extremes of functionals 1. Functional extremes The basic problem of the variational method is with respect to the functional extremes. For instance, in 1696 J. Bernoulli raised the brachistochrone problem as follows, which significantly promoted the variational method. It is to determine a curve passing through two fixed points A and B in a plane perpendicular to the earth s surface such that a point mass moving along the curve under the influence of gravity travels from A and B in the shortest possible time (all possible resistances such as friction are neglected). From the viewpoint of distance, the straight line connecting A and B is the shortest one. However, in the initial period, the dropping could not gain fast speed, so that the time spent would not be the shortest. In Fig. 1.1, A is the origin, Ax is a horizontal axis and Ay is a vertical axis downwards. Let y = y(x) be a smooth curve connecting points A(0, 0) and B(a, b). The point mass slips downwards along this curve. Since the curve is smooth, the mass does not feel a force along the tangential direction of the curve. The normal force changes the direction of the velocity but not its magnitude. Since the initial velocity is zero, the speed of the mass when it moves to a point M(x, y) is where g is the acceleration of gravity. v = 2gy, (1.1.3)

4 4 Mathematics for Physicists A(0, 0) x M(x, y) y B(a, b) Fig Let S be the length of the curve, and dt the differential with respect to time. Then, v = ds 1+y dt = 2 dx. (1.1.4) dt Therefore, 1+y dt = 2 dx 1+y 2 = dx. (1.1.5) v 2gy Consequently, the time T needed for the mass to slip along the curve y = y(x) froma to B is a 1+y 2 T [y(x)] = dx. (1.1.6) 2gy 0 Thus, the formal statement of the barchistochrone problem is that it is to find a curve y = y(x) satisfying the boundary conditions y(0) = 0, y(a) = b (1.1.7) such that the functional T [y(x)] reaches its minimum. The functional extremes defined below are similar to those of functions. Before doing so, the concept of ε-neighborhood of curve y = y(x) is defined. Definition 3. The ε-neighborhood of a curve y = y(x) defined on [,x 1 ] means all the possible curves y = y 1 (x) satisfying the condition y 1 (x) y(x) ε (1.1.8) within the whole interval [,x 1 ] (Fig. 1.2). It is also termed that y 1 (x) has an ε-proximity of zero order to y(x). For C 1 class

5 Variational Method 5 y y = y 1 (x) y = y (x)+ε y = y (x) y = y (x) ε x 1 Fig functions, the ε-neighborhood of y = y(x) requires not only (1.1.8), but also the following inequality: y 1 (x) y (x) ε. (1.1.9) In this case, it is said that y 1 (x) has an ε-proximity of first order to y(x). Similarly, the concept of the ε-proximity of n-th order can be defined. Given a curve y = y 0 (x), the set including all the curves y 0 (x) that have the ε-proximity of n-th order is called the ε-neighborhood of n-th order of y 0 (x). Definition 4. Let J[y(x)] be a functional of a certain function class {y(x)}. Suppose that function y 1 (x) belongs to function class {y(x)} which has an ε-proximity of a certain order to y 0 (x), where ε>0. If the functional value at y 0 (x) is not less than that at any such y 1 (x), i.e., J[y 0 (x)] J 1 [y 1 (x)], (1.1.10) then it is said that the functional J[y(x)] reaches its relative maximum at y 0 (x). If (1.1.10) is always valid when the set {y(x)} is the ε-neighborhood of zero order of the function y 0 (x), then J[y(x)] is said to reach its strong relative maximum at y 0 (x), or simply strong maximum. If (1.1.10) is valid when the set {y(x)} is the ε-neighborhood of first order of the function y 0 (x), then J[y(x)] is x

6 6 Mathematics for Physicists said to reach its weak relative maximum at y 0 (x), or simply weak maximum. In a similar way, if the symbol in (1.1.10) is changed to be, the conceptions relative minimum, strong minimum and weak minimum can be defined. In the following, all of the maxima and minima will be referred to as extremes. The function y 0 (x) that makes the functional J reaches its extremes is termed as extremal curves. 2. Basic variational lemma For later usage, let us now introduce the basic variational lemma. Lemma 1. Let f(x) be a continuous function on [,x 1 ]. For any function η(x) belonging to C 2 class functions and being zero at two points x = and x = x 1, the integral f(x)η(x)dx =0, (1.1.11) then f(x) 0. Proof. We use reduction ad absurdum. Suppose that at point ξ( <ξ<x 1 ), f(ξ) 0, e.g., f(ξ) > 0. Since f(x) isacontinuous one, there must be a neighborhood of ξ, ξ 1 <ξ<ξ 2.Inthis neighborhood f(x) > 0. Now we construct a function as follows. 0 x ξ 1, η(x) = (x ξ 1 ) 4 (x ξ 2 ) 4 ξ 1 x ξ 2, 0 ξ 2 x x 1. It meets all the conditions for the function η(x) in the lemma. As a matter of fact, it is easily verified that η( )=η(x 1 ) = 0, and the first and second derivatives of (x ξ 1 ) 4 (x ξ 2 ) 4 with respect to x exist and are zero at x = ξ 1 and x = ξ 2. Outside the interval [ξ 1,ξ 2 ], η(x) is always zero. Hence the function itself and its first and second derivatives are continuous on the whole [,x 1 ]. For such an η(x), ξ2 f(x)η(x)dx = f(x)η(x)dx >0, (1.1.12) ξ 1 i.e., the integral is not zero. This contradicts to the premise (1.1.11). Hence there should be f(x) =0.

