Variational Calculus & Variational Principles In Physics

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1 Variational Calculus & Variational Principles In Physics Michael A. Carchidi Noember 3, 9 1. A Typical Problem In The Calculus Let R be the set of real numbers and let f : R R be a differentiable function defined on R. In a typical calculus problem, you are gien the function f : R R oer an interal a x b, and you are asked to determine the maximum or minimum alues of f. One method of solution, as deeloped in the calculus, is to determine all a x b for which f (x) =or f (x) is not defined (the so-called critical points of f), call these, x, x 3,..., x n, and then computed {f(a),f( ),f(x ),f(x 3 ),...,f(x n ),f(b)} and then while f min =min{f(a),f( ),f(x ),f(x 3 ),...,f(x n ),f(b)} f max =max{f(a),f( ),f(x ),f(x 3 ),...,f(x n ),f(b)}. Example For Reiew Only Determine the maximum and minimum alue for the function f : R R, defined by f(x) =3x 4 8x 3 +84x 96x +1 oer the interal x 5. Tosolethis,wefirst compute f (x) =1x 3 84x +168x 96

2 andthenwesole 1x 3 84x +168x 96 =. This leads to x = {1,, 4} as the critical points and each of these is in the region x 5. Then we form the following table of alues. x y = f(x) Comment 1 Endpoint 1 36 Critical Point 31 Critical Point 4 63 Critical Point 5 4 Endpoint This shows that the minimum alue for f is 63 at x =4, while the maximum alue for f is +1 at x =.Aplotoff(x) ersus x for x 5 along with the fie points in this table is shown in the following figure. Plot of f(x) =3x 4 8x 3 +84x 96x +1 ersus x for x 5

3 Example 1. - For Reiew Only Determine the maximum and minimum alue for the function f : R R, defined by f(x) =max{x x +3, 1x x 1} oer the interal x 6.5. Tosolethis,wefirst note that f(x) =max{x x +3, 1x x 1} is the same as f(x) = x x +3, for x 3 1x x 1, for 3 x 6 1 so that f (x) = x, for <x<3 1 x, for 3 <x<6 1. Then f (x) =yields x =1or x =5, while f (x) not define yields x =3,since f (x) has a jump discontinuity at x =3. This leads to x = {1, 3, 5} as the critical points and each of these is in the region x 6.5. Then we form the following table of alues. x y = f(x) Comment 3 Endpoint 1 4 Critical Point 3 Critical Point 5 4 Critical Point Endpoint This shows that the minimum alue for f is at x =3, while the maximum alue for f is 4 at either x =1or x =5. Notethattheextremealuesoff (maximum or minimum) are uniue, but optimal points (i.e., where these extreme alues occur) need not be uniue. A plot of f(x) ersus x for x 6.5 along with the fie 3

4 points in this table is shown in the following figure. Plot of f(x) =max{x x +3, 1x x 1} ersus x for x 6.5. Some Typical Problems In The Calculus of Variations We now present a number of typical problems that arise in the calculus of ariations. In this section, we shall only state the problems. We shall sole these problems later in the chapter. We begin by stating some problems that contain no constraints except for boundary conditions. Example.1: A Geodesic Problem In The Plane Gien two points (,y 1 ) and (x,y ) in the xy plane, we want to determine the function y = y(x) that connects these points and has the shortest length. Using the formula for arc length from the calculus, we see that we are searching for a function y = y(x) so that the boundary conditions y( )=y 1 and y(x )=y (1a) are satisfied and for which L[y] = 1+(y ) dx where y = dy dx (1b) 4

5 is as small as possible. Of course we know that the answer to this problem is the straight line µ y y 1 y(x) =y 1 + (x ) x and L min = (x ) +(y y 1 ). Example.: A Geodesic Problem On A Surface Gien two points (u 1, 1 ) and (u, ) that lie on a surface described by the ector euation r = r(u, ) we want to determine the function = (u) that connects these points and has the shortest length. Using the formula for arc length from the calculus, we hae ds = dr dr with and so ds = = u t u t r r du + u d r u r u dr = r r du + u d du + r r du + u d r u r dud + r r d or u ds = t r u r r u + u r r + r ( ) du where = d/du. Thus we are searching for a function = (u) so that the boundary conditions (u 1 )= 1 and (u )= (1c) 5

6 are satisfied and for which L[] = Z u u 1 is as small as possible. u t r u r u + Example.3: The Minimum Work Problem r u r r + r ( ) du (1d) Aparticleistomoealongapathfromthepoint(,y 1 ) to the point (x,y ) in the xy plane. There is a force F = F(x, y) =F 1 (x, y) b i + F (x, y) b j that acts on the particle as it moes. We seek to determine that path for which theworkdonebytheforceisassmallaspossible.since we hae W [y] = W = Z (x,y ) (,y 1 ) F dr = Z (x,y ) (,y 1 ) F 1 (x, y)dx + F (x, y)dy (F 1 (x, y)+f (x, y)y )dx where y = dy dx. Thus we seek to find that function y = y(x) such that the boundary conditions are satisfied and for which is as small as possible. y( )=y 1 and y(x )=y (a) W [y] = (F 1 (x, y)+f (x, y)y )dx (b) Example.4: The Minimum Surface Area of Reolution Problem Here we start with a function y = y(x) that connects the points (,y 1 ) and (x,y ) in the first uadrant and we want to reole the cure represented by this function about the positie x axis to form a surface of reolution. We seek that 6

7 function which gies the surface that has the smallest possible lateral area. Using the expression from calculus that gies this area as S[y] = πy 1+(y ) dx where y = dy dx, we see that we seek to find that function y = y(x) such that the boundary conditions y( )=y 1 and y(x )=y (3a) are satisfied and for which S[y] = πy 1+(y ) dx, (3b) is as small as possible. Example.5: The Brachistochrome Problem Suppose that a bead (of mass m) can slide under its weight (without friction) along a wire in the shape of a planar cure from the point (,h) to the point (a, ). We seek to find that cure (described by the function y = y(x)) forwhich the time of trael is as small as possible, gien that the particle starts from rest. To compute the time of trael, we use conseration of energy, which states that mgh = 1 m + mgy or =g(h y). Buttheelocityoftheparticleasitslidesalongthecuredescribedbyy = y(x) is = ẋ b i + ẏ b j with ẏ = dy dx dx dt = y (x)ẋ = y ẋ so that = ẋ b i + ẋy b j. Then = ẋ + ẋ (y ) =(1+(y ) )ẋ and hence or =(1+(y ) )ẋ =g(h y) ẋ = dx dt = u t 7 g(h y) 1+(y ).

