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B E v 0 Fig. 4.4 Choose the reference system in such a way that the origin coincides with the initial position of the charge P (0), the axis x 3 has the same direction and orientation as B, and the axis x 2 is orthogonal to both E and B, so that E 1 ≥ 0
(Fig. 4.4). If v(0) = v 0 , we need to study the following problem: ¨ x 1 = E 1 + ˙ x 2 ω, x 1 (0) = 0, ˙ x 1 (0) = v
0 1 , (4.107) ¨ x 2 = − ˙x 1 ω, x 2 (0) = 0,
˙ x 1 (0) = v 0 2 , (4.108)
¨ x 3 = E 3 , x 3 (0) = 0, ˙ x 3 (0) = v
0 3 , (4.109) where we set e m
i = E i , eB mc = ω.
(4.110) If B = 0 the motion is in a uniform electric field, and the generic trajectory is a parabola. Suppose B = / 0 and integrate once; this yields ˙ x
= E 1 t + ωx 2 + v 0 1 , (4.111) ˙ x 2 = −ωx 1 + v
0 2 , (4.112) ˙ x 3 = E 3 t + v
0 3 . (4.113) 146 The dynamics of discrete systems. Lagrangian formalism 4.8 Equation (4.112) can be used in (4.107) to obtain an equation for the only variable x 1 : ¨ x 1 + ω 2 x 1 = E 1 + ωv
0 2 . (4.114) Set
C = E 1 ω = c
E 1 B . (4.115)
Then the integral of equation (4.114) is given by x 1 (t) = −D cos(ωt + α) + 1 ω
0 2 ), (4.116) and after imposing the initial conditions, we find D cos α = 1 ω (C + v 0 2 ), (4.117)
D sin α = 1 ω v 0 1 , (4.118)
from which D and α are easily computed. Equation (4.111) now yields x 2
− Ct − 1 ω v 0 1 . (4.119)
Therefore the projection of the point onto the plane normal to the magnetic field moves in a circular trajectory, with radius D = 1
(v 0 1 ) 2 + (C + v 0 2 ) 2 1/2
(4.120) and frequency ω, around the centre 1 ω
0 2 ), −Ct − 1 ω v 0 1 . The latter moves uniformly according to E ×B with velocity C defined by (4.115). The projection motion is circular and uniform if E 1 = 0 (implying C = 0). In this case equation (4.120) defines the Larmor radius. The motion in the x 3 -coordinate is due exclusively to the electric field: x 3 (t) = 1 2 E 3 t 2 + v 0 3 t. (4.121)
Note that in correspondence with the zeros of ˙ x 1 , given by ωt n +α = (2n+1) π/2, one finds ˙ x 2 (t n ) = ± ωD−C, and hence if ωD > C, the motion in the x 2 -direction is periodically inverted and the projection of the trajectory onto the plane x 3 = 0 self-intersects (if ωD = C, it forms cusps). As an exercise, let E 3
0 3 = 0 and prove that for ω → 0 equations (4.116), (4.119) reproduce the motion in a uniform force field. 4.9 The dynamics of discrete systems. Lagrangian formalism 147 4.9
Symmetries and conservation laws. Noether’s theorem The invariant properties of a system with respect to the action of a group (with one or more parameters) are called symmetries. In a Lagrangian system, to such symmetries there correspond conservation laws, which are first integrals of the motion of the system. We shall see as an example that conservation of momenta corresponds to the invariance of the Lagrangian with respect to coordinate translations, conservation of the angular momentum corresponds to the invariance of the Lagrangian with respect to rotations, and so on. The rigorous mathematical formulation of this relation between symmetries and conservation laws is the content of Noether’s theorem. Consider a Lagrangian system with l degrees of freedom; for simplicity, we assume the system to be independent of the time t. Let L(q, ˙q) be its Lagrangian. D efinition 4.3 An invertible coordinate transformation q = f(Q) is admis- sible for a given system if and only if the Lagrangian is invariant under the transformation, and hence if L(q, ˙q) = L(Q, ˙ Q).