7 Variational Method 7 Please note that the conclusion of the lemma stands under the conditions that the function η(x)isanyc 2 class one being zero at x = and x = x 1 and satisfying (1.1.11). If some of the conditions are not met, one cannot gain the conclusion f(x) = 0 from only (1.1.11). Lemma 2. Suppose a function f(x, y) is continuous in region D. For any function η(x, y) belonging to C 2 class functions and being zero at the boundary of the region, the integral f(x, y)η(x, y)dxdy =0, (1.1.13) then f(x, y) =0. D Proof. Suppose that at point (a, b) within the region f(a, b) > 0. Then it must be that in a circle K of radius ρ centered on (a, b), f(x, y) > 0. Now we construct a function as follows η(x, y) = { 0, [(x a) 2 +(y b) 2 ρ 2 ] 4, (x a) 2 +(y b) 2 ρ 2, (x a) 2 +(y b) 2 <ρ 2. Obviously, this function η(x, y) meets the conditions required in the lemma. However, we have f(x, y)η(x, y)dxdy = f(x, y)η(x, y)dxdy > 0. D K This contradicts to (1.1.13). The conclusion is that f(x, y) should be zero everywhere on D. Apparently, lemmas 1 and 2 are for 1- and 2-dimensional cases, respectively. In the cases of triple and multiple integrals, there are similar lemmas The Variational of Functionals and the Simplest Euler Equation The variational of functionals Let a functional be of the following form: J[y(x)] = F (x, y, y )dx, (1.2.1)

8 8 Mathematics for Physicists where F is a continuous function of three arguments x, y, y, and has continuous partial derivatives of second order. In (1.2.1), y(x) C 2 is in turn a function of x. Before discussing how to calculate such kind of functionals, the concept of variational of the functionals should be given, since it is of importance in searching the extremal curve of the functionals. This concept is similar to the derivative in studying the extremes of functions. Suppose that a function y(x) varies slightly, and becomes y(x) + δy(x), where δy(x) means a function, instead of δ multiplied by y(x). The function δy(x) is termed as the variational of y(x). Let us see what is the increment of the functional (1.2.1) provided that both y and y + δy are its admissible functions. Hereafter, we denote δy =(δy(x)) for the sake of convenience. The increment is defined as where J = J[y + δy] J[y] = = [F (x, y + δy, y + δy ) F (x, y, y )]dx [ ] F F x1 δy + y y δy dx + (ε 1 δy + ε 2 δy )dx, (1.2.2) lim ε δy 0,δy 1 =0, 0 lim ε δy 0,δy 2 =0. (1.2.3) 0 The first integral in the second line of (1.2.2) is called the variational of the functional J[y] at point y(x), denoted as δj: δj = ( ) F F δy + y y δy dx. (1.2.4) Apparently, δj is linear with respect to δy. Let us inspect the difference of the increment and variational, J δj. Denoting δy = { max δy(x), δy (x) }, (1.2.5) x x 1

9 Variational Method 9 we have J δj = (ε 1 δy + ε 2 εy )dx max( ε 1 + ε 2 )(x 1 ) δy. (1.2.6) It is seen from (1.2.3) that J δj is the higher-order infinitesimal of δy. Thus, the variational δj is named as the linear main part of functional increment J. This coincides with the case in differential calculus, where the differential of a function is the linear main part of the infinitesimal increment of the function. For example, let J[y] = x 1 (y 2 + y 2 )dx. Then J = J[y + δy] J[y] = (2yδy +2y δy )dx + [(δy) 2 +(δy ) 2 ]dx. (1.2.7) In this equation, the first integral is linear to δy, and the second one is a higher-order infinitesimal of δy. Thus δj = (2yδy +2y δy )dx. (1.2.8) Surely, this result can be achieved directly from (1.2.4). The differential of a function f(x) can be written as its derivative with respect to a parameter α: df = d dα f(x + α x) α=0. (1.2.9) Inthesameway,thevariationalδJ of a functional J[y] can be written as its derivative with respect to a parameter α, namely, δj = d dα J[y + αδy] α=0. (1.2.10) As a matter of fact, d x1 ( ) F dα J[y + αδy] F α=0 = δy + y y δy dx = δj. (1.2.11) The variational of a functional (1.2.4) can also be put into the following form: δj = δfdx. (1.2.12)

10 10 Mathematics for Physicists In this way, the variational of a functional is written as an integral. The integrand F is regarded as a functional depending on functions y and y at fixed x. By comparison of (1.2.12) and (1.2.4), it is seen that δf = F F δy + y y δy. (1.2.13) This reminds us that the differential of a binary function f = f(x, y) is df = f f xdx + ydy. Evidently, the first order variational has the same form as the first order differential. One needs only distinguish that the former is the function of a function, while the latter is the function of an argument. We use letter d to present differential and Greek letter δ to present variational. When a functional F depends on a certain function y, the partial derivative F/ y with respect to y is manipulated just as that of a function f with respect to one of its arguments x. Next, let us see the second variational. When δj is regarded as a functional, and δf is also a functional in the same sense as F,we have δ 2 J = Application of (1.2.13) leads to δ 2 F = δf y δy + δf y δy δ 2 F dx. (1.2.14) = 2 F y 2 (δy) F y y δy δy + 2 F y 2 (δy ) 2. This can be compared to the second differential of a binary function: d 2 f = 2 f x 2 (dx) f x y dxdy + 2 f y 2 (dy)2 ( = dx x +dy ) 2 f. y It is seen that the manipulation of variation is just the same as that of function differentials. Similarly, it is easy to show that the variationals of the product of two functions and of the inverse of a function are δ(f 1 F 2 )=F 2 δf 1 + F 1 δf 2 (1.2.15)

11 Variational Method 11 and ( ) 1 δ F = 1 δf, (1.2.16) F 2 respectively. For functionals, one has variationals with the same forms as these two equations. We show them in the following. Suppose that there are two functionals and J 1 = b1 a 1 F 1 (x 1,y 1,y 1)dx 1 b2 J 2 = F 2 (x 2,y 2,y 2 )dx 2. a 2 The variational of their product is δ(j 1 J 2 )= = = b2 a 2 dx 2 b2 a 2 dx 2 b2 b1 a 1 dx 1 δ(f 1 F 2 ) b1 a 2 F 2 dx 2 a 1 dx 1 (F 2 δf 1 + F 1 δf 2 ) b1 a 1 dx 1 δf 1 + b2 a 2 δf 2 dx 2 b1 a 1 dx 1 F 1 = J 2 δj 1 + J 1 δj 2. (1.2.17) The variational of the inverse of a functional being of the form of (1.2.1) is as follows. ( ) 1 1 δ = J x1 F (x, y + δy, y + δy )dx 1 F (x, y, y )dx 1 = J + δj 1 J = 1 ( ) 1 J 1+δJ/J 1. We denote { δy max δy, δy }, ε 1 = max x x 1 { } F ε 2 = max x x 1 y. x x 1 { F y },