8 Note that the positie suare root is computed since we expect ẋ> as the particle slides down the wire. Thus we hae and hence dt = T [y] = u t 1+(y ) g(h y) dx Z a u t 1+(y ) g(h y) dx. Therefore we seek a function y = y(x) so that the boundary conditions y() = h and y(a) = (4a) are satisfied and for which T [y] = Z a u t 1+(y ) g(h y) dx (4b) is as small as possible. For example, if y is the straight line connecting (,h) and (a, ), weget y = y(x) =h(1 x/a) and or T [y] = Z a u t T [y] = u 1+( h/a) Z a g(h h(1 x/a)) dx = t 1+h /a gh/a Z a 1 x dx = u u t 1+h /a ghx/a dx t 1+h /a gh/a a which finally reduces to s (a + h T [y] = ). gh We shall see that the minimum-time path is not a straight line and thus the aboe expression is NOT going to be the minimum time. We now present two problems that contain extra constraints in addition to boundary conditions. 8

9 Example.6: The Hanging Chain Problem A uniform flexible chain (haing linear mass density ρ) hangs in static euilibrium under its own weight between two supports. The chain has a fixed length of L andweseekthatshapey = y(x), which minimizes the total potential energy of the chain. Toward this end we assume that the chain hangs between the two points ( a, ) and (a, ), where or course a <L,with L[y] = Z a a Since the potential energy of the chain is we see that U = mgy cm = mg 1 m 1+(y ) dx = L = constant. Z chain Z y(x)dm = g y(x)ρds chain Z a U[y] =ρg y 1+(y ) dx where y = dy a dx. Thus we see a function y = y(x) so that the boundary conditions y( a) =y(a) = are satisfied along with the extra constraint, L[y] = Z a a 1+(y ) dx = L = constant (5a) (5b) for which is as small as possible. Z a U[y] =ρg y 1+(y ) dx a (5c) Example.7: The Maximum Area Problem Gien a simply closed cure in the xy plane that has a fixed perimeter P, we seek that cure which encloses the maximum area. Note that a simple closed cure is one that DOES NOT intersect itself. For example, a figure 8 is not a 9

10 simple closed cure since it intersects itself at the center. A simple closed cure can be written in parametric form as r = r(t) =x(t) b i + y(t) b j and its perimeter can be computed using I I Z t P = dr = dx + dy = ẋ + ẏ dt C C t 1 where t 1 6= t and r(t 1 )=r(t ). The area enclosed is computed using I I A = 1 r dr = 1 ydx) = C C(xdy 1 t 1 (xẏ yẋ)dt. Thus we seek two functions x = x(t) and y = y(t) such that the boundary conditions x(t 1 )=x(t ) and y(t 1 )=y(t ) (6a) are satisfied along with the constraint, P [x, y] = Z t t 1 Z t ẋ + ẏ dt = P = constant (6b) and for which isaslargeaspossible. A[x, y] = 1 Z t t 1 (xẏ yẋ)dt (6c) 3. The Simplest Problems In The Calculus of Variations If the first fie examples aboe (.1 -.5), we are gien integrands of the form f(y )= 1+(y ) and and u f(u,, u )= t r u r u + r u r r + r ( ) f(x, y, y )=F 1 (x, y)+f (x, y)y, f(y )=πy 1+(y ) 1

11 and which we represent collectiely as f(y, y )= u t 1+(y ) g(h y) f = f(x, y, y ). We are also gien that the function y = y(x) must contain the two points (,y 1 ) and (x,y ), and we seek to find such a function y = y(x) so that the boundary conditions y( )=y 1 and y(x )=y (7a) are satisfied and for which J[y] = f(x, y, y )dx (7b) is as large (or as small) as possible. The expression J[y] is called a functional. Unlike a function, which takes as input, a real number and returns as output a real number, a functional takes as input, a differentiable function y = y(x) (along with fixed real numbers and x ) and it returns as output a real number. 4. The Euler-Lagrange Euation Our goal is to determine that function y = y(x) that satisfies the boundary conditions y( )=y 1 and y(x )=y (8a) for which the functional J[y] = f(x, y, y )dx (8b) is an extremum (maximum or minimum). 4.1 The Ordinary Calculus Problem Toseehowwemightsolethis,letusreiewone method on how we might sole the ordinary calculus problem inoling a differentiable function y = f(x) and finding an extremum for it. Toward this end, we assume that we hae a point for which f is a relatie extremum, and let us call this point x c. Let us also assume that this point is where f has a relatie minimum alue. The proof for a relatie 11

12 maximum alue is ery similar. Consider now the two points x ±α = x c ±α, where α is ery small and positie. These two points are to the left and right of x c. Since x c is a point where f has a relatie minimum, for α small enough, we must hae or f(x +α )=f(x c + α) f(x c ) and f(x α )=f(x c α) f(x c ), f(x c + α) f(x c ) and f(x c ) f(x c α). Diiding by α, we get f(x c + α) f(x c ) α and f(x c ) f(x c α) α for all positie α and near zero. Taking the limit as α approaches zero, we then get f(x c + α) f(x c ) lim α + α which yield or f (x c )=. and f (x c ) and f (x c ) f(x c ) f(x c α) lim α + α Note that you should show that the same result follows een if we assume α to be negatie throughout. Thus we find that a necessary condition for x c is that f (x c )=,iff(x c ) is to be a relatie minimum for the function f. A similar argument leads to f (x c )=, if f(x c ) is to be a relatie maximum for the function f as well. Thus we find that if f has a relatie extrema at x = x c,thenf (x c )=. 4. The Variational Calculus Problem To sole the ariational calculus problem we assume that y = y(x) is the function which satisfies the boundary conditions y( )=y 1 and y(x )=y 1

