Calculus of variations - Lecture 11

Transcription

1 Calculus of variations - Lecture 11 1 Introuction It is easiest to formulate the problem with a specific example. The classical problem of the brachistochrone (1696 Johann Bernoulli) is the search to fin the path that a falling object must take to travel between two points in minimum time. Figure 1 shows the geometry. The path we seek is given by y(x) in the 2-D plane. The arc length of an element on this path is; (x 1, y ) 1 Path1 M Path2 (x, y ) 2 2 Figure 1: The geometry, showing various paths in orer to fin the one where a falling object moves in minimum time between the points l = x 2 + y 2 = x 1 + (y ) 2 The time to travel over the element is, t = l/v (x), where V (x) is the velocity. The total time between the points is; x (y ) t = x 2 x 1 V (x) We are to fin the function, y(x), which results in a minimum of t. The motion is in a gravitational fiel which etermines V (x). This particular problem may be reuce by using conservation of energy. Assume that the gravitational fiel is constant which prouces a force proportional to the istance. The potential energy taken from the gravitational fiel increases the kinetic energy of the object. Total energy is conserve. Therefore; 1

2 (1/2)mV 2 + Mgy = constant The mass of the object, M, starts from rest where the initial potential energy is Mgy 1 = constant. (1/2)V 2 = Mg(y 1 y) Substitution for V into the equation for the time; t = 1 x 2 x 1 + (y ) 2 2g x y1 1 y 2 General Statement of the problem Now we use the above example to write a general statement of the problem. Suppose a function of the form, L(x, y(x), y (x)). This is embee in the integral; x 2 I = x L x 1 We wish to fin the function, y(x), which causes I to have an extremum, ie either a maximum, minimum, or inflection point. Suppose we allow the solution to have the form, y (x), an provie the variation in the path by another function, η(x). Write this as; y(x) = y (x) + λ η(x) The constant, λ, etermines the amount of the arbitrary function η(x) inclue in the correct function we seek. Note that η(x 1 ) = η(x 2 ) = as the path has fixe en points. These are bounary conitions. In general, I, is a function of λ so λ I =. x2 I λ = = x [ L y x 1 y λ + y L y λ ] Then y y = η(x) an λ λ = η (x). Put this back in the equation for λ I an integrate the secon term by parts. x 2 x L y x 1 y λ = [ L y η(x)] x 2 x 1 x 2 x x 1 x ( L y ) η(x) The surface term vanishes so the final result for the integral is; x2 I λ = = x [ L x 1 y x ( L y )]η(x) 2

3 The function η(x) is arbitrary, so the integran itself must vanish. [ L y x ( L y )] = This is the Euler-Lagrange equation. It is a ifferential equation whose solution is the function, y(x). There are several forms of this equation which can be obtaine for special cases. In the case where L = L(x, y) L y = L = f(x) In the case where L = L(x, y ) x [ L y ] = L y = constant L = cy + b In the case where L = L(y, y ) Use the ientity; f x [y y f] = y [ f y x ( f y )] f x Then if f x = an y L [y x y L] = [y y L L] = constant 3 Brachistochrone For this problem, L = 1 + y 2 y1 y L is inepenent of x. Thus from the last section; 3

4 [y L y L] = constant Substitution where the constant is efine by c; y 2 (1 + y 2 ) 1 + y 2 y 1 y = c y 2 = 1 c2 (y 1 y) c 2 (y 1 y) The equation may be solve by integration using a constant a; x = y1 y y a (y1 y) Let (y 1 y) = a sin 2 (θ/2). Then after integration; y = y 1 a sin 2 (θ/2) = y 1 (a/2)(1 cos(θ) This is the parametric equation of a cycloi which is prouce by a point of a circle rolling on a flat surface. 4 Geoesics Suppose we want the arc of minimum length connecting 2 points on a surface in 3-D. In this case we suppose two parameters to obtain the parametric equations; x = x(u, v) y = y(u, v) z = z(u, v) As an example, the spherical surface has r = constant, with variables, θ an φ. A surface is efine by the function. g(x, y, z) = an the arc length is; l 2 = x 2 + y 2 + z 2 As the surface epens on the variables,(u, v), we can expan x to be; x = u x u + x v v There are similar expressions for y an z. These are substitute into the expression for the square of the elemental length; 4