12 12 Mathematics for Physicists Then δj x 1 (ε 1 + ε 2 ) δy, i.e., δj is an infinitesimal with one order higher than δy. Thus, by expanding Taylor series to the first order term, we have [ ( 1+ δj J ( ) 1 δ = 1 J J The conclusion is that ) 1 1] = 1 (1 δjj ) J 1 δ ( ) 1 J = 1 J 2 δj = 1 δj. (1.2.18) J The simplest Euler equation 1. Euler equation Theorem 1. Let y(x) be the extremal curve of functional (1.2.1). Then the variational of the functional at y = y(x) is δj =0. (1.2.19) Proof. Because we are talking about the necessary condition that the extremal curve y(x) must satisfy, we select a function y* witha special form to compare J[y] withj[y ]. For example, y (x) =y(x)+αδy(x). (1.2.20) Here α is a small parameter, so that y (x) isofgivenε-proximity to y(x), and δy(x) isanyc 2 class function. By the premise, the function of α, J[α] =J[y ]=J[y + αδy], takes its extreme at α = 0. Therefore, (1.2.10) leads to δj = d dα J[y + αδy] α=0 =0. (1.2.21) That is to say, the variational of J[y] aty = y(x) isδj =0. Theorem 2. Suppose that y(x) is the extremal curve of functional (1.2.1). Then the function y = y(x) necessarily observes the following differential equation: F y d dx F y = 0 (1.2.22) or F y y y + F yy y + F xy F y =0, (1.2.23)

13 Variational Method 13 where F yy is the second partial derivative of F with respect to y and y, and so on. Proof. Now we take a function of a more special form y = y + αδy so as to compare J[y] withj[y ]. It is required that y and y take the same values at points x = and x = x 1, i.e., y ( )= y( ),y (x 1 )=y(x 1 ), which results in δy( )=0, δy(x 1 )=0. (1.2.24) Now we take integration by parts for the second integrand in (1.2.4). This brings δj = + (F y δy + F y δy )dx = F y δy(x) x 1 The first term is zero by (1.2.24). As a result, δj = (F y d dx F y )δydx. (1.2.25) (F y d dx F y )δydx. (1.2.26) Finally, it is known by the basic variational lemma that the extremal curve y(x) obeys the differential equation (1.2.22). The ordinary differential equation of second order (1.2.22) or (1.2.23) is called the Euler equation of extremal problem of functional (1.2.1). Its general solution contains two arbitrary constants. Usually, when discussing functional extremes, there may be some additional conditions on the values of the admissible function at boundaries and x 1. For example, the two ends are fixed, which means that y( )=y 0, y(x 1 )=y 1, (1.2.27) where,y 0,x 1,y 1 are constants. By the boundary conditions (1.2.27), one is able to determine the two constants in the solution of the Euler equation. Thus the extremal curve of functional (1.2.1) is obtained.

14 14 Mathematics for Physicists 2. Two special cases In this subsection the following two specials cases of Euler equation are presented. The first case is that F is independent of y, i.e., F = F (x, y ). Because F y = 0, Euler equation becomes d dx F y =0. (1.2.28) This brings the first integral F y (x, y )=C 1. (1.2.29) This is a differential equation not explicitly containing y. From this equation y can be solved, and then its integration brings the solution. Sometimes, properly selecting parameters is helpful. Example 1. Find the extremal curve of functional 1+y 2 J = dx x υ x which satisfies the boundary conditions y( )=y 0, y(x 1 )=y 1. Solution. Because F is independent of y, the first integral of the Euler equation is F y = Its integration brought to y x 1+y 2 = C 1. x 2 +(y C 2 ) 2 =1/C 2 1, (C 1 0). This is a set of circles centered at the y axis. The undetermined constants C 1 and C 2 can be solved from boundary conditions. The second case is that F is independent of variable x, i.e., F = F (y,y ). Because F xy = 0, the Euler equation, with the help of (1.2.23), becomes y F y y + y F y y F y =0. (1.2.30)

15 Variational Method 15 On the other hand, d dx (F y F y )= y (y F y y + y F yy F y )=0. (1.2.31) Subsequently, (1.2.30) has the first integral: F y F y = C 1. (1.2.32) Integrating once more gives the possible extremal curves. The integral constants are determined by boundary conditions. Example 2. In the previous section, we have restated the barchistochrone problem as to find the extremal curves of the functional a 1+y 2 T [y(x)] = dx 0 2gy which meets boundary conditions y(0) = 0, y(a) =b. Please find the solution. Solution. Because F is independent of x, the Euler equation has the first integral (1.2.32), which leads to 1+y 2 2gy y 1 2gy y 1+y 2 = C. With the denotation C 1 =1/2gC 2, it is simplified to be y(1 + y 2 )=C 1. We solve this equation with parameter method. Let y =cot(θ/2). Then the equation becomes y = C 1 1+y 2 = C 1 sin 2 θ 2 = C 1 (1 cos θ). 2 As a consequence, dx = dy y = C 1 sin(θ/2) cos(θ/2)dθ cot(θ/2) = C 1 (1 cos θ)dθ. 2 Integrating the last expression gives x = C 1 2 (θ sin θ). Hence, the required curve is x = C 1 2 (θ sin θ)+c 2, y = C 1 (1 cos θ). 2

16 16 Mathematics for Physicists In terms of the boundary conditions y(0) = 0, one obtains C 2 =0. This is a class of cycloids with their radius being C 1 /2. The constant C 1 can be determined by the condition of the cycloid s value at point B. Hence, the solved barchistochrone is a cycloid through points A and B. In the end of this section, we stress one point. Just as that the root of equation f (x) = 0 is merely the necessary condition of y = f(x) taking its extremes, the solutions of the Euler equation are the necessary but not sufficient condition of functional (1.2.1) taking its extremes. Nevertheless, in practical applications to the problems raised from such fields as engineering, mechanics, physics and so on, the solved extremal function y = y(x) often happens to be the one that makes the functional J[y(x)] take its extremes. In the following sections, the equations obtained through the variational method are also the necessary conditions of functionals taking their extremes The Cases of Multifunctions and Multivariates Multifunctions When a functional depends on more than one admissible function, the corresponding Euler equation can also be achieved in a similar way as above. Here we discuss the case of two admissible functions: x J[y(x),z(x)] = F (x, y, y,z,z )dx. (1.3.1) Suppose that y(x),z(x) C 2, and there exist all possible second partial derivatives of F with respect to its two admissible functions. If functional (1.3.1) takes its extremes at y(x) andz(x), let us find the necessary conditions that y(x) andz(x) mustsatisfy. The following routine is in fact the same as the case of one admissible function. This time, the comparison curves are taken as y = y + αδy and z = z + αδz, whereα is a small parameter and δy( )=δy(x 1 )=δz( )=δz(x 1 )=0. (1.3.2)