(4.122) Example 4.9 If a Lagrangian has a cyclic coordinate (see Remark 4.7), it follows that it is invariant under translations in this coordinate. Example 4.10 Rotations around the origin q 1
1 cos α + Q 2 sin α,
q 2 = −Q 1 sin α + Q 2 cos α
are admissible for the Lagrangian L (q
1 , q
2 , ˙
q 1 , ˙ q 2 ) = m 2 ˙ q 2 1 + ˙ q 2 2 − V
q 2 1 + q 2 2 corresponding to the plane motion of a point particle of mass m in a central force field. D efinition 4.4 A one-parameter s ∈ R family of invertible transformations q = f (Q, s) is called a one-parameter group of transformations if it satisfies the following properties: (a) f (Q, 0) = Q, for every Q; (b) for every s 1 , s
2 ∈ R, f(f(Q, s 1 ), s
2 ) = f (Q, s 1 + s
2 ). If for every s ∈ R the transformation q = f(Q, s) is admissible, then the group is called admissible. Note that (a), (b) imply that if q = f (Q, s), then Q = f (q, −s).
148 The dynamics of discrete systems. Lagrangian formalism 4.9 T
of transformations q = f (Q, s), the Lagrange equations associated with L admit the first integral I(q, ˙q) given by I(q, ˙q) = l i =1 ∂L ∂ ˙ q i ∂f i ∂s (q, 0). (4.123) Proof
The invariance property of the Lagrangian implies that if q(t) is a solution of Lagrange’s equations (4.75), then Q(t, σ) = f (q(t), σ) is also a solution, ∀ σ ∈ R. This means that d dt
˙ Q L(Q, ˙ Q) = ∇ Q L(Q, ˙ Q),
∀σ ∈ R, (4.124)
where ˙ Q = (∂/∂t)Q(t, σ). In addition, the definition of an admissible transform- ation yields 0 =
∂ ∂σ L(Q, ˙ Q) = ∇ Q L · ∂Q ∂σ + ∇ ˙ Q L · ∂ ˙
Q ∂σ , (4.125) and using equation (4.125) and multiplying equation (4.124) by ∂Q/∂σ we find d dt
˙ Q L · ∂Q ∂σ = 0. (4.126)
For σ = 0 this is exactly the invariance of I(q, ˙q) along the motion. Example 4.11 If the Lagrangian L(q, ˙q) admits the translations q k = Q k + s as transforma- tion group, the coordinate q k is cyclic and I(q, ˙q) = p k is a constant of the motion. Example 4.12 If a Lagrangian L(q, ˙q), where q ∈ R
3 , admits the rotations around the axis q 1 :
1 = Q
1 , q 2 = Q
2 cos s + Q 3 sin s,
q 3 = −Q 2 sin s + Q 3 cos s
as transformation group, the function I(q, ˙q) = ∂L ∂ ˙
q 2 q 3 − ∂L ∂ ˙ q 3 q 2 = p 2 q 3 − p 3 q 2 is a constant of the motion, coinciding with the component of the angular momentum along the axis q 1 . 4.9 The dynamics of discrete systems. Lagrangian formalism 149 Example 4.13 The Lagrangian of a point particle constrained to move on a surface of revolution around the z-axis, with no active forces acting on it, is equal to (see Example 4.1) L(u, v, ˙ u, ˙v) =
m 2 1 + (f (u)) 2 ˙ u 2 + u
2 ˙v 2 . The coordinate v is cyclic, and therefore the conjugate kinetic momentum p v
2 ˙v is a constant of the motion. Note that p v is equal to the component L 3 of the
angular momentum p v = m(x 1 ˙ x 2 − x
2 ˙ x 1 ). Remark 4.10 There are transformations which, although not admissible, leave the equations of motion invariant. In this case there are no associated first integrals, but the study of the equations can help to establish interesting properties of the motion, without explicitly solving them. Example 4.14 Consider a Lagrangian system L(q, ˙q) = l i =1 m ˙ q 2 i 2 − V (q),
(4.127) where V is a homogeneous function of q 1 , . . . , q l of degree d: V (αq 1
l ) = α
d V (q
1 , . . . , q l ).