13 and for which J[y] = f(x, y, y )dx is as large (or as small) as possible. We then consider a function where α is small and y(x, α) =y(x)+αη(x) y(, α) =y 1 and y(x, α) =y which reuires that η( )=η(x )=. We also note that y(x, ) = y(x), which is the function we seek. Then J[y(x, α)] = J(α) = f(x, y(x, α),y (x, α))dx is a function of α. Tofind the extrema, we then want to reuire that Computing J/ α, we hae J(α) α =. α= J(α) α = α = = = f(x, y(x, α),y (x, α))dx α f(x, y(x, α),y (x, α))dx ( x x α + y(x, α) + y(x, α) α ( x () + y(x, α) η(x)+ y (x, α) y (x, α) α y (x, α) η (x) ) dx ) dx since y(x, α) =y(x)+αη(x) and y (x, α) =y (x)+αη (x) lead to y(x, α) α = η(x) and y (x, α) α = η (x). 13

14 Thus we hae J(α) α = = ( y(x, α) η(x)+ y(x, α) η(x)dx + ) y (x, α) η (x) dx Using integration by parts on the second integral, we find that y (x, α) η (x)dx = = y (x, α) η(x) x y (x, α) η (x)dx. (9) d dx y (x, α) η(x)dx y η(x ) (x, α) y x=x η( ) (x, α) x=x1 Z x d η(x)dx dx y (x, α) = since η( )=η(x )=,andsowehae Z x y (x, α) η (x)dx = Putting this into Euation (9), we hae or simply Then J(α) α J(α) α = J(α) α d dx d dx y(x, α) η(x)dx = = α= y (x, α) y (x, α) d dx ( y(x, α) d dx y (x, α) ( y(x) d dx y (x) y (x, α) ) ) η(x)dx η(x)dx. η(x)dx. η(x)dx. η(x)dx Setting this eual to zero and using the fact that η(x) is any function that satisfies η( )=η(x )=, we must conclude that y(x) d dx 14 y (x) =

15 or simply d dx y y =. This is the Euler-Lagrange euation and it represents a necessary condition for the function y = y(x) which minimizes (or maximizes) J[y]. To summarize, we see that a function y = y(x) which satisfies the boundary conditions y( )=y 1 and y(x )=y (1a) and for which J[y] = f(x, y, y )dx (1b) is as large (or as small) as possible must also satisfy the differential euation d dx y y =. (1c) 4.3 Another Form For The Euler-Lagrange Euation Starting with the function f = f(x, y, y ),wenotethat df dx = dx x dx + dy y dx + dy y dx = x + y y + y y. But from Euation (1c) we hae y = d dx y and so we find that df dx = x + d dx y y + y y = x + d y dx y or just d y dx y f + x =. (1d) 15

16 5. The Euler-Lagrange Euation: Two Special Cases We now consider two special cases of the Euler-Lagrange Euation d dx y y =. 5.1 The function f does not explicitly depend on the dependent ariable y If the function f(x, y, y ) does not explicitly depend on the dependent ariable y, then f(x, y, y )=f(x, y ) and y = (11a) in which case the Euler-Lagrange Euation (1c) reduces to d = or dx y y = constant. (11b) 5. The function f does not explicitly depend on the independent ariable x If the function f(x, y, y ) does not explicitly depend on the independent ariable x, then f(x, y, y )=f(y, y ) and x = (1a) in which case the Euler-Lagrange Euation (1d) reduces to d y dx y f = or y y f = constant. (1b) 6. Solutions to Some Typical Problems In The Calculus of Variations We now present the solutions to some of the problems discussed in Section aboe. 16

17 Example 6.1: A Geodesic Problem In The Plane Gien two points (,y 1 ) and (x,y ) in the xy plane, we want to determine the function y = y(x) that connects these points and has the shortest length. Using the formula for arc length from the calculus, we see that we are searching for a function y = y(x) so that the boundary conditions y( )=y 1 and y(x )=y (13a) are satisfied and for which L[y] = 1+(y ) dx (13b) is as small as possible. For here we hae f(x, y, y )= which leads to / y =and hence which leads to y = 1+(y ) = f(y ) y 1+(y ) = constant s y A 1+(y ) = A or y = 1 A = B and hence y(x) =Bx + C. To determine B and C, we put in the boundary conditions y( )=y 1 and y(x )=y, and these result in B = y µ y 1 y y 1 and C = y 1 x x and hence µ y y 1 y(x) =y 1 + (x ). x We also hae s µ y y L min = 1+(y ) 1 dx µ y y 1 dx = 1+ =(x )s 1+ x x 17

18 or simply as expected. L min = (x ) +(y y 1 ) Example 6.: A Geodesic Problem On A Sphere Gien two points (ϕ 1, θ 1 ) and (ϕ, θ ) that lie on the sphere described by r = r(ϕ, θ) =r cos(ϕ)sin(θ) b i + r sin(ϕ)sin(θ) b j + r cos(θ) b k we want to determine the function θ = θ(ϕ) that connects these points and has the shortest length. Using the formula for arc length discussed earlier with u = ϕ and = θ, wehae with and so that and Then u u ds = t r ϕ r ϕ + r ϕ r r θ θ + θ r (θ θ ) dϕ r ϕ = r sin(ϕ)sin(θ)b i + r cos(ϕ)sin(θ) b j r θ = r cos(ϕ)cos(θ)b i + r sin(ϕ)cos(θ) b j r sin(θ) b k, r ϕ r ϕ = r sin (θ) and r θ r θ = r. r ϕ r θ = ds = r sin (θ)+r (θ ) dϕ = r sin (θ)+(θ ) dϕ. where θ = dθ/dϕ. Thus we are searching for a function θ = θ(ϕ) so that the boundary conditions θ(ϕ 1 )=θ 1 and θ(ϕ )=θ (14a) are satisfied and for which L[θ] =r Z ϕ ϕ 1 sin (θ)+(θ ) dϕ. (14b) 18