17 Variational Method 17 Substituting y and z into (1.3.1), we represent J as being a function of the parameter α: J(α) = F (x, y + αδy, y + αδy,z+ αδz, z + αδz )dx. J(α) takesitsextremesatα = 0. Therefore, We regard J (α) α=0 = (F y δy + F y δy + F z δz + F z δz )dx =0. δj = (F y δy + F y δy + F z δz + F z δz )dx (1.3.3) as the variational of functional (1.3.1). It is seen that the necessary conditions that (1.3.1) takes for its extremes at y(x) andz(x) are that its variational at these two functions should be zero. Now we take integration by parts for the second and fourth terms in (1.3.3) and use (1.3.2) to get δj =[F y δy + F z δz] x 1 [( + F y d ) ( dx F y δy + F z d ) ] dx F z δz dx = ( F y d dx F y ) δydx + Specially, when δz =0,wehave ( F z d ) dx F z δzdx =0. (1.3.4) (F y d dx F y )δydx =0. That is, by the basic variational lemma, the function in the parenthesis should be zero. When δz = 0, we obtain another equation. In summary, the functions y(x) andz(x) that make functional (1.3.1) take its extremes should satisfy the following differential equations of second order: F y d dx F y =0, F z d (1.3.5) dx F z =0.

18 18 Mathematics for Physicists These equations are combined with given boundary conditions, so that the extremal curves can be solved. For example, in an extremal problem with fixed boundaries, the conditions are y( )=y 0, y(x 1 )=y 1, (1.3.6) z( )=z 0, z(x 1 )=z 1. In the present case, one can image that x is a function of y and z, and there is a curve in a three-dimensional Cartesian system such that it in fact reflects the two functions y(x) andz(x). Example 1. Find the extremal curves of the functional J = π/2 0 (y 2 + z 2 +2yz)dx, which satisfies the boundary conditions ( π ) ( π ) y(0) = 0, y =1; z(0) = 0, z = Solution. F = y 2 + z 2 +2yz. By (1.3.5), the corresponding Euler equations are { y z =0, z y =0. To solve the equations, the function z is eliminated: y (4) y =0. Its general solution is y = C 1 e x + C 2 e x + C 3 cos x + C 4 sin x. By z = y,weget z = C 1 e x + C 2 e x C 3 cos x C 4 sin x. Now using the boundary conditions, the constants are C 1 =0, C 2 =0, C 3 =0, C 4 =1. Finally, the required curves are y =sinx, z = sin x.

19 Variational Method 19 If there are M admissible functions and one variable: y α (x), (α = 1, 2,...,M), we use {y α (x)} to denote the function set. The functional is expressed as x J[{y α (x)}] = F (x, {y α (x)}, {y α (x)})dx. (1.3.7) Then it is natural to extend (1.3.5) to the following M equations: F d y α dx Multivariates F y α =0, (α =1, 2,...,M). (1.3.8) We turn to the cases of multivariates. For simplicity without loss of generality, we take a functional of a binary function u(x, y) asan example: J[u] = F (x, y, u, u x,u y )dxdy. (1.3.9) D It involves double integrals. Here u(x, y) C 2 (D) andd is a given region in the xy plane. The function F is assumed to have second partial derivatives with respect to all of its argument and admissible functions. Suppose that u = u(x, y) has been the extremal surface that makes the functional (1.3.9) take its extremes. In order to find the necessary conditions it meets, we once more select comparison function as before in the following form: u = u(x, y)+αη(x, y), where η(x, y) is an arbitrary C 2 class function and satisfies boundary condition η(x, y) C =0,andα is a small parameter. Evidently, u and u take the same value at the boundary. Substituting u into (1.3.9), we have J[u ]= F (x, y, u + αη, p + αη x,q+ αη y )dxdy = J(α). D

20 20 Mathematics for Physicists For the sake of simplicity, here we have denoted p = u x and q = u y.so,theextremalsurfaceu(x, y) should make J (0) = 0. As a result, J (0) = F u η + F p η x + F q η y )dxdy D = D + D ( F u x F p ) y F q η(x, y)dxdy [ x (F pη)+ ] y (F qη) dxdy =0. (1.3.10) The last line can be, by use of Green s formula, transformed to integration along the boundary line: [ x (F pη)+ ] y (F qη) dxdy = (F p ηdy F q ηdx). (1.3.11) D By the boundary condition η(x, y) C = 0, this integration is zero. Hence, using the basic variational lemma leads to F u x F p y F q =0. (1.3.12) This is the Euler equation that u(x, y) mustsatisfy.itcontainsthe partial derivatives of undetermined function, so that is a partial differential equation. Example 2. Functional J[u] = D C (u 2 x + u 2 y)dxdy takes its extreme when its admissible function meet u = 2 u x u y 2 =0. This is a well-known Laplace equation. Solving u(x, y) from this equation accompanied by boundary condition u(x, y) C = u 0 (x, y) is

21 Variational Method 21 the first boundary value problem or Dirichlet problem in region D: u = 2 u x u =0, (x, y) D, y2 u(x, y) C = u 0 (x, y). In summary, the idea of proving the above formulas is as follows. The functional is written as a function depending on a parameter α. Then the derivative with respect to this parameter is set to be zero to obtain the extremes. One may also take more than one parameter such as α, β and so on, if necessary, and then take derivatives with respect to every parameter to obtain extremes of the functional. If there are N variables x i, (i =1, 2,...,N), we use {x i } to denote them. The function is y = y(x 1,x 2,...,x N )=y({x i }). The functional is J[y] = R F ( x 1,x 2,...,x N,y, y x 1, y x 2,..., y x N ) dx 1 dx 2 dx N, (1.3.13a) whichisinshortwrittenas ( { }) y J[y] = F {x i },y, d N x. (1.3.13b) R x i It should not be difficult to extend (1.3.12) to be the following equations: F y F =0. x i=1 i ( y/ x i ) (1.3.14) Finally, if there are N variables x i, (i =1, 2,...,N)andM functions y α ({x i }), (α =1, 2,...,M), the functional will be ( { }) yα J[{y α ({x i })}] = F {x i }, {y α ({x i })}, d N x. (1.3.15) R x i In this case, combination of (1.3.8) and (1.3.14) results in the following Euler equations: F N F =0, y α x i=1 i ( y α / x i ) (α =1, 2,...,M), (1.3.16) which contain M equations.