5 l 2 = P u 2 + 2Q u v + R v 2 P = ( x u )2 + ( y u )2 + ( u z )2 Q = 2 x u v + 2 y u v + 2 z u v R = ( x v )2 + ( y v )2 + ( v z )2 The length of the arc is then; u 2 I = u P + 2Q v + R v 2 u 1 Derive the Euler-Lagrange equations with the integran efine by; L = P + 2Q v + R v 2 It is useful here to focus the example on spherical coorinates. In this case; x = a sin(v) cos(u) y = a sin(v) sin(u) x = a cos(v) P = a 2 sin 2 (v) Q = R = a 2 L = a sin 2 (v) + v 2 L is inepenent of u ( u was x in the previous evelopment an v was y). The Euler- Lagrange equation is; v L v L = constant Use the above value for L, ifferentiate, an collect terms. v 2 = a sin 4 (v) sin 2 (v) 5

6 Solve the oe by integration. v 2 u = v v 1 1 a sin4 (v) sin 2 (v) sin(u B) = A cot(v) In this equation, A an B are constants. The expression can be shown to represent an arc in a plane which passes through the center of the sphere - a great circle. 5 Lagrange Multipliers Suppose a function, f(x, y, z). For f to be an extrememum; F = f f f x + y + x y z z = f s = an f must not change in any of the 3 irections. Thus; f x = f y = f z = Suppose the variables are subject to a constraint, g(x, y, z) =. One coul solve this expression for one of the variables, an substitute into f reucing the egrees of freeom in f. Then using g as a constraint, consier H = f + λg. H = f + λg H = [ f x + λ g ]x + [ f x y + λ g ]y + [ f y z + λ g z ]z Because of the constraint, f x f y f z Choose the constraint parameter λ so that g = z z(x, y). [ f z + λ g z ] = [ f z + λ] = [ f x + λ g ]x + [ f x y + λ g y ]y = Since x an y are inepenent; [ f x + λ g x ] = 6

7 y = yo y. A B. Figure 2: Constraining the search for a maximum in f by the function y = y. Contours represnet the lines of equal magnitue x f y + λ g y = The choice of λ provies H =. Thus we have the equations; [ f x + λ g x ] = [ f y + λ g y ] = [ f z ] + λ = For a set of N constraints, we expect the equations; f + N x i k=1 λ k g k x i = The λ k are the Lagrange multipliers an they o not have to be explicitely etermine. Figure 2 shows the result of constraining the function, f, by the equation y = y when looking for a maximum. The maximum at A shifts to the maximum at B along the constraint curve. 6 Example A cable of length L hangs between two points at the same height on walls a istance,, apart. This is illustrate in Figure 3. Note that the gravitational potential energy (PE) must minimize in the equilibrium position. Let λ be the mass per unit length of the cable. 7

8 x y Figure 3: An example of a hanging cable PE = g(λ l) y(x) = gλ l 2 = x 2 [1 + ( y x )2 ] PE = gλ xy[ 1 + y 2 ] x( l x ) y L = y[ 1 + y 2 ] The length of the cable is the constraint. L = l = x 1 + y 2. Let G = 1 + y 2, an employ a Lagrange multipier to inclue the constraint. H = L + ηg Note that L = y G an H is inepenent of x. The Euler-Lagrange equation is then; y H y H = constant Substitute for H, write the constant as c, an collect terms. y = ±[(1/c 2 )(y + ǫ) 2 1] 1/2 The slope changes irection as the position varies x. Integrate the equation to fin the position, x. x = c[cosh 1 ( y + η c ) cosh 1 (η/c)] x /2 8

