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g
B = ( α, –b) α, b > 0 Fig. 9.2
which in our case can be written f (1 + f
2 ) = c
2 , (9.18) and hence (c 2 /f ) − 1 −1/2
df = dx 1 . Using the substitution f = c 2 sin
2 ϕ/2 this
yields x 1 = k(ϕ − sin ϕ), f = k(1 − cos ϕ), (9.19) where the positive constant k must be determined by imposing the condition that the point passes through (α, −β). It follows that the brachistochrone curve is an arc of a cycloid. An excellent illustration of the use of the Euler equations is given by the problem of determining the ‘shortest path’ connecting two points on a Riemannian manifold M . If (u 1 , . . . , u l ) are local coordinates, consider the curve t → u(t) = (u 1 (t), . . . , u l (t)) on the manifold, with 0 < t < 1. If g ij (u 1 , . . . , u l ) is the Riemannian metric given on M , the length of the curve u is given by the functional (u) = 1
l k,j
=1 g kj (u 1 (t), . . . , u l (t)) ˙
u k (t) ˙ u j (t) dt. (9.20) By Theorem 9.1 a curve makes the length (u) stationary if and only if it is a solution of the Euler equations (9.12), where F = l k,j
=1 g kj ˙ u k ˙ u j , i.e. a 9.2 Analytical mechanics: variational principles 307 solution of d dt g ij ˙ u j g kj ˙ u k ˙ u j = (∂g
kj /∂u
i ) ˙
u k ˙ u j 2 g kj ˙ u k ˙ u j , (9.21) where i = 1, . . . , l and we have adopted the convention of summation over repeated indices. Take the natural parameter s = s(t) = t 0 g kj ˙ u k ˙ u j dt on the curve. We then find d dt = g kj ˙ u k ˙ u j d ds , and substituting this into (9.21) gives ∂g ij ∂u k du k ds du j ds + g ij d 2 u j ds 2 = 1 2 ∂g kj ∂u i du k ds du j ds , i.e. g ij d 2 u j ds 2 + 1 2 ∂g ij ∂u k + ∂g ik ∂u j − ∂g kj ∂u i du k ds du j ds = 0. Multiplying both terms by g ni (the elements of the inverse matrix of (g ij )) and summing over i, we find the geodesic equation (1.68). We have proved the following. T heorem 9.2 Among all paths connecting two fixed points on a Riemannian manifold, the geodesics keep the length functional (9.20) stationary. Remark 9.3 In reality we have not proved that the geodesics make the length functional attain its minimum. Indeed, this is generally false. Consider as an example a pair of points not diametrically opposed on a sphere; they are connected by two arcs of a maximal circle. Both these arcs make the length functional stationary, but only one of them realises the minimal distance. If the two points are dia- metrically opposed there are infinitely many geodesics of equal length connecting them. A more interesting case is the case of a flat bidimensional torus (Fig. 9.3). In this case, it is easy to verify that, given any two points, there exist infinitely many geodesics connecting them. Only one of them minimises the length. However, it can be proved (see, for example, Dubrovin et al. 1991a, chapter 5) that for any given pair of points on a Riemannian manifold, sufficiently close to each other, the shortest path connecting them is unique and it is given by a geodesic.