19 is as small as possible. Here we effectiely hae f(ϕ, θ, θ )= sin (θ)+(θ ) and ϕ = which means that θ θ f = constant. or, writing the constant as 1/A and putting in the expression for f, wehae θ θ sin sin (θ)+(θ ) = 1 (θ)+(θ ) A or sin (θ) sin (θ)+(θ ) = 1 A. Soling for θ we get θ = A sin 4 (θ) sin (θ) =sin(θ) or dϕ = dθ sin(θ) A sin (θ) 1 = csc (θ)dθ A csc (θ) = A sin (θ) 1 Setting u =cot(θ), wehaedu = csc (θ)dθ, and then we hae dϕ = du A 1 u. csc (θ)dθ A 1 cot (θ) and hence or Z ϕ = 1 A 1 u du = A 1 u sin 1 + B A 1 ϕ = sin 1 A 1 cot (θ) A 1 + B and so, setting C = A 1, wefind that ϕ = sin 1 C cot (θ) + B C 19

20 or or This leads to or simply C cot (θ) C =sin(b ϕ) C cot (θ) =C sin (B ϕ). cot (θ) =C cos (B ϕ) cot(θ) =C cos(b ϕ) where B and C are computed from the boundary conditions θ(ϕ 1 )=θ 1 and θ(ϕ )=θ. Torecognizethiscurewewritethisas cos(θ) =C sin(θ)cos(b ϕ) or or cos(θ) =C sin(θ)(cos(b)cos(ϕ)+sin(b)sin(ϕ)) r cos(θ) =Cr sin(θ)cos(ϕ)cos(b)+cr sin(θ)sin(ϕ)sin(b). But x = r sin(θ)cos(ϕ), y = r sin(θ)sin(ϕ) and z = r cos(θ), andsowefind that z = Cxcos(B)+Cy sin(b) which is the euation of a plane passing through the center of the sphere (the origin). The path along the sphere which minimizes (or maximizes) arc length is then the cure that lies on both the sphere and this plane and this cure is a circle of intersection between the sphere and this plane, and this is known as a great circle on the sphere sincethiscirclehasthesamecenterasthatofthesphereand henceithasthesameradiusasthatofthesphere,whichisthelargestradiusa circle that is drawn on a sphere could hae.

21 Example 6.3: The Minimum Work Problem Aparticleistomoealongapathfromthepoint(,y 1 ) to the point (x,y ) in the xy plane. There is a force F = F(x, y) =F 1 (x, y) b i + F (x, y) b j that acts on the particle as it moes. We seek to determine that path for which the work done by the force is as small as possible. As seen earlier, this problem reduces to finding a function y = y(x) such that the boundary conditions are satisfied and for which y( )=y 1 and y(x )=y W [y] = is as small as possible. Here we hae (F 1 (x, y)+f (x, y)y )dx f(x, y, y )=F 1 (x, y)+f (x, y)y and placing this into the Euler-Lagrange euation, we get d dx y y = or But and so we hae df (x, y) dx df (x, y) dx F 1(x, y) y = F (x, y) x y F (x, y) y =. + F (x, y) y y F (x, y) x + F (x, y) y F 1(x, y) y F (x, y) = y y y or simply F (x, y) x F 1(x, y) y 1 =

22 which is an algebraic euation, not a differential euation. We see then that this may not hae a solution which also satisfies the boundary conditions y( )=y 1 and y(x )=y, since we hae no arbitrary constants to adjust. Note also that if the force F is conseratie, which means that F (x, y) F = x F 1(x, y) bk = y then any function connecting the points (,y 1 )and(x,y ) will do and the work is independent of the path. Of course, this is as expected. Example 6.4: The Minimum Surface Area of Reolution Problem As seen earlier, for this problem, we seek to find that function y = y(x) such that the boundary conditions are satisfied and for which is as small as possible. (1b) we hae y( )=y 1 and y(x )=y S[y] = πy 1+(y ) dx, Thus f(x, y, y )=πy 1+(y ), and using Euation y y f = constant y yy 1+(y ) y 1+(y ) = A π or y(y ) y y(y ) 1+(y ) = A π or y 1+(y ) = A π = B Soling for y,weget y = y B B and so dx = Bdy y B

23 which leads to x + C = Z µ Bdy y y B = B cosh 1. B Thus we find that µ x + C y = B cosh B with B and C obtained by reuiring the boundary conditions µ µ x1 + C x + C y 1 = B cosh and y = B cosh B B. 7. The Brachistochrome Problem: A Detailed Analysis A bead of mass m is to slide without friction in a constant graitational field g. The bead starts from rest at the point (,h) in an xy plane and slides without friction along some path to the point (a, ), where x is the horizontal direction and y is the ertical direction. It is assumed that a> and h>, and the problem is to determine the path that allows the bead to accomplish the transit from (,h) to (a, ) in the least possible time. It should be noted that there is no physical ground at y =so that motion in which y< is allowed, and as we shall see, is possible. We had seen that the transit time T is gien as or simply with f(x, y, y )= T [y] = T [y] = u Using Euation (1b) we find that Z a u Z a t 1+(y ) g(h y) t 1+(y ) g(h y) f(x, y, y )dx, so that dx, (15) x = (16) f y y = c 1. (17) 3

24 Putting Euation (16) into Euation (17) leads to u t 1+(y ) g(h y) y y = c 1 1+(y ) g(h y) or 1 g(h y)(1 + (y ) ) = c 1, which reduces to 1+(y ) = with A = 1 gc 1 Soling Euation (18a) for y leads to y = ± s A 1 gc 1(h y) = A h y (18a) = constant (18b) s h y 1=± A (h y), h y or, s h y dy = ±dx A (h y) which gies Z s Z h y A (h y) dy = ± dx = ±x + c. If we let, u = h y, thenudu = dy, sothat or x c = Z Z s u A u udu = ±x + c, u µ u A u du = u A u + A sin 1 A and so (after putting in the fact that u = h y), we hae h y x c = h y A (h y)+a sin 1 A 4