9 Invert this equation to obtain; y = c cosh[(x/c) + cosh 1 (η/c)] η/c At the mipoint the slope is zero, so /2c = cosh 1 (η/c). Then let c = γ so that η = γ cosh(/2γ) Fin the value of γ using the constraint on the length. L/2 = /2 x 1 + y 2 = The above is integrate to give; /2 x [1 + sin 2 (/2 x)2 ( γ )] 1/2 L/2 = γ sinh(/2γ) 7 Electrostatics - Minimizing energy The electric fiel is obtaine from the potential function, φ(x, y, z), by partial ifferentiation, E = φ. The energy ensity containe in the fiel is; W (V olume) = ǫ 2 E E The fiel is obtaine by the charge ensities as they arrange themselves on a conucting surface to obtain an equi-potential. This prouces an energy minimum subject to constraints of the electrostatic problem. Thus; W = (ǫ/2) τ φ 2 Look for an extrememum of this integral. The integran is L = L(φ x, φ y, φ z ). Now introuce a constraint η(x, y, z) an efine; V i = L i + λ η i The bounary conition is that the potential is constant, η i = on the conucting surfaces, i = x, y, z. The extrememum is obtaine by W =. This results in the Euler-Lagrange η equation which in this case is Laplace s equation. 9

10 2 V i x 2 i = Then as a simple example, choose a set of functions, V = V (r/b) p with p < so that the potential converges as r. We are to minimize W with respect to p. Substitute; W = V 2 1 ǫ 2b 2p W = 2πV 2 1 p 2 b 2(2p + 1) b r 2 r(p r p 1 ) 2 Ω W = (2πV 2 p(p + 1) p 1 b) 2p + 1 = The solutions are p =, 1 so V = V 1 (r/b) 1. In this case we have the exact solution since the function is an exact solution to Laplace s equation. W = V 2 1 b/2 However suppose we ha chosen a function of the form, V = V 1 e α(r b)/b. This oes not satisfy Laplace s equation but oes satisfy the bounary conition. Put this form into the integral to fin the value of W. W = πv 2 1 ǫb α [1 + (1/2α2 )(2α + 1)] W α = (πǫv 2 1 b) [ln(α) 1/α 1/4α2 ] = α 1.85 W 1.23(V 2 1 b/2) Compare this to the exact answer above. This technique provies an approximation metho which will be explore further later. 8 Example A capacitor is charge, then ippe into a ielectric flui. The flui is rawn up into the capacitor. No issipation is assume, so energy is conserve. A cross section of the geometry is shown in Figure 4. The with of the capacitor out of the page is a. Assume the separation of the plates, s is the length of the plates, L, so a simple parallel plate capacitor is assume (ie the fringe fiels are neglecte). The flui ensity is ρ, an has ielectric constant, ǫ = ǫ r ǫ. The capacitor is isconnecte from the charging voltage before ipping into the 1

11 s L ε h Figure 4: The geometry of the capacitor problem flui. The charge on the capacitor plates oes not change, but reistributes as the voltage on the plates changes as the flui rises between the plates. Energy then is rawn from the capacitor to raise the flui against gravity. First fin the potential energy of the flui raise to height, h. Potential Energy (PE) = m = ρ asy PE = asρgh 2 /2 h m g y The capacitance of a simple parallel plate capacitor is C = A ǫ/s where A is the plate area, s their separation, an ǫ the ielectric constant. The energy store in the capacitor is W C = (1/2)CV 2 = (1/2)Q 2 T /C, where V is the voltage, an Q T the store charge. In this problem the capacitor is charge an isconnecte from the voltage. The total charge on the plates remains on the plates but reistributes as the flui is rawn up into the plates. However, the voltage between the plates changes. We use the energy W = (1/2)Q 2 /C. The capacitor is viewe as 2 capacitors in parallel - one with ielectric, ǫ = ǫ r ǫ, an one with ielectric, ǫ. The energy store in this final capacitor system is; W fc = (1/2)[ Q2 1 C 1 + Q2 2 C 2 ] 11