308 Analytical mechanics: variational principles 9.2
3
2
1
2
3
Q Fig. 9.3 The curves γ 1 , γ
2 and γ
3 are geodesics connecting the two points P and Q on the torus. The minimal length is attained by γ 1 . Note that the three geodesics are not homotopic. We now consider the problem of seeking the stationary points of a functional in the presence of a constraint. We illustrate this for the case of the functional ϕ(q) =
t 1 t 0 F (q(t), ˙q(t), t) dt, (9.22) with constraint t 1 t 0 Φ (q(t), ˙q(t), t) dt = c, (9.23) where
Φ is a function with the same properties as F , and c is a constant. The problem can be solved by writing Euler’s equations for the function G = F + λ Φ . These, together with (9.23), yield the unknown q(t) as well as the Lagrange multiplier λ. Example 9.4 Among all plane closed curves of fixed perimeter, find the curve which encloses the maximal area (isoperimetric problem). We seek the curve in the parametric form x 1 = f (t), x 2 = g(t), 0 < t < 2π. 9.2 Analytical mechanics: variational principles 309 The constraint is 2π 0 (f 2 + g
2 ) 1/2 d t = ,
(9.24) and the functional to be studied is x 2
x 1 , i.e. 2π 0 g(t)f (t) d t. (9.25) Hence the function under investigation is G = gf + λ(f 2 + g 2 ) 1/2 , and we can write the equations ∂G ∂f
2 + g
2 ) −1/2 = c 1 , d d t ∂G ∂g = ∂G ∂g = f . By integrating the second equation we obtain g − c 1 = −λf (f 2 + g
2 ) −1/2 , f − c 2 = λg (f
2 + g
2 ) −1/2 . Squaring and summing, we finally obtain the equation of a circle: (g − c
1 ) 2 + (f − c
2 ) 2 = λ 2 . (9.26) The constants c 1 and c
2 do not play an essential role, as their variation only produces a translation. The multiplier λ is determined by (9.24), as we must have that 2πλ = . To complete the solution of the problem, we must prove that by perturbing the circle, which we assume to be of radius 1, keeping the same length of the resulting curve, the enclosed area is reduced. We write the equation of the circle of radius 1 in the form x = x 0 (ϕ) and the equations of the perturbed curves in the form x(ϕ) = x
0 (ϕ)(1 + f (ϕ)), 0 < ϕ < 2π, (9.27)
where f is 2π-periodic and such that f 1, f
1, and f = max 0≤ϕ≤2π
|f(ϕ)|. We hence consider only perturbed curves which enclose a ‘starred’ domain (i.e. a domain which contains all the radii ensuing from one of its points, suitably chosen). Indeed, it is easy to realise that if a domain is not a star domain, we can modify the curve preserving its length, but enlarging the enclosed area, so we exclude such domains from our analysis. Since x 0 (ϕ) is the unit vector tangent to the circle, the length of the curve (9.27) is given by (f ) =
2π 0 [(1 + f ) 2 + f
2 ] 1/2 d ϕ.
310 Analytical mechanics: variational principles 9.2 We can impose the condition (f ) = 2π up to order higher than f 2 and
f 2 , by writing [(1 + f ) 2 + f 2 ] 1/2 1 + f + 1 2 f 2 (using √ 1 + x
1 + 1 2 x − 1 8 x 2 ). We find
2π 0 f + 1 2 f 2 d ϕ = 0. (9.28) Since f is periodic, we can consider its Fourier expansion (Appendix 7): f (ϕ) = a 0 + ∞ n =1 (a n cos nϕ + b n sin ϕ),
(9.29) f (ϕ) =
∞ n =1 ( −na
n sin nϕ + nb n cos nϕ),
(9.30) and hence 2π 0
d ϕ = 2πa
0 , 2π 0 f 2 d ϕ = π
∞ n =1 n 2 (a 2 n + b 2 n ), and equation (9.28) implies the relation a 0 = − 1 4 ∞ n =1 n 2 (a 2 n + b 2 n ). (9.31)
We now compute the area enclosed by the perturbed curve: A(f ) =
1 2 2π 0 (1 + f )
2 d ϕ = π + 2π 0 f + 1 2 f 2 d ϕ. (9.32) Again using equation (9.28) we can estimate the variation A(f ) − π =
1 2 2π 0 (f 2 − f 2 ) d ϕ. (9.33) We now have 2π 0 f 2 d ϕ = π 2a 2 0 + ∞ n =1 (a 2 n + b
2 n ) , where a 2 0 can be ignored. Indeed, it follows from equation (9.31), or from equation (9.28), that the average of f (i.e. a 0 ) is of the same order as ||f || 2 , 9.2 Analytical mechanics: variational principles 311 and hence that a 2 0 is of the order of the error, and can be ignored. We can re-interpret equation (9.33) as A(f )