q = αQ, t = βτ,
dq dt = α β dQ dτ , where β = α 1−d/2 , transforms the Lagrangian (4.127) as follows: L Q,
dτ = α 2 β 2 l i =1 1 2 m dQ i dτ 2 − α
d V (Q) = α d L
dQ dτ . (4.128) Since these two functions are proportional, the equations of motion are invariant. Hence, if q(t, q 0 , ˙q 0 ) is a solution of Lagrange’s equations associated with (4.127), Q(τ, Q 0
Q 0 ) is also a solution, and therefore Q α (d/2)−1 t, 1 α q 0 , α −d/2 ˙q 0 = 1 α q(t, q 0 , ˙q 0 ) is a solution and the two trajectories are called similar. If d = 2 one again finds that the period of motion is independent of the amplitude for harmonic oscillators. For the case that d = −1, corresponding to a
150 The dynamics of discrete systems. Lagrangian formalism 4.10 Newtonian potential, and considering for simplicity circular orbits, we find that if T is the period and l is the orbit length, the ratio l 3/2
/T is constant for similar trajectories (as stated by the third law of Kepler). Example 4.15 Among the transformations that modify the Lagrangian but not the equations of motion are the gauge transformations for the vector and scalar potentials A, ϕ of the electromagnetic field (4.94), (4.96). Let f (x, t) be an arbitrary regular function, and set A = A +
∇f, ϕ = ϕ
− 1 c ∂f ∂t . (4.129) Then the definitions of the fields B, E through equations (4.94) and (4.96) are invariant. What changes is the Lagrangian (4.105), which is transformed into L = T
− e ϕ − 1 c v · A −
e c df dt . (4.130) As we remarked (see Remark 4.8), this generates the same motions as the former Lagrangian. 4.10 Equilibrium, stability and small oscillations Consider an autonomous system of differential equations of first order in R n : ˙x = w(x), (4.131)
where w is a regular vector field defined on R n . Lagrange’s equations (4.40) can be written in the form (4.131), where x represents the vector (q, ˙q) in the phase space. Indeed, after setting the equations in normal form (4.41), ¨ q = χ(q, ˙q), it is enough to introduce in the phase space the field w = ( ˙q, χ(q, ˙q)) to obtain equation (4.131). D efinition 4.5 A point x 0 is an equilibrium point if the constant function x(t) = x 0 is a solution of the system of differential equations (4.131). P roposition 4.3 A point x 0 is an equilibrium point if and only if the vector field w at the point is zero: w(x 0 ) = 0. Proof It is trivial that at an equilibrium point, the vector field w is zero: from the definition of an equilibrium point it follows that w(x 0 ) = ˙x(t) = 0. Conversely, if w(x 0 ) = 0 then x(t) = x 0 is a solution of the system for all t. The definition of stability of equilibrium is analogous to the definition given in Chapter 3 for a single point. 4.10 The dynamics of discrete systems. Lagrangian formalism 151 D
0 is (Lyapunov) stable if for every neighbourhood U of the equilibrium point there exists a neighbourhood U such
that, for any initial condition x(0) in U , the corresponding solution x(t) is in U for every t > 0. If this stability condition does not hold, then the equilibrium is called unstable. Remark 4.11 Using spherical neighbourhoods, we can equivalently define stability as follows: for every ε > 0 there exists a number δ > 0 such that, for any initial condition x(0) such that |x(0) − x 0 | < δ, |x(t) − x 0 | < ε for every t > 0. Instability can be characterised by the condition that there exists an > 0 such
that for any fixed δ > 0 there exists an initial condition x(0) in |x(0) − x 0 | < δ
for which |x(t) − x 0 | > for some t > 0. It is evident from the definitions that we are referring to stability in the future, but it is possible to consider the analogous concept in the past by inverting the direction of time. Example 4.16 Consider a system of linear equations in R n ˙x = Ax, where A is a real diagonalisable n × n matrix, with constant coefficients. Suppose that the eigenvalues λ 1 , . . . , λ n of A are all distinct and non-zero. Then the general integral of the equation is given by x(t) =
n j =1 c j e λ j t u j , where u 1 , . . . , u n are the eigenvectors of A. The constants c j (complex in general) are fixed by the initial conditions. Obviously x = 0 is an equilibrium position, and it is easy to verify that it is stable if the real parts of all the eigenvalues are non-positive: Re λ j ≤ 0, j = 1, . . . , n (simply use the linear transformation that diagonalises the matrix A). The analysis of the equilibrium stability for systems with one degree of freedom is carried out in Chapter 3 (Section 3.4). We now consider the corresponding problem for autonomous Lagrangian systems with several degrees of freedom. As we saw (Section 4.6), if V (q) is the potential energy, the equilibrium equations are ∂V ∂q