22 22 Mathematics for Physicists 1.4. Functional Extremes under Certain Conditions In many functional extremal problems, the admissible functions themselves may also be exerted some restrictions, which is similar to the conditional extreme value problems of functions. In this section, we discuss two types of conditional extreme value problems of functionals Isoperimetric problem This was originated from the following geometric problem. One is seeking a curve l through points A and B with fixed length, which makes a trapezoid with curve sides ABCD take its maximum area. Suppose y = y(x) is the required solution. It is easy to see that the problem is attributed to finding the extreme of a functional S[y] = y(x)dx, (1.4.1) which satisfies the boundary conditions y( )=y 0,y(x 1 )=y 1 and constraint x 1+y 2 dx = l. Because it is required that the perimeter of the trapezoid with curve sides ABCD is fixed, this problem is called isoperimetric problem. The general statement of the problem is as follows. Among all the admissible curves of the functional x J 1 = G(x, y, y )dx = l (constant), (1.4.2) find one which makes another functional x J = F (x, y, y )dx (1.4.3) take its extremes. This type of problem is termed as an isoperimetric problem. Equation (1.4.2) is termed as an isoperimetric condition. Similar to the conditional extreme problem of multivariate functions, there is a theorem for the isoperimetric problem. This theorem utilizes the method of Lagrange multiplier, and makes the variational problem of conditional extremes to be one without the conditions.

23 Variational Method 23 y y = y (x) B (x 1, y 1 ) A (, y 0 ) D Fig Theorem 1 (Lagrange). If a curve y = y(x) is the extremal one of (1.4.3), but not that of functional J 1, then there must be a constant λ which makes y = y(x) be the extremal one of the functional H(x, y, y )dx, (1.4.4) where Hence, y= y(x) satisfies Euler equation Proof. We take comparison function C H = F + λg. (1.4.5) H y d dx H y =0. (1.4.6) y (x) =y(x)+αη 1 (x)+βη 2 (x), (1.4.7) x where α and β are small parameters, and η 1 (x)andη 2 (x)aregivenc 2 class functions which satisfy η 1 ( )=η 1 (x 1 )=η 2 ( )=η 2 (x 1 )=0. Substituting y (x) into (1.4.2) and (1.4.3), respectively, one gains and J(α, β) = J 1 (α, β) = x x F (x, y + αη 1 + βη 2,y + αη 1 + βη 2)dx (1.4.8) G(x, y + αη 1 + βη 2,y + αη 1 + βη 2)dx = l. (1.4.9)

24 24 Mathematics for Physicists Let y(x) be the extremal curve of functional J[y] under the condition (1.4.2). It is seen from (1.4.7) that as α = β =0,y = y. Thus,the binary function J(α, β) takesitsextremesasα = β =0 under the condition J 1 (α, β) =l. By the method of Lagrange multiplier, we have J α + λ J 1 α =0, (α = β =0), (1.4.10) J β + λ J 1 β =0 where λ is to be determined. Substitution of J and J 1 into these two equations leads to [(F y + λg y )η 1 +(F y + λg y )η 1 ]dx =0 (1.4.11) [(F y + λg y )η 2 +(F y + λg y )η 2]dx =0 Taking integration by parts for the second terms in the two equations, and noticing η 1 ( )=η 1 (x 1 )=η 2 ( )=η 2 (x 1 ) = 0, we obtain [ (F y + λg y ) d ] dx (F y + λg y ) η 1 dx = 0 (1.4.12) and [ (F y + λg y ) d ] dx (F y + λg y ) η 2 dx =0. (1.4.13) By assumed condition, y(x) doesnotmakej 1 take its extremes, so that does not satisfy the equation G y d G dx y = 0. Therefore, it is possible to select a η 2 (x) which yields ( G y d ) dx G y η 2 dx 0. (1.4.14) Subsequently, one gains from (1.4.13) that λ = ( x1 ( F y d ) F dx y η 2 dx G y d ). (1.4.15) G dx y η 2 dx

25 Variational Method 25 Because η 2 and η 1 are independent of each other, λ is independent of η 1. Similarly, λ is also independent of η 2. Therefore, λ must be a constant unrelated to both η 1 and η 2. Finally, applying the basic variational lemma to (1.4.12) leads to F y + λg y d dx (F y + λg y )=0. (1.4.16) Equation (1.4.16) is an ordinary differential equation of second order containing parameter λ. Its general solution comprises three undetermined constants C 1, C 2 and λ, which could be determined by the isoperimetric condition (1.4.2) and given boundary conditions, e.g., y( )=y 0 and y(x 1 )=y 1. Example 1. Find the extremal curve of the functional S = ydx, y( )=y 0,y(x 1 )=y 1, which satisfies the isoperimetric condition 1+y 2 dx = l. Solution. We take an auxiliary functional J = (y + λ 1+y 2 )dx. Let H = y + λ 1+y 2. Then its Euler equation is H y d dx H y =0. Because H does not involve x, it has the first integral y + λ 1+y 2 λy 2 1+y 2 = C 1. We employ parameter method to solve the differential equation. Let y =tant. Then, y C 1 = λ cos t, dx = dy λ sin tdt y = = λ cos tdt, x = λ sin t + C 2. tan t Thus, the required curve meets the parameter equation x C 2 = λ sin t, y C 1 = λ cos t.