25 or h y x c = (h y)(a h + y)+a sin 1 A Putting in the condition, y = h when x =,weget h h c = (h h)(a h + h)+a sin 1 A which leads to c =,andso If we now let h y x = (h y)(a h + y)+a sin 1 A.. (19) then h y A =sin(θ) or y = h A sin (θ), x = A sin (θ)(a A sin (θ)) + A sin 1 (sin(θ)) = A sin (θ)a cos (θ)+a θ = A sin(θ)cos(θ)+a θ or Therefore we hae and x = ±A µ θ 1 sin(θ) x = ± A (θ sin(θ)) 1 cos(θ) y = h A sin (θ) =h A If we let ϕ =θ, thenwehaeforx and y, = ± A (θ sin(θ)). = h A (1 cos(θ)). x = ± A (ϕ sin(ϕ)) (a) 5

26 and y = h A (1 cos(ϕ)) (b) Now by the construction of the problem, we see that x and so the positie sign is to be used in Euation (a), and if we call the constant A /, thepositie constant B, wenowhae x = B(ϕ sin(ϕ)) (c) and y = h B(1 cos(ϕ)) (d) for ϕ β, whereϕ = β when the bead is at the point x = a, y =. This says that β is determined using the euations a = B(β sin(β)) and =h B(1 cos(β)) so that B(β sin(β)) B(1 cos(β)) = a h resulting in 1 cos(β) β sin(β) = h a which gies the alue of β. Onceβ is known then either (1a) B = h 1 cos(β) or B = a β sin(β) (1b) gies the alue of B. Thus the solution to the Brachistochrome problem is gien by ϕ sin(ϕ) x = a (a) β sin(β) and 1 cos(ϕ) cos(ϕ) cos(β) y = h h = h (b) 1 cos(β) 1 cos(β) for ϕ β, whereβ is the solution to F 1 (β) 1 cos(β) β sin(β) = h a. (c) 6

27 AplotofF 1 (β) ersus β is shown in the figure below Note that lim β + F 1(β) = lim β + Plot of F 1 (β) ersus β 1 cos(β) = lim β sin(β) β + β / = lim β 3 /6 β β Wemaynowsolefortheminimumtransittimeintermsofβ by first placing Euation (18a) into Euation (15) and getting T min = Z a u t 1+(y ) Z a g(h y) dx = u Then Putting in Euations (a,b) we get T min = A g Z a = Aa h g dx h y = 1 cos(β) β sin(β) t A /(h y) g(h y) dx = A g Z β Z β a ³ 1 cos(ϕ) β sin(β) dϕ h ³ 1 cos(ϕ) 1 cos(β) dϕ = Aaβ h g A g Z a 1 cos(β) β sin(β) dx h y But A = B = u t (h b) 1 cos(β) 7

28 and so T min = = a aβ 1 cos(β) h g β sin(β) β β sin(β) s h 1 cos(β) u ut 1 cos(β) g(h b) or s aβ 1 cos(β) T min =. β sin(β) gh But from Euation (1a) we hae and so or with 1 cos(β) β sin(β) = h a T min = T min = u and so 1 cos(β) h aβ s β sin(β) β sin(β) ag t aβ g(β sin(β)) = F (β) F (β) = β β sin(β) s a g = β sin(β) a (3) AplotofF (β) ersus β is shown below APlotofF (β) ersus β for < β < along with the β 8

29 Note that lim F (β) = lim β + β + β + β = lim β sin(β) β β3 /6 = lim β + s 6 +. β 7.1 Some Examples of The Brachistochrome Problem Let us now consider some examples or plots for the particle s path. Suppose that h =m, and a =1m. Then Euation (c) becomes, 1 cos β = which gies β ' β sin β as the smallest solution. Note we choose the smallest solution since it will lead to the least time as seen by the plot of F (β) ersus β. Using Euations (a,b) we get x = ϕ sin ϕ β sin β and y = and a plot of y ersus x yields the figure below. cos ϕ cos β 1 cos β The Particle s Path using h =m, a =1m Using Euation (3), we get the minimum time of T min '.6943 seconds. The straight-line time is s (a + h T line = ) u = t (1 + ) =.7143 seconds gh (9.8)() 9

30 which is larger than.6943 seconds. Suppose that h =1m, and a =m. Then Euation (c) becomes, 1 cos β β sin β = 1 which gies β ' as the smallest solution. Using Euations (11a,b) we get ϕ sin ϕ cos ϕ cos β x = and y = β sin β 1 cos β. and so a plot of y ersus x yields the figure below The Particle s Path Using h =1m, a =m Using Euation (3), we get the minimum time of T min '.86 seconds. Note here that the particle s path dips below the horizontal leel at y =.Thestraightline time is s (a + h T line = ) u = t ( +1 ) = 1.1 seconds gh (9.8)(1) which is larger than.86 seconds. Suppose that h = a. Then Euation (c) becomes, 1 cos β β sin β =1 which gies β '

31 as the smallest solution. Using Euations (a,b) we get x = h ϕ sin ϕ β sin β and y = h cos ϕ cos β 1 cos β and so a plot of y ersus x (in units of h) yieldsthefigure below The Particle s Path Using h = a Using Euation (3), we get the minimum time of T min '.583 seconds when h =1meter. The straight-line time is s (a + h T line = ) u = t (1 +1 ) =.6389 seconds gh (9.8)(1) which is larger than.583 seconds. Suppose that h =1manda =5m. Then Euation (c) becomes, 1 cos β β sin β = 1 5 which gies β ' as the smallest solution. Using Euations (a,b) we get x =5 ϕ sin ϕ β sin β and y = cos ϕ cos β 1 cos β 31

32 and so a plot of y ersus x yields the figure below The Particle s Path Using h =1manda =5m Using Euation (3), we get the minimum time of T min ' seconds. Note that once again, the particle s path dips way below the horizontal leel at y =. The straight-line time is s (a + h T line = ) u = t (5 +1 ) =.335 seconds gh (9.8)(1) which is larger than seconds. Suppose that h =1manda =1m. Then Euation (c) becomes, 1 cos β β sin β = 1 1 which gies β ' as the smallest solution. Using Euations (a,b) we get x =1 ϕ sin ϕ β sin β and y = cos ϕ cos β 1 cos β 3