= 0, i = 1, . . . , l. (4.132) Let q be a solution of equations (4.132). We now prove the following stability criterion, for the case of smooth constraints. T heorem 4.5 (Dirichlet) If q is an isolated minimum of the potential energy, the corresponding configuration is one of stable equilibrium. 152 The dynamics of discrete systems. Lagrangian formalism 4.10 Proof
The hypotheses imply that q solves equations (4.132), and hence that it is an equilibrium configuration. In addition, there exists a neighbourhood A ⊂ R l
q in which V (q) > V (q), ∀ q =
/ q. We can choose V (q) = 0. Consider now any neighbourhood B ⊂ R 2l
in the phase space, and for any ε > 0, define the energy sublevel set Ω ε = {(q, ˙q)|T (q, ˙q) + V (q) < ε}. Recall that T (q, ˙q) ≥ a
0 | ˙q
2 | for some constant a 0 > 0 (Theorem 4.1). Con- sequently Ω ⊂ Ω = {(q, ˙q) | a 0 | ˙q
2 | + V (q) < } ⊂ M ∩ N where M = {(q, ˙q) | | ˙q| < ( /a 0 ) 1/2 }, N = {(q, ˙q) | V (q) < }. Since by hypothesis, the diameter of M ∩ N tends to zero when → 0, we can find ε so small that Ω ε ⊂ Ω ⊂ B ∩ (A × R l ). On the other hand, because of conservation of energy, every trajectory originating in Ω ε must remain in Ω ε . This yields the stability condition (Definition 4.6). C orollary 4.2 For any holonomic system with fixed smooth constraints, for which the active forces are only due to gravity, the stable equilibrium config- urations occur in correspondence with isolated minima of the height of the centre of mass. Example 4.17 We refer to Fig. 3.5. The isolated minima of x 2 , x 4 of V (x) correspond to positions of stable equilibrium. Consider for example the point (x 2 , 0) in the phase space, and consider a generic neighbourhood U . Define e max
∈ (e 2 , e 3 ) in such a way that the trajectory with energy e max
is entirely lying in U . The region determined by this trajectory contains all trajectories with energy in the interval (e 2 , e
max ), and hence the definition of stability holds. We now consider the motion near configurations of stable equilibrium. Rewrite the Lagrangian of the system as L(Q, ˙ Q) =
1 2 l i,j =1 a ij (Q) ˙
Q i ˙ Q j − V (Q), (4.133) where there appear the vectors Q = q −q, ˙Q = ˙q, with q an isolated minimum of V (q). As we have seen, it is always possible to choose the initial conditions in such a way that the trajectory in the phase space remains in a fixed neighbourhood of (q, 0). Select now a neighbourhood so small that inside it one can neglect terms of degree greater than two in the expansion of the function L(Q, ˙ Q). Hence replace equations (4.133) with the quadratic approximation L(Q, ˙
Q) = 1 2 l i,j
=1 a ij ˙ Q i ˙ Q j − l i,j =1 V ij Q i Q j , (4.134) 4.10 The dynamics of discrete systems. Lagrangian formalism 153 where we set a ij = a ij (q),
V ij = ∂ 2 V ∂Q i ∂Q j (q).
(4.135) Denoting by A and V the symmetric matrices of the coefficients a ij and V
ij , respectively, the Lagrangian (4.133) can be written in matrix notation as L(Q, ˙ Q) =
1 2 ( ˙ Q T A ˙ Q − Q
T V Q),
(4.136) and the associated Lagrange equations are linear: A ¨ Q + V Q = 0. (4.137) Assuming that the matrix V is also positive definite, we can prove the following. T heorem 4.6 If A, V are symmetric and positive definite, there exists a linear transformation in R l which decouples equations (4.137) into l harmonic oscil- lations, called normal modes of the system and whose frequencies are called fundamental frequencies of the system. Proof We follow the standard procedure to find the general integral of a system of linear ordinary differential equations with constant coefficients. Hence we seek a solution of (4.137) of the form Q = we iλt
, (4.138)
where w is a vector in R l to be determined and λ ∈ C. Substituting (4.138) into (4.137) we find e iλt
(V − λ
2 A) w = 0
and we must therefore study the generalised eigenvalue problem det(µA
− V ) = 0. (4.139)
Accounting for multiplicity, this system has l solutions µ 1 , . . . , µ l corresponding to the eigenvectors w 1 , . . . , w l . We prove that in this case the l roots µ 1 , . . . , µ l are positive. The method consists of reducing (4.137) to diagonal form, by a sequence of linear transformations. The choice of each such transformation must obey the criterion of symmetry conservation of the matrices of coefficients. Since A is a symmetric, positive definite matrix, there exists a unique symmet- ric, positive definite matrix whose square is equal to A, which we denote by A 1/2
154 The dynamics of discrete systems. Lagrangian formalism 4.10 (the square root of A). Indeed, since A is symmetric, there exists an orthogonal matrix S which diagonalises A: SAS
−1 = SAS
T = ⎛ ⎜ ⎜ ⎜ ⎝ α 1 0 . . .