12 C 1 = ahǫ s C 2 = a(l h)ǫ s There is a constraint that Q T = Q 1 +Q 2. Write the equation for the conservation of system energy as; W = (1/2) Q2 T s alǫ asρgh2 2 + (1/2) aǫ s [ Q2 1 + Q2 2 hǫ r (L h) ] Solve for the equilibrium height, h, subject to the constraint Q T = Q 1 + Q 2. One coul use this constraint with the efinitions of the capacitance an conservation of energy to reuce the above equation into a variable of h alone, an then set the erivative of the energy to zero to fin h. We solve the problem here using Lagrange multipliers. Define the constraint conition as f = Q T Q 1 Q 2 =, multiply this by a Lagrange multiplier λ an subtract from the energy equation above. Then take the 4 partial erivatives; W = asρgh + (1/2) s h aǫ [ Q2 1 ǫ r h 2 W = Q Ts λ = Q T alǫ W = sq 1 + λ = Q 1 ahǫ ǫ r W sq = 1 + λ = Q 2 a(l h)ǫ ǫ r Q 2 1 (L h) 2 ] = Solve these equations. The partials with respect to the Q s introuce the conservation of charge. The final result is; h = W C ǫ ρ(volume)g [ǫ r 1] 9 Least Action Suppose the motion of a boy between two points, A an B. The motion can be along any path which keeps the energy conserve, see Figure 5. In general the time require to travel the ifferent paths is not hte same. From Newton s laws; t [m i r i t ] = F i The inex i represents the i th particle, with m i the mass, r i its isplacement, an F i the force. Now let δ r i be a isplacement of the particle from its path. Then; 12

13 B Path 2 A Path1 Figure 5: Possible paths for a particle to travel in space-time between A an B i t [m i r i t ] δ r i = i F i δ r i This can be manipulate into the form (use δ[ r i t ] = r i t t [ i m i r i t δ r i] δ i (1/2) m i v 2 i i m i v 2 i (δt) t (δt) t ); = i F i δ r i The change in potential energy is, δv 1/2 m i vi 2. The final result is; i t [ i = i m i r i t δ r i] = δ(t V ) + 2T t δt F i δ r i, an the kinetic energy is T = T + V = constant so δt = δv. Integrate the power over time ; t 2 t 1 t t [ i m i r i t r i] = t 2 Since all paths arrive at the same time; t2 δ t (2T) = t 1 t 1 t [δ(2t) + 2T t (δt)] This is the principle of least action. It states that among all paths between 2 points in space time, the path which is chosen is the one in which the time is an extrememum. 13

14 1 Example Suppose a particle moves subject to no applie forces. There will then be no potential energy, an the kinetic energy enerqy is constant. δt = δv = The position as a function of time is x(t). Therefore using the form for the velocity u = x t, the Euler-Lagrange equation is; T x t ( T u ) = T = (1/2)µ 2 m u t = u = constant t u = µ 11 Example - Fermat s principle Suppose a path for light in a meium has a vlocity, u(y). The time for a ray to travel from (x 1, y 1 ) to (x 2, y 2 ) is; t = (x 2,y 2 ) (x 1,y 1 ) l/u = (x 2,y 2 ) (x 1,y 1 ) x (1 + y,2 ) u This is the brachistochrone problem with Euler-Lagrange equation; y f y f = c with c a constant. Substitution for the value of f gives; 1 u 1 + y 2 = c Here, rays of equal time have the same phase. If we think about all possible paths, there is a large number with nearly the same phase about the stationary point. Thus in some sense relate to Quantum Theory, the probability of the stationary path is the largest. 14