33 and so a plot of y ersus x yields the figure below The Particle s Path Using h =1manda =1m Using Euation (3), we get the minimum time of T min '.147 seconds and that once again, the particle s path dips way below the horizontal leel at y =.The straight-line time is s (a + h T line = ) u = t (1 +1 ) = seconds gh (9.8)(1) which is larger than.147 seconds. Note that another solution to Euation (c), 1 cos β β sin β = 1 1 is β ' Using Euations (a,b) we get x =1 ϕ sin ϕ β sin β and y = cos ϕ cos β 1 cos β 33

34 and so a plot of y ersus x yields the figure below The Particle s Path Using the Larger Value of β Note how the particle goes back up to its original height of 1 meters before coming back down. It is clear that the larger alue of β in the solution of Euation (c) shouldnotbeused. NotealsothatusingEuation(3),wegetthetimeofthis motion to be T min '.9583 seconds which is larger that the alue of.147 seconds. It should be clear that the particle s path will dip below the horizontal leel at y =when dy/dϕ >, which occurs when dy dϕ = d h h dϕ 1 cos ϕ = h 1 cos β sin ϕ 1 cos β or when sin ϕ <, which occurs when ϕ > π. Therefore, going back to the plot of F 1 (β) = 1 cos β β sin β and finding when its alue at β = π, weget F 1 (π) = 1 cos π π sin π = π. > Therefore when h b a < π 34

35 the particle s path will not dip below the horizontal leel at y =,andwhen h b a > π it does dip below the horizontal leel at y =.Infactwhen h b a the path should be tangent to the horizontal leel at y =. The following example shows this to be true. Suppose that h =meters, a = π meters, and b =.Then Euation (c) becomes, = π 1 cos β β sin β = π which gies β = π as the smallest solution. Using Euations (a,b) we get x = π ϕ sin ϕ = x = π β sin β µ ϕ sin ϕ = ϕ sin ϕ π sin π and µ 1 cos ϕ 1 cos ϕ y = = =1+cosϕ 1 cos β 1 cos π and so a plot of y ersus x yields the figure below The Particle s Path Using h =manda = π m 35

36 Using Euation (3), we get the minimum time T min = u t aπ g(π sin π) = s aπ g ' 1.3 seconds. The straight-line time is s (a + h T line = ) gh u u = t (π + ) (9.8)() = seconds which is larger than 1.3 seconds. 8. Functional Dependent On Seeral Dependent Variables Suppose a functional is dependent on seeral dependent ariables such as J[y 1,y,y 3,...,y n ]= whichweabbreiateby J[y k ]= f(x, y 1,y,y 3,...,y n,y 1,y,y 3,...,y n)dx f(x, y k,y k)dx. (4) If y 1 = y 1 (x), y = y (x),..., y n = y n (x) are the functions which satisfy the boundary conditions y k ( )=y k1 and y k (x )=y k for k =1,, 3,...,n, and which maximize or minimize J, then we assume y k (x, α) =y k (x)+αη k (x) are small deiations from these, with η k ( )= and η k (x )= for k =1,, 3,...,n.Thisleadsto J(α) = f(x, y k (x, α),y k(x, α))dx 36

37 and we hae J(α) α = α = = = = f(x, y k (x, α),yk(x, α))dx α f(x, y k(x, α),yk(x, α))dx ( α n x x + X y k (x, α) nx k=1 nx k=1 k=1 y k (x, α) η k(x)+ y k (x, α) x y k(x, α) η k(x) y k (x, α) η k(x)+ yk(x, α) η k(x) + yk(x, α) dx dx. yk(x, ) α) dx x Using integration by parts on the second integral, we hae y k (x, α)η k(x)dx = yk (x, α)η k(x) = d dx x y k(x, α) d dx η k (x)dx y k (x, α) since η k ( )=and η k (x )=for k =1,, 3,...,n.Thuswehae Then J(α) α J(α) α = n X k=1 nx = α= k=1 y k (x, α) d dx y k (x) d dx for all choices of η k (x), leadsonlyto y k (x) d dx yk(x) y k(x, α) y k(x) = η k (x)dx. η k (x)dx = η k (x)dx or d = (5) dx yk y k for k =1,, 3,...,n. These are the Euler-Lagrange Euations. 37

38 It is important to point out that although the euations d dx y y = and d y dx y f + x = are euialent when f is a function of only one dependent ariable, y, i.e., f = f(x, y, y ),ITISNOTTRUEthat d = dx yk y k and d yk f + dx yk x = are euialent when f is a function of more than one dependent ariable, y, i.e., f = f(x, y k,yk). What we can do, howeer, is derie an additional euation that follows from the n euationsgienineuation(5).todothiswecompute Replacing each of df dx = x + n X y k ia Euation (5), we then hae df dx = x + n X k=1 k=1 by d y k dy k dx + y k dx = n x + X d dx k=1 d dx yk dyk y k yk yk dyk dx dx + y k dyk dx 38

39 and so we find that d X n y k dx k=1 If / x =, then we may say that y k f + x =. (6) nx y k k=1 yk f = constant. (7) 9. Extra Constraints and Lagrange Multipliers In Ordinary Calculus In this section, we inestigate the Lagrange Multiplier method for soling the extrema problems from ordinary calculus inoling a function of n ariables with m<neuality constraints. That is, we want to max or min z = f(,x,x 3,...,x n ) (Objectie Function) subject to g 1 (,x,x 3,...,x n )=b 1 (Constraint #1) g (,x,x 3,...,x n )=b (Constraint #) g 3 (,x,x 3,...,x n )=b 3 (Constraint #3). g m (,x,x 3,...,x n )=b m. (Constraint #m) where all functions f and the g i s are known and the b i s are known constants. Note that in general, we hae m<n. To introduce the Method of Lagrange Multipliers, we consider the following example, which we sole two different ways. Example 9.1 Consider the following problem. minimize z = f(,x,x 3 )=x 1 + x + x 3 (Objectie Function) subject to +x + x 3 =3 (Constraint #1) 3 +7x + x 3 =5 (Constraint #) One method of solution reuires that we sole the two constraint euations for (say) and x in terms of x 3, and then place this into the objectie function, 39