0 0 α 2 . . .
0 .. . .. . . . . .. . 0 0 . . . α l ⎞ ⎟ ⎟ ⎟ ⎠ , (4.140) where α 1 , . . . , α l are precisely the eigenvalues of A. Since A is positive definite, the eigenvalues are all positive, and we can define A 1/2 = S T ⎛ ⎜ ⎜ ⎜ ⎝ √ α 1 0 . . . 0 0 √ α 2 . . . 0 .. . .. . . . . .. . 0 0 . . . √ α l ⎞ ⎟ ⎟ ⎟ ⎠ S. (4.141) It is easily verified that A 1/2 is symmetric and positive definite and that (A 1/2 ) 2 = A. Moreover, A −1/2
= S T ⎛ ⎜ ⎜ ⎜ ⎝ 1/ √ α 1 0 . . . 0 0 1/ √ α 2 . . .
0 .. . .. . . . . .. . 0 0 . . . 1/ √ α l ⎞ ⎟ ⎟ ⎟ ⎠ S is also symmetric. Through the change of variables Y = A 1/2
Q (4.142)
equation (4.137) becomes ¨ Y + A −1/2 V A
−1/2 Y = 0,
(4.143) and hence (4.139) is equivalent to det(A −1/2
V A −1/2
− µ) = 0. (4.144)
Evidently A −1/2
V A −1/2
is symmetric and positive definite; it follows that its eigenvalues µ 1 , . . . , µ l are real and positive. We conclude (see Example 4.16) that the configuration q is of stable equilibrium for the linearised system. Setting C = A
−1/2 V A
−1/2 , (4.145) 4.10 The dynamics of discrete systems. Lagrangian formalism 155 if W is an orthogonal matrix, diagonalising C, so that W T CW = ⎛ ⎜ ⎜ ⎜ ⎝ µ 1 0 . . . 0 0 µ 2 . . .
0 .. . .. . . . . .. . 0 0 . . . µ l ⎞ ⎟ ⎟ ⎟ ⎠ , (4.146) and if we define Y = W X,
(4.147) equation (4.143) becomes ¨ X +
⎛ ⎜ ⎜ ⎜ ⎝ µ 1 0 . . . 0 0 µ 2 . . .
0 .. . .. . . . . .. . 0 0 . . . µ l ⎞ ⎟ ⎟ ⎟ ⎠ X = 0.
(4.148) This equation represents l independent harmonic oscillations with frequency ω i =
µ i , i = 1, . . . , l (normal modes). The linear transformation yielding the normal modes is hence given by X = W
T A 1/2 Q. (4.149)
Remark 4.12 Recall that if C is a real symmetric × matrix with eigenvalues (µ 1 , . . . , µ ), the orthogonal matrix W diagonalising C can be constructed as follows: ortho-
normal column vectors w (1)
, . . . , w ( )
such that (C − µ
j )w (j) = 0 can be easily determined. The matrix W = (w (1) , . . . , w ( ) ) is orthogonal and W T
⎛ ⎜ ⎜ ⎜ ⎝ µ 1 0 . . . 0 0 µ 2 . . .