26 26 Mathematics for Physicists The parameter t can be eliminated to have (x C 2 ) 2 +(y C 1 ) 2 = λ 2. This is a class of circles. The constants C 1,C 2 and λ can be determined by the given isoperimetric condition and boundary conditions Geodesic problem The problem is to find two functions y(x) andz(x), which make the functional J = F (x, y, y,z,z )dx (1.4.17) take its extremes and satisfy an additional condition G(x, y, z) =0. (1.4.18) In mechanics, this problem is the so-called constraint problem. From aspect of geometry, it is to find a curve on the surface (1.4.18) making the functional (1.4.17) take its extremes. Note that here are one argument x and two functions y(x) andz(x). Image that in the three-dimensional Cartesian system, there is a surface where x is a function of y and z. The required extremal curve is on the surface, and in fact contains two functions y(x) andz(x). A natural idea in solving this problem is taking z as a function of x and y from (1.4.18). Then, this function is substituted into (1.4.17). As a result, we achieve an ordinary variational problem which seeks a function y(x) without the additional condition. Now we follow this routine to find the equations that the extremal curves y(x) andz(x) must satisfy. Suppose G z 0. According to the existence theorem of implicit functions, the function z = ϕ(x, y) can be deduced from (1.4.18) z = ϕ(x, y). It is then substituted into (1.4.17) to get J = F (x, y, y,ϕ,ϕ x + ϕ y y )dx. (1.4.19) For the sake of simplicity, we denote F (x, y, y )=F(x, y, y,ϕ,ϕ x + ϕ y y ). So, y(x) should meet the Euler equation F y d dx F y =0. (1.4.20)

27 Variational Method 27 Because Fy = F y + F z ϕ y + F z (ϕ xy + ϕ yy y ), d dx F y = d dx F y + ϕ d y dx F z + F z (ϕ (1.4.21) xy + ϕ yy y ), (1.4.20) becomes ( F y + ϕ y F z d ) dx F z d dx F y =0. (1.4.22) On the other hand, ϕ y = z y = Gy G z. Subsitituting it into (1.4.22), one obtains ( 1 F y d ) ( G y dx F y = 1Gz F z d ) dx F z. (1.4.23) The two sides of the equation should always be identical. Therefore, both should equal to a common constant λ. Thusonegets d dx F y [F y + λ(x)g y ]=0, d (1.4.24) dx F z [F z + λ(x)g z ]=0. These are the differential equations of second order that the extremal functions y(x) and z(x) ought to satisfy. In summary, we achieve the following theorem. Theorem 2. Suppose that the functions y(x) and z(x) make the functional (1.4.17) take its extremes and satisfy (1.4.18). Then there must be a proper factor λ(x), which makes y(x) and z(x) satisfy the Euler equations of the functional J = x 1 H(x, y, y,z,z )dx: d dx H y H y =0, (1.4.25) d dx H z H z =0, where H = F + λ(x)g. Note the formal difference between (1.4.24) and (1.4.25). The first term of the former is written as H y instead of F y. They are actually

28 28 Mathematics for Physicists the same. The reason is that in the constraint condition (1.4.18), G does not contain y and z. Example 2 (geodesic problem). Find the shortest distance between two fixed points A(,y 0,z 0 )andb(x 1,y 1,z 1 )onasurface ϕ(x, y, z) =0. Solution. It is known that the distance between two points is expressed by l = 1+y 2 + z 2 dx. Hence, the problem is to find the minimum of the functional l under the condition ϕ(x, y, z) = 0. We take an auxiliary function l = [ 1+y 2 + z 2 + λ(x)ϕ(x, y, z)]dx. Its corresponding Euler equations are λ(x)ϕ y d y =0, dx 1+y 2 + z 2 λ(x)ϕ z d z =0. dx 1+y 2 + z 2 From these two equations and the constraint condition ϕ(x, y, z) =0, the factor λ(x) and functions y = y(x) andz = z(x) canbesolved. The undetermined constants in the general solutions can be found under the boundary conditions y( )=y 0,y(x 1 )=y 1 and z( )= z 0,z(x 1 )=z 1. In mechanics, the constraint problem often manifests a more general form as follows. One wants to find the extremes of a functional containing n admissible functions J[y 1,y 2,...,y n ]= F (x; y 1,y 2,...,y n ; y 1,y 2,...,y n )dx (1.4.26) under m constraint conditions ϕ k (x, y 1,y 2,...,y n )=0, k =1, 2,...m, (m <n). (1.4.27)

29 Variational Method 29 This problem is attributed to solve Euler equations of the functional [ ] x1 m J = F + λ k (x)ϕ k dx. (1.4.28) The equations are [ d dx F y i F yi + m k=1 λ k (x) ϕ k y i k=1 ] =0, i =1, 2,...,n. (1.4.29) 1.5. Natural Boundary Conditions The Euler equations that the extremal curves or surfaces should satisfy are ordinary or partial differential equations. We have shown that some boundary conditions are needed in order to explicitly determine the extremal functions. In the examples above, the boundary conditions are that the extremal functions take fixed values at the boundaries, e.g., (1.2.27). Correspondingly, the variational of the extremal functions at the boundaries are zero, see, for example, (1.2.24). In practical applications, the boundary conditions may be beyond this kind. Even the boundaries themselves may vary. For instance, let us see the simplest functional J[y] = F (x, y, y )dx. (1.5.1) Generally speaking, its lower and upper integral limits,x 1 and the values of its admissible functions at the boundaries y( )and y(x 1 ) may all vary. That is to say, the boundaries vary. Here we only discuss a simplest case: the integral limit,x 1 in (1.5.1) remain unchanged, while the values of y( )andy(x 1 )mayvary.inthe aspect of geometry, the two ends of the admissible curve can vary along the lines going through x = and x = x 1 and parallel to the y axis, see Fig The basic condition of taking extremes of a functional is that its variational is zero. δj =0. (1.5.2) This is our start point in the following discussion.