40 thereby yielding a function of a single ariable. This leads to and which reduces to Then " x1 x # = " 11 5x3 4 + x 3 z =(11 5x 3 ) +( 4 + x 3 ) + x 3 z = x 3 +3x 3 = f(x 3 ). f (x 3 )= x 3 = yields x 3 =1 and the fact that f (x 3 )=6> for all x 3, we know that this is a local minimum. If fact, it is a global minimum as well. Then so that " x1 x # " 11 5x3 = 4 + x 3 # = # " 11 5(1) 4 + (1) (,x,x 3 )=(5,, 1) is the point yielding z min = =47. Notethattheaboecriticalpoint cannot be obtained by simply soling # = " 5 (,x,x 3 ) = (,x,x 3 ) x = (,x,x 3 ) x 3 =. This is because not all of, x and x 3 are independent, since they are related ia the euality constraints, and hence df = (,x,x 3 ) d + (,x,x 3 ) x dx + (,x,x 3 ) x 3 dx 3 = does not imply that (,x,x 3 ) = (,x,x 3 ) x = (,x,x 3 ) x 3 =. # 4

41 The method of Lagrange Multipliers introduces for each constraint a multiplier and defines a new objectie function L = x 1 + x + x 3 + λ 1 (3 ( +x + x 3 )) + λ (5 (3 +7x + x 3 )). In this way, x, x 3 along with λ 1 and λ can be treated as independent and the critical point(s) can be obtained by soling the set of euations L = L = L = L = L = x x 3 λ 1 λ To sole this unconstrained problem, we thus write with L = λ 1 3λ = L x = x λ 1 7λ = L x 3 = x 3 λ 1 λ = L λ 1 = 3 ( +x + x 3 )= L λ = 5 (3 +7x + x 3 )= L = x 1 + x + x 3 + λ 1 (3 ( +x + x 3 )) + λ (5 (3 +7x + x 3 )). This leads to the following system of 5 euations and 5 unknowns = λ 1 +3λ x = λ 1 +7λ x 3 = λ 1 + λ +x + x 3 = x + x 3 = 6. To sole this system of 5 euations and 5 unknowns, we use the first three euations to sole for, x and x 3 in terms of λ 1 and λ.thisleadsto = λ 1 +3λ, x = λ 1 +7λ 41 & x 3 = λ 1 + λ.

42 These are then placed into the last two euations to yield two euations for the two unknowns λ 1 and λ. This leads to λ1 +3λ λ1 +7λ λ1 + λ 3 = and or simply and λ1 +3λ λ1 +7λ λ 1 +3λ =1 18λ 1 +59λ = 1. λ1 + λ the solution to these then gies λ 1 =58and λ = 16. Putting these into the euations for, x and x 3 leads to (,x,x 3 )=(5,, 1) which agrees with our earlier results. Consider now the extrema problem inoling n decision ariables and m euality constraints (with m<n), = max or min z = f(,x,x 3,...,x n ) (Objectie Function) subject to g 1 (,x,x 3,...,x n )=b 1 (Constraint #1) g (,x,x 3,...,x n )=b (Constraint #) g 3 (,x,x 3,...,x n )=b 3 (Constraint #3). g m (,x,x 3,...,x n )=b m. (Constraint #m) where all functions f and the g i s are known and the b i s are known constants. The m euality constraints indicates that not all the n ariables are independent. In fact, if all the euality constraints are independent, then only n m of the n ariables are independent. This means that we cannot determine the critical points by simply soling (,x,x 3,...,x n ) x j = 4

43 for j =1,, 3,...,n. To get around this problem, we introduce a Lagrange Multiplier λ i for each constraint, and define the function L(x, λ) =f(x)+ mx i=1 λ i (b i g i (x)) (8a) where x (,x,x 3,...,x n ) and λ (λ 1, λ, λ 3,...,λ m ). Then, to determine the critical points to the objectie function f(x), wesolethesystemofn + m euations L x j = and L λ i = for j =1,, 3,...,n and i =1,, 3,...,m. This leads to the n + m euations for j =1,, 3,...,n and for i = 1,, 3,...,m. constraint euations. (x) x j mx i=1 λ i g i (x) x j = (8b) b i g i (x) = (8c) Note that the last set of m euations are just the m 1. Extra Constraints and Lagrange Multipliers In Variational Calculus The Lagrange Multiplier method discussed aboe for ordinary calculus problems can be extended to problems in ariational calculus and perhaps the simplest waytoshowthisisbysolingthehangingcableproblemintroducedaboe. Example 1.1: The Hanging Chain Problem - A Solution Auniformflexible chain (haing linear mass density ρ) hangs under its own weight between two support. The chain has a fixed length of L and we seek that shape y = y(x), which minimizes the total potential energy of the chain. Toward this end we assume that the chain hangs between the two points ( a, ) and (a, ). We had seen that to sole this problem, we seek a function y = y(x) so that the boundary conditions y( a) =y(a) = (9a) 43

44 are satisfied along with the extra constraint, L[y] = Z a a 1+(y ) dx = L = constant (9b) for which Z a U[y] =ρg y 1+(y ) dx (9c) a is as small as possible. The Lagrange Multiplier method considers the functional or J[y] = Z a a µz a ρgy 1+(y ) dx + λ 1+(y ) dx L a J[y] = Z a a (ρgy + λ) 1+(y ) dx λl. Now λ and L are constants, but in some cases, it may be necessary to allow λ to be a function of x. Forhere,sinceλL is a constant, let us consider just and using J[y] = Z a a (ρgy + λ) 1+(y ) dx f(x, y, y )=(ρgy + λ) 1+(y ) in the Euler-Lagrange Euation (1b), we hae with x = y y f = constant yielding or (ρgy + λ)y y 1+(y 1+(y ) (ρgy + λ) ) = A ρgy + λ 1+(y ) = A. Soling for y,weget (ρgy + λ) A y = A 44