− π −π ∞ n =1 (n 2 − 1)(a
2 n + b 2 n ). (9.34) We can conclude that the perturbation causes a decrease in the area, as soon as one of the Fourier coefficients with index n > 1 is different from zero. We must still examine the case that f = a 0 + a
1 cos ϕ + b 1 sin ϕ, when the perturbation A(f ) − π is of order greater than two. To evaluate (f) we must consider the expansion √ 1 + x 1 + 1 2 x − 1 8 x 2 + 1 3! 3 8 x 3 − 1 4! ( 5 16 x 4 ) , which yields (keeping terms up to fourth order) [(1 + f ) 2 + f 2 ] 1/2 1 + f + 1 2 f 2 − 1 2 f f 2 + 7 24 f 4 − 1 4 f 2 f 2 . To compute the integral of this expression we must take into account the fact that a 0 is of the same order as a 2 1 and b 2 1 , and hence many terms can be discarded. Furthermore it is easy to compute 1 2π 2π 0 cos 4 ϕ d ϕ = 3 8 , 1 2π 2π 0 sin 2 ϕ cos
2 ϕ d ϕ = 1 8 . Finally, the condition (f ) = 2π can be written as −a 0
− 1 4 (a 2 1 + b 2 1 ) = 1 4 (a 2 1 + b 2 1 ) + 5 64 (a 2 1 + b 2 1 ) 2 , or, to the same order of approximation, −a 0 = 1 4 (a 2 1 + b 2 1 ) + 9 64 (a 2 1 + b 2 1 ) 2 . The area relative variation is then given by A(f )
− π π = 2a 0 + a
2 0 + 1 2 (a 2 1 + b 2 1 ), yielding, to fourth order, A(f )
− π π = − 7 32 (a 2 1 + b 2 1 ) 2
312 Analytical mechanics: variational principles 9.3 9.3
Hamilton’s variational principle: Lagrangian form The analogy between the Euler equations (9.12) and the Lagrange equations (4.75) is evident. The latter ones are also called the Euler–Lagrange equations, and we can regard them as the equations characterising when the functional A(q) = t
t 0 L(q, ˙q, t) dt, (9.35) called the Hamiltonian action, is stationary in the class Q of perturbed motions, defined by (9.8). These motions are called motions with synchronous perturbations (to stress the fact that we are not altering the time-scale). We can summarise what we have just discussed in the following statement. T heorem 9.3 (Hamilton principle) The natural motion is characterised by the property that the Hamiltonian action is stationary in the class of synchronous perturbations which preserve the configurations of the system at the initial and final time. Remark 9.4 Recall that L = T −V . We can then state that the natural motion makes the time average of the difference between the kinetic and potential energy stationary. We stress the fact that the Hamilton principle is a characterisation of the motion, in the sense that it can be regarded not only as a consequence of the Lagrange equations, but it can also be assumed as the fundamental postulate of mechanics, from which the Lagrange equations can be immediately deduced. We now examine a series of examples in which we find that the Hamiltonian action is not only stationary, but even minimised along the natural motion. Example 9.5: motion of a free point particle in the absence of forces It is sufficient to recall the problem solved in Section 9.1, removing the condition that the point is constrained on a line. Example 9.6: motion of a point mass under gravity Choose the reference frame in such a way that the natural motion has equations x ∗
(t) = v 01 t, x ∗ 2 (t) = 0, x ∗ 3 (t) = v
03 t +
1 2 gt 2 (9.36)
(axis x 3 oriented along the descending vertical, initial velocity v 0 = (v
01 , 0, v
03 )).
The synchronous perturbations are defined by x 1 (t) = v 01 t + η 1 (t),
x 2 (t) = η 2 (t),
x 3 (t) = v 03 t +
1 2 gt 2 + η
3 (t),
(9.37) with η
i ∈ C
2 [0, θ], η i (0) = η
i (θ) = 0, i = 1, 2, 3, for a given θ > 0. 9.3 Analytical mechanics: variational principles 313 The variation of the Hamiltonian action A = θ 0 1 2 mv 2 + mgx
3 d t (9.38) can be easily computed: δA = 1
m θ 0 i ˙ η 2 i d t, (9.39)
and is positive for every non-zero perturbation (we can add that it is of order 2 with respect to the perturbation, in the sense that for fixed η 1 , η
2 , η
3 multiplied by α, it follows that δA = O(α
2 )).