15 12 Hamilton s principle Consier a system of particles subject to constaints. Develop the equations of motion using least action. t [m i r i t δ r i] = Fi δ r i In the above δ r i is an arbitrary isplacement subject to constraints. The equation of least action is obtaine. t [m i r i t δ r i] = δ(t V ) + 2T t (δt) If paths are trace which have equal times then the last term on the right vanishes. m i r i t δ r i t t = δ t t t (T V ) t However, if all times are the same then δ t(t V ) =. This is Hamilton s principle, t an L = T V is the Lagrangian. In general the Lagrangian is a function of the coorinates, q i an the time erivatives, q i. The Euler Lagrange equations for the Lagrangian, L, is; L g i t [ L q i ] = Now L is inepenent of the parameter, t. Thus we obtain; t [ q L q L] = The potential is only a function of the coorinates so V q qi T = 2T qi =. Thus; Then look at L = q i T qi L = 2T T + V = T + V = Total Energy. Now efine; H = p i q i L In the above p i = T q i. The function H is the Hamiltonian of the system. Then; H pj = q j + p i q i p j L qi q i p j = q i H = T qi q i L = T + V H is the system energy. Then look for the extrememum of the integral; 15

16 t 2 I = t L = t [ p i q i H] t 1 Here p i an q i are consiere inepenent variables. Introuce the constraint; q i H p i =. This is inclue by Lagrange multipliers, λ k. We obtaine the moifie Lagrangian; f = p i q i H + λ k [ q k H p k ] This prouces a set of Euler Lagrange equations; f p f = i t ṗ i f q i t f q i = Which give the following equations; q i H λ 2 H p k = i p i p k H q i λ k 2 H q i p k t [p i + λ i ] = We fin that 2 H = so choose the solution, λ p i p k =. This prouces the relation k between conjugate variables. ṗi = H q i q i = H pi This system of equations constitute Hamilton s formulation of the laws of motion. It is now useful to compare Hamilton s principle to the principle of least action. In Hamilton s principle, all paths are travele in the same time. In the principle of least action the energy is kept constant over the given path. 13 Example A particle is constraine to move in a verticle circle uner the gravitational force. The potential energy, PE, is; 16

17 PE = mgr[1 + sin(θ)] The kinetic energy, KE, is; KE = (1/2)mv 2 = (1/2)m[ṙ 2 + r 2 θ] The constraint is that r = R a constant. Construct the Lagrangian, L = T V. L = (1/2)m[ṙ 2 + r 2 θ] mgr[1 + sin(θ)] The Lagrangian has variables, L(r, θ, ṙ, θ). Introuce the constraint using the Lagrange multiplier. The equations of motion are; L = L + λ(r R ) L r t [ L ṙ ] = L θ t [ L θ ] = This prouces the equations; mg[1 + sin(θ)] + λ t [mṙ] = mgr cos(θ) t [m θr 2 ] = r = r = ṙ = R θ + t cos(θ) = λ = mg[1 + sin(θ)] On the other han, since L is inepenent of the parameter t; L ṙ L ṙ = constant This just results in the conservation of energy. The total energy is then; W = (1/2)mr 2 θ2 mgr [1 + sin(θ)] As this now results in a first orer oe, the equation can be easily integrate. 17

18 s m θ M x Figure 6: The geometry of a block sliing of a moveable wege 14 Example A box on mass,m, slies own an incline plane of length, L, an mass, M, which is itself is able to slie horizontally. Choose the coorinates along the sloping ege of the slie to be s, an the horizontal irection of its motion to be, x. The kinetic energy of the slie is (1/2)Mẋ 2. The velocity of the box is; v = ( x t t s cos(θ))ˆx s t sin(θ) ŷ The total kinetic energy is then; KE = (1/2)M[ x t ]2 + (1/2)m[( x t s t )2 + ( s t sin(θ))2 ] The potential energy is, PE = mgs sin(θ) The Lagrangian is L = T V = (1/2)(M + m)( x t )2 m x t s cos(θ) + (1/2)m(s t t )2 + mgs sin(θ) Apply the Euler Lagrange formulation for the minimization to obtain the two equations; st [(M + m)(x t ) ms t cos(θ)] = t [ mx t cos(θ) + mgs] mg sin(θ) = t Integrate the first equation above which represents conservation of momentum. (M + m)( x t ) ms t = C 1 18