30 30 Mathematics for Physicists From (1.2.25) it is known that δj = F y δy(x) x 1 + ( F y d ) dx F y δydx =0. (1.5.3a) Please note that the Euler equations obtained before is based on the condition (1.2.24) where the variationals of the extremal curve at ends are zero. It is not so in the present case, as the curve ends may vary. Nevertheless, the admissible curves include those having fixed ends as well as those having unfixed ends. Therefore, it can be certain that as long as y(x) is the extremal curve of the functional (1.5.1), it meets the Euler equation. Assuming that we have found the right extremal curve y = y (x) withfixedendsy( )andy(x 1 ), these boundary conditions can be applied to compose a new variational problem with fixed boundaries. Then obviously, y (x) is surely the solution of the new problem. That is to say, y (x) certainly meet the Euler equation F y d dx F y =0. In other words, whether the ends are fixed or not, the extremal curve always meets the Euler equation. Hence, the second term in (1.5.3a) is zero, leaving the first term. By (1.5.2), we have δj =(F y δy) x1 (F y δy) x0 =0. (1.5.3b) Because the changes of the extremal curve at the two ends (δy) x0 and (δy) x1 are independent of each other, see Fig. 1.4, it should x=x0 =0, F y x=x1 =0. (1.5.4) F y These are the conditions that the function y(x) should obey at ends x = and x = x 1, called natural boundary conditions. A special case is that one of the ends, say the left one, is fixed, i.e., y( )=y 0. (1.5.5) The right end may vary along the line x = x 1. Then we have δy(x) = 0 and varying δy(x 1 ). By (1.5.3) the natural boundary

31 Variational Method 31 y y+δy (δy) x1 (δy) x0 y (x,) x 1 Fig The integral limit and x 1 remain unchanged, while the values of y()andy(x 1)mayvary. conditions degrade to F y x=x1 =0. (1.5.6) Example 1. In Subsection 1.2.2, we have obtained the general solution of the brachistochrone problem is a class of cycloids: x = C 1 (θ sin θ)+c 2, y = C 1 (1 cos θ). Now let the left end of the cycloid fixed, y(0) = 0, while the right end varies along the line x = x 1. Then, from y(0) = 0 one still gets C 2 = 0. The constant C 1 is to be determined by (1.5.6). Because F = 1+y 2 2gy, y F y x=x1 = =0. 2gy(1 + y 2 ) x=x1 x It follows that y (x 1 ) = 0. Hence, the required cycloid should be vertical to the line x = x 1, see Fig By y (x 1 ) = 0, one easily find that the point B corresponds to θ = π, whichresultsinx 1 = C 1 π and C 1 = x 1 π. Finally, the required curve is expressed as x = x 1 π (θ sin θ), y = x 1 (1 cos θ). π

32 32 Mathematics for Physicists O x=x 1 x y B Fig The initial point of the cycloid is at the origin, while its other end B may vary along the line x = x 1. Then it can be certain that the cycloid is vertical to the line x = x 1 at point B. If both ends, x 1, as well as the function values at the ends y( )=y 0 and y(x 1 )=y 1 can vary, i.e., the two ends (,y 0 )and (x 1,y 1 ) can vary arbitrarily, it is a more complicated case. We do not intend to introduce the derivation here, but simply present the final results. In such a case, the natural boundary conditions that the extremal curve should meet are as follows. F y x=x0 =0,F y x=x1 =0, (1.5.7a) (F y F y ) x=x0 =0, (F y F y ) x=x1 =0. (1.5.7b) The Euler equation should be solved under these boundary conditions. In other words, the two integration constants in the general solution and the positions,x 1 all should be determined by Eqs. (1.5.7). If the two ends (,y 0 )and(x 1,y 1 ) of the admissible curve vary along two given lines g 0 ( )andg 1 (x 1 ) but not arbitrarily, i.e., then, one has y 0 = g 0 ( ), y 1 = g 1 (x 1 ), (1.5.8) δy 0 = g 0 ()δ, δy 1 = g 1 (x 1)δx 1. (1.5.9) Equations (1.5.8) are just constraint conditions. In this case, the admissible function should meet the following natural boundary conditions: [F +(g 0 y )F y ] x=x0 = 0 (1.5.10a)

33 Variational Method 33 and [F +(g 1 y )F y ] x=x1 =0. (1.5.10b) They are termed as transversality conditions, manifesting the relations between the derivatives of the extremal function at the ends and constraint lines g 0 ( )andg 1 (x 1 ). In solving Euler equations, Eqs. (1.5.9) and (1.5.10) are employed to determine the integral constants and the positions and x Variational Principle The application of the variational method in physics is called the variational principle. The variational principle was firstly induced from natural phenomena, and then expressed in the form of a principle. It is usually considered as a starting point for deducing the equations that a matter should obey when it moves or exists. This principle belongs to the category of scientific hypothesis, and it manifests different forms in different scientific disciplines. An example is the Fermat s principle in optics: light propagates from point A to B along such a path that it takes least time. Suppose light propagates along a path y = y(x) in a plane. We denote light speed by v(x, y) atpoint(x, y) in the media. Then, similar to the case of the brachistochrone problem, it is derived that the time needed when light propagates from A(,y 0 )tob(x 1,y 1 ) along path y = y(x) is 1+y 2 T [y] = v(x, y) dx. (1.6.1a) Thus, following Fermat s principle, the problem becomes the one to find a curve which makes the functional T [y] take its minimum. If we let time t be an argument, every point x, y along the path is a function of time: x = x(t),y = y(t). Then (1.6.1a) is brought to be ẋ2 t1 +ẏ T [x(t),y(t)] = 2 dt. (1.6.1b) t 0 v(x, y) In the following, we will introduce respectively the variational principles in classical mechanics and quantum mechanics.

34 34 Mathematics for Physicists Variational principle of classical mechanics In classical mechanics, the variational principle mainly means Hamilton s least-action principle, or simply Hamilton s principle. In a mechanical system, a Lagrangian is defined as its kinetic energy T minus potential U: L = T U. (1.6.2) Here T and U may be functions of generalized coordinates and momenta which vary with time. Let us construct a functional: t1 J = L dt. (1.6.3) t 0 Hamilton s principle says that among all possible movements of the system within a period of time (compatible to constraint conditions, if any), the system necessarily takes one that makes the variational of the functional J be zero: δj =0. (1.6.4) Or, the functional reaches its extremes. Usually, the extreme is a minimum. Hence, the principle is also called the least-action principle. TheJ defined by (1.6.3) is termed as the action. In the following, we take a conservative field as an example of applying the Hamilton s principle. Suppose that there is a given particle system. The coordinates of the particles are denoted as (x i,y i,z i ),i =1, 2, 3,n. The i-th mass point has mass m i, and is applied a force F i (i =1, 2,...n) which is generated by the potential U(t, x 1,y 1,z 1,...,x n,y n,z n )of the system: F i,x = U, F i,y = U, F i,z = U. (1.6.5) x i y i z i The movement of the particle system is described by Newton equations. This means that the coordinates of each mass point as functions of time x = x i (t),y = y i (t),z = z i (t) observe the following