Example 9.7: the harmonic oscillator Choose the reference frame in such a way that we can write x ∗
(t) = a sin ωt, x ∗ 2 (t) = x
∗ 3 (t) = 0 (9.40) and consider the variations x 1
1 (t),
x 2 (t) = η 2 (t),
x 3 (t) = η 3 (t),
(9.41) with η
i chosen as in the previous problem. Since L = 1 2 mv 2 − 1 2 mω 2 i x 2 i , we find δA = 1 2 m θ 0 i ( ˙
η 2 i − ω 2 η 2 i ) d t + m
θ 0 ( ˙ x ∗ 1 ˙ η 1 − ω 2 x ∗ 1 η 1 ) d t. (9.42)
One integration by parts in the second integral yields − θ 0 (¨ x ∗ 1 + ω 2 x ∗ 1 )η 1 d t = 0,
and hence we can conclude that δA evaluated along the natural motion is of order 2 with respect to the perturbation (implying that A is stationary). Finally we note that an integral of the type θ 0 ( ˙ η 2 − ω 2 η 2 ) d t can be estimated using |η(t)| =
t 0 ˙ η(τ ) d τ ≤ √ t ⎡ ⎣ t 0 ˙ η 2 (τ )
d τ ⎤ ⎦ 1/2
. 314 Analytical mechanics: variational principles 9.4 We thus find θ 0 ( ˙ η 2 − ω 2 η 2 ) d t ≥ 1 − 1 2 ω 2 θ 2 ⎡ ⎣ θ 0 ˙ η 2 (t) d t ⎤ ⎦ 1/2
. Hence we can conclude that δA > 0 if θ < √ 2/ω, i.e. the Hamiltonian action has a minimum when computed along the natural motion, provided that we impose a restriction on the length of the time interval over which it is computed. As an exercise, compute δA for η 1 = α sin 2 πt/θ, η
2 = η
3 = 0 and note that δA 0 for θ
2π/3ω. 9.4
Hamilton’s variational principle: Hamiltonian form As we have explicitly observed, so far we have based our analysis of variational principles on the Lagrangian formalism. This is convenient for the ease with which one can then define the synchronous perturbations in the space of Lagrangian coordinates. Passing to the Hamiltonian formalism, we need only to express the action in the canonical variables (p, q): A(p, q) = t 1
0 [p · ˙q − H(p, q, t)] dt, (9.43) where ˙q = ˙q(p, q, t), but we must define the variations in the phase space. This is naturally done by perturbing q ∗ k (t) and in turn p ∗ k (t), in such a way that the formal relation p k = ∂L/∂ ˙
q k is preserved. However it is more convenient to introduce independent variations for q k and p k : q k (t) = q
∗ k (t) + η k (t),
k = 1, . . . , , p k (t) = p ∗ k (t) + ζ k (t), k = 1, . . . , , (9.44)
with η k , ζ k ∈ C
2 [t 0 , t 1 ], η k (t 0 ) = η k (t 1 ) = 0, k = 1, . . . , , where q ∗ k
p ∗ k (t) denote the solutions of the Hamilton equations. In this way we can define perturbed curves in phase space (Fig. 9.4), which in general are not admissible trajectories for the system (consider e.g. the trivial case = 1 with p = m ˙ q and take ζ(t) = / m ˙ η(t)).
The class of trajectories (9.44) is therefore larger than the class of synchronous perturbations. If we prove that the functional A is stationary along the solutions of the Hamilton equations with respect to this more extended class of perturbations, it follows that it is also stationary within the more restricted class of synchronous perturbations. This is the idea in the proof of the following theorem.
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