0 .. . .. . . . . .. . 0 0 . . . µ l ⎞ ⎟ ⎟ ⎟ ⎠ , As an example, if C = 2 1 1 2 , µ
1 = 3, µ
2 = 1,
w (1)
= 1/ √ 2 1/ √ 2 , w (2) = 1/ √ 2 −1/ √ 2 , W = 1/ √ 2 1/ √ 2 1/ √ 2 −1/
√ 2 . Example 4.18 Consider a point particle of mass m moving under the action of its weight on a surface of parametric equations x = (x(q
1 , q
2 ), y(q
1 , q
2 ), z(q
1 , q
2 )).
156 The dynamics of discrete systems. Lagrangian formalism 4.10 The Lagrangian of the system is given by L(q 1 , q 2 , ˙
q 1 , ˙ q 2 ) = 1 2 m E(q 1 , q
2 ) ˙
q 2 1 + 2F (q 1 , q 2 ) ˙
q 1 ˙ q 2 + G(q 1 , q
2 ) ˙
q 2 2 − mgz(q 1 , q 2 ), where E, F and G are the coefficients of the first fundamental form of the surface. A point (q 1 , q 2 ) is an equilibrium point for the system only if it is a critical point of z = z(q 1 , q 2 ). The Lagrangian of the linearised equations is L = 1
m E ˙ Q 2 1 + 2F ˙
Q 1 ˙ Q 2 + G ˙ Q 2 2 − 1 2 mg z 11 Q 2 1 + 2z 12 Q 1 Q 2 + z 22 Q 2 2 , where Q = q −q, E, F , G are the coefficients of first fundamental form evaluated at q, and z 11
∂ 2 z ∂q 2 1 (q 1 , q 2 ) ,
z 12 = ∂ 2 z ∂q 1 ∂q 2 (q 1 , q 2 ) , z 22 = ∂ 2 z ∂q 2 2 (q 1 , q 2 ) .
The fundamental frequencies of the system, ω 1 and ω 2 , are the solutions of the eigenvalue problem with characteristic polynomial det
ω 2 E F F G − g z 11 z 12 z 12 z 12 = 0. On the other hand, denoting by e, f and g the coefficients of the second fundamental form of the surface (see Appendix 3) evaluated at (q 1 , q 2 ), one
verifies that e = z
11 , f = z 12 , g = z 22 . For example, e = z 11 ∂x ∂q 1 ∂y ∂q 2 − ∂x ∂q 2 ∂y ∂q 1 (E G − F
2 ), but in (q 1 , q
2 ) we have ∂z ∂q
= ∂z ∂q 2 = 0,
and therefore EG − F 2 = ∂x ∂q 1 2 + ∂y ∂q 1 2 ∂x ∂q 2 2 + ∂y ∂q 2 2 − ∂x ∂q 1 ∂x ∂q 2 + ∂y ∂q 1 ∂y ∂q 2 2 = ∂x ∂q 1 ∂y ∂q 2 − ∂x ∂q 2 ∂y ∂q 1 , implying e = z 11 .
4.10 The dynamics of discrete systems. Lagrangian formalism 157 The principal curvatures k 1 and k
2 of the surface (Appendix 3) at the equilib- rium point are the solutions of the eigenvalue problem of the first fundamental form with respect to the second, i.e. the roots of the characteristic polynomial det k
F F G − e f f g = det k E F F G − z 11 z 12 z 12 z 22 = 0. It follows that the principal curvatures are directly proportional to the square of the fundamental frequencies of the linearised equations k 1
ω 2 1 g , k 2 = ω 2 2 g . (4.150)
We now compute the fundamental frequencies for the case that l = 2, and that the matrix A is diagonal: A = α
0 0 α 2 , α 1 , α
2 > 0,
(4.151) and of course V is symmetric and positive definite. The Lagrangian of the linearised motion is then given by L 2 (Q, ˙ Q) =
1 2 α 1 ˙ Q 2 1 + α 2 ˙ Q 2 2 − V 11 Q 2 1 − 2V
12 Q 1 Q 2 − V 22 Q 2 2 , (4.152) and the matrix (4.145) is C =
⎛ ⎜ ⎜ ⎜ ⎝ V 11 α 1 V 12 √ α 1 α 2 V 12 √ α 1 α 2 V 22 α 2 ⎞ ⎟ ⎟ ⎟ ⎠ . The eigenvalue equation is µ 2 − V 11 α 1 + V 22 α 2 µ +
V 11 V 22 − V
2 12 α 1 α 2 = 0. We