19 Assume the system starts at rest an C 1 = then a secon integration gives; (M + m)x ms cos(θ) = C 2 Choose x = s = when t = so that C 2 =. Then integrate the secon equation above using the initial conitions. ( x t Eliminate x t cos(θ) + s) gt sin(θ) = t an integrate again to obtain; s = [ M + m M = m sin 2 (θ) ]gt2 2 sin(θ) The position, x, can be obtaine by substitution in an above equation. The time for the box to slie to the bottom when s = L is; t = 2L g sin(θ) [(M + m) sin2 (θ) ] M + m 15 Relativistic Lagrangian formulation We wish the Lagrangian which reprouces the relativistic equations of motion. We choose; L = mc 2 [1 1 β 2 ] V ( r) In the above β is the particle velocity in units of the velocity of light, c, an the potential is V ( r). The potential itself is a scalar function an the argument r simply means it epens on all 3 coorinates. Apply the Euler Lagrange equations; t [ L ẋ i ] L x i = Note that ẋ i = v i an r = x i ˆx i. Substitute into the Euler Lagrange equations to obtain; v/c 2 t [mc2 ] + V 1 β 2 t [γmc β] = V = F The relativistic momentum is γmc β, an the time change of the momentum is the force. 19

20 16 Eigenfunction problem Consier fining the extrememum of I by varying the function f(x). x 2 I = x [τf 2 µf 2 ] + a 1 f 2 + a 2 f 2 x 1 x 2 The function, f, must satisfy the conition, xσf 2 (x) = 1. This is a normalization conition. the other functions, τ(x) an σ(x) satisfy normal ifferentiable conitions, an a i x 1 are non-negative constants. The above can be written; x 2 I = x [τf 2 ] µf 2 + x 1 x (af2 )] The function f is change to inclue the constraint by f f + λ σf 2, where λ are the Lagrange multipliers. Apply the Euler Lagrange equation to obtain; x [τf + (µ + λσ)f = This results in a Sturm-Liouville equation in self ajoint form. Suppose we choose τ = x, µ = η 2 /x, an σ = x. Make the substitution, z = λx an x 1 =. The result is Bessel s equation with solution, J n (z); z 2 2 f z 2 + z f z + (z2 n 2 )f = < z < x 2 λ 17 Ritz metho Assume an eigenvalue problem an arrange the eigenvalues, λ 1 < λ 2 <. Note that the eigenfunction, f n is associate with eigenvalue, λ n. The eigenfunction, f n minimizes the integral, I, subject to a normailzation constraint. Suppose we expan the function, φ(x), which will be taken as a trial solution, in terms of f n. φ = c n f n c n = x 2 xσ f n φ Require, c 1 = c 2 = = c k 1 =. The normalization requirement is; x 2 xσf 2 = x 2 c n xσφf = c 2 n = 1 2

21 Substitute into I without the normalization constraint. I = x 2 c n x [τφ f n µφf n ] Integration by parts an ignoring the surface terms; I = x 2 c n xφ[ x (τf n ) + µf n] Now f n satisfy the oe of the eigenvalue problem. Thus I = x 2 I = c 2 n λ n xφλ n σf n However, c n = for n < k an λ n > λ k so that I is a minimum when λ = λ k. Thus the minimization of I yiels an eigenvalue λ k 18 Example Suppose the equation; y + λy = the solution is y = A sin( λx) + B cos( λx). The bounary conitions are chosen to be y() = 1 an y(1) =. Thus A =, an λ = π/2. For the lowest eigenvalue; y 1 = cos(πx/2) Then take a trial function, y = 1 x a, as a potential solution. This function satisfies the bounary conitions. We want to choose the best value for a. The normalization conition requires that; 1 xy n = 1 Put this back into the value of I an integrate. 21

22 I = 1 1 xy[ x y ] xy 2 x The term in the enominator is the normalization integral. The result is; I = Then take I a (2a + 1)(a + 1) 2(2a 1) = which gives; a = 1 ± 1 + (4)5/4 2 [ ].7247 a = The negative value cannot satisfy the bounary conition. Thus I = Note that the true value is (π/2) 2 =