35 Variational Method 35 equations: m i ẍ i = F i,x = U, x i m i ÿ i = F i,y = U, y i (1.6.6) m i z i = F i,z = U. z i For this system, the potential is U and its kinetic energy is T = 1 n m i (ẋ 2 i +ẏi 2 +żi 2 )=T(ẋ 1, ẏ 1, ż 1,...,ẋ n, ẏ n, ż n ). 2 i=1 Let the initial positions x 1 (t 0 ),y 1 (t 0 ),z 1 (t 0 ),...,z n (t 0 ) and final positions x 1 (t 1 ),...,z n (t 1 ) be fixed. We consider the following two functionals: and J 2 = J 1 = t 0 t1 t 0 t1 t 0 T (ẋ 1,...ż n )dt U(t 1,x 1,...,z n )dt. The variational of the first functional is t1 ( T δj 1 = δẋ T ) δż n dt. t 0 ẋ 1 ż n Integrating by parts for every term and noticing δx 1 (t 0 )=δx 1 (t 1 )= = δz n (t 0 )=δz n (t 1 )=0, one obtains t [( ) ( ) ] d T d T δj 1 = δx δz n dt t 0 dt ẋ 1 dt ż n t = (m 1 ẍ 1 δx m n z n δz n )dt. t 0 For the second functional, t ( U δj 2 = δx U ) δz n dt. x 1 z n

36 36 Mathematics for Physicists It follows from Newton equations (1.6.6) that δj 1 = δj 2, or δ(j 1 J 2 )=0. In summary, if functions x 1 (t),y 1 (t),z 1 (t),...,x n (t),y n (t),z n (t) describe the movement of particle system within time period t 0 t t 1, then these functions make the functional t1 t1 J 1 J 2 = (T U)dt = Ldt (1.6.7) t 0 t 0 take its extremes. In another word, the system obeys Hamilton s principle. Reversely, once the functional of the system is written in the form of (1.6.7), the equations must be (1.6.6) that particles in the system obey, so that the functional takes its extremes. In the above sections, these equations of motion were called the Euler equations. When applied in physics, they are usually called the Euler-Lagrange equations, or even further simplified as the Lagrange equations. Usually, kinetic energy and potential are the functions of generalized coordinates and momenta. If there are n pairs of generalized coordinates and momenta, which are represented by {q i (t)} and { q i (t)}, respectively, then according to the cases of multifunctions discussed in Subsection 1.3.1, the Lagrange equations can be written following (1.3.8) as d L L =0, (i =1, 2,...,n). (1.6.8) dt q i q i Kinetic energy does not explicitly contain time t, and usually it is a quadratic function of generalized momenta. If potential U does not contain time t either, L will not either. In Subsection we have discussed a special case that if a functional did not contain an argument, there would exist a first integral. Its integration constant E here is just the total energy of the system, and equals the kinetic energy plus potential: E = T + U. Therefore, that the potential does not explicitly contain time means the conservation of the system s total energy. Furthermore, if

37 Variational Method 37 potential does not explicitly contain one generalized coordinate q i, then by (1.6.8) one knows that the derivative of Lagrange function with respect to the generalized momentum corresponding to this coordinate is a constant: L = P i. (1.6.9) q i This constant P i is proportional to the generalized momentum. Hence, that the potential does not explicitly contain a coordinate q i means the conservation of the momentum corresponding to this coordinate. Furthermore, we assume that the movement of the particles is restricted by m constraint conditions: ϕ j (t, x 1,y 1,z 1,...,x n,y n,z n )=0, j =1, 2,...,m. (1.6.10) By utilizing Hamilton s principle, it is able to derive the equations of motion of the particle system. The trajectory of every mass point x i = x i (t), y i = y i (t), z i = z i (t) should make the functional t1 [ ] t1 1 n J = (T U)dt = m i (ẋ 2 i t 0 t 0 2 +ẏ2 i +ż2 i ) U dt i=1 take its extremes. Hence, this is a variational problem of functional extreme under certain conditions. Let us put down an auxiliary functional t1 J = 1 n m m i (ẋ 2 i 2 +ẏ2 i +ż2 i ) U + λ j (t)ϕ j dx. t 0 i=1 The Euler equations related to this functional m i ẍ i = U m + λ j (t) ϕ j x i x i m i ÿ i = U y i + m i z i = U z i + j=1 m j=1 m j=1 λ j (t) ϕ j y i λ j (t) ϕ j z i j=1 (i =1, 2 n)

38 38 Mathematics for Physicists are the differential equations of the movement of this particle system. Now we generalized Hamilton s principle to an arbitrary system. Given the Lagrangian of the system, the corresponding action is constructed by (1.6.3). Usually, the action is denoted as S: t1 S = L dt. (1.6.11) t 0 The equation of motion of the system is determined by taking the action s variation to be zero, δs =0.Mostoften,δS = 0 will make the action S reach its minimum. This is called the least-action principle. The function L in (1.6.11) is usually called the Lagrangian density. In an arbitrary system, the Lagrangian may not be kinetic minus potential as in (1.6.7). We will see an example in relativistic mechanics. Example 1. A particle is confined on the surface of a sphere with radius R, and moves by gravity. Find its equations of motion. Solution. Since it moves on a spherical surface, it is convenient to adopt spherical polar coordinates (r, θ, ϕ). The sphere center taken as the origin, the kinetic energy of the particle is and its potential is T = 1 2 m(ṙ2 + r 2 θ2 + r 2 sin 2 θ ϕ 2 ), Subsequently, its Lagrangian is U = mgz = mgr cos θ. L = 1 2 m(ṙ2 + r 2 θ2 + r 2 sin 2 θ ϕ 2 ) mgr cos θ. (1.6.12) It is required to be moving on the spheric surface, i.e., r = R, which is a constraint condition: G(t, r, θ, ϕ) =r R =0.

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