45 which leads to dx = Ady (ρgy + λ) A which leads to Soling for y, we get x + B = A ρgy + λ ρg cosh 1. A y(x) = A ρg cosh µ ρg A (x + B) λ ρg. Setting y( a) =y(a) =reuires that B =so that y(x) = A µ ρgx ρg cosh λ A ρg and leads to Thus we hae or simply y(a) = A µ ρga ρg cosh λ A ρg = λ = A cosh y(x) = A µ ρgx ρg cosh A µ ρga. A A µ ρga ρg cosh A cosh(cx) cosh(ca) y(x) = C To determine C, we still hae to reuire that which leads to L = Z a a L = Z a a 1+(y ) dx where C = ρg A. Z a 1+sinh (Cx)dx = cosh(cx)dx = sinh(ca) a C 45

46 Therefore we find that the solution to the hanging chain problem is y(x) = C is the solution to the euation cosh(cx) cosh(ca) C sinh(ca) = L Ca a. A plot of the function sinh(z)/z is shown in the figure below z A Plot of the function sinh(z)/z Since recall that a <Lso that L/a >1, we see that there will always be a solution to the euation sinh(ca) = L Ca a > 1. Example 1.: The Maximum Area Problem Gien a simply closed cure in the xy plane that has a fixed perimeter P, we seek that cure which encloses the maximum area. Note that a simple closed cure is one that DOES NOT intersect itself. For example, a figure 8 is not a simple closed cure since it intersects itself at the center. A simple closed cure can be written in parametric form as r = r(t) =x(t) b i + y(t) b j 46

47 and its perimeter can be computed using I I Z t P = dr = dx + dy = ẋ + ẏ dt C C t 1 where t 1 6= t and r(t 1 )=r(t ). The area enclosed is computed using I I A = 1 r dr = 1 ydx) = C C(xdy 1 t 1 (xẏ yẋ)dt. Thus we seek two functions x = x(t) and y = y(t) such that the boundary conditions x(t 1 )=x(t ) and y(t 1 )=y(t ) are satisfied along with the constraint, P [x, y] = L[x, y] = 1 Z t t 1 t 1 Z t Z t ẋ + ẏ dt = P = constant and for which A[x, y] = 1 Z t (xẏ yẋ)dt t 1 is as large as possible. To sole this we introduce the Lagrange multiplier λ and define Z t Z t (xẏ yẋ)dt λ ẋ + ẏ dt or where = t 1 L[x, y] = ½ 1 (xẏ yẋ) λ ẋ + ẏ ¾ dt Z t t 1 t 1 f(x, y, ẋ, ẏ)dt f = 1 (xẏ yẋ) λ ẋ + ẏ Then yields d dt ẋ x = and d dt d 1 dt y λ ẋ ẋ 1 + ẏ ẏ = ẏ y = 47

48 and These reduce to ẏ + d dt µ d 1 dt x λ ẏ ẋ 1 =. + ẏ ẋ λẋ ẋ = and ẋ d + ẏ dt or, after integrating with respect to time t, y + which says that or λẏ ẋ + ẏ = λẋ ẋ + ẏ = c λẏ and x ẋ + ẏ = c 1 (y c ) +(x c 1 ) = λ ẋ ẋ + ẏ + λ ẏ ẋ + ẏ = λ (x c 1 ) +(y c ) = λ which is the euation of a circle of radius λ and center (c 1,c ), and these can be obtained using the boundary conditions along with the constraint, x(t 1 )=x(t ) and y(t 1 )=y(t ) P [x, y] = Z t t 1 ẋ + ẏ dt = P = constant. 11. Variational Principles in Physics - Least Time In Optics Many fundamental principles in physics are based on ariational principles. For example, Fermat s principle in optics states that when light traels between two point A and B in space, the path taken by the light will be that path which minimizes the time of trael. Conseuently, if light is traeling through a medium haing index of refraction n(x, y, z), thenthespeedoflightthrough thismedium is gien by c = n(x, y, z) 48

49 where c is the speed of light in acuum. Therefore, if r = r(t) is the path following by the light from point A to point B, then = c n(x, y, z) = resulting in n(x, y, z) dt = dr = c where ds = dr is arc length. Thus we hae Z B dr dt = dr dt n(x, y, z) ds c T = 1 c A n(x, y, z)ds (3a) as the time of trael which is to be minimized. For planar motion, this is simplified to T [y] = 1 Z xb n(x, y) 1+(y c ) dx x A (3b) and can be minimized using the methods discussed aboe. In fact, using the Euler-Lagrange euations, we find that d n(x, y)y n(x, y) 1+(y d+(y ) y ) = along with the boundary conditions gies the path followed by the light. After a little algebra, this euation reduces to the non-linear differential euation n(x, y) n(x, y)y + y n(x, y) (1 + (y ) )=. (31) x y Of course if n is not an explicit function of y, thenwehae µ n(x) 1+(y y ) = constant which reduces to y 1+(y ) = 1 An(x) or dx dy = A n (x) 1 (3a) 49

50 for a constant A. On the other hand, if n is not an explicit function of x, thenwe hae µ y n(y) 1+(y y ) n(y) 1+(y ) = constant or n(y) 1+(y ) = 1 A which reduces to dx = dy A n (y) 1 (3b) for a constant A. 1. Variational Principles in Physics - Least Action In Mechanics Hamilton s principle in mechanics inoling a system of N particles with n degrees of freedom described by the generalized coordinates 1,, 3,..., n and corresponding generalized speeds 1,, 3,..., n can be stated as follows. Suppose that such a system can moe from an initial configuration α at time t α to another configuration β at time t β. The action of the system between these two timesisdefined by Z tβ A = L( k, k,t)dt (33a) t α where L = T V is the Lagrangian of the system. Hamilton s principle of least action simply states that a mechanical system eoles in time so that A is as small as possible. Of course, using the ideas discussed aboe, we may now say that this leads to the set of Lagrange euations for k =1,, 3,...,n. d dt L k L k = (33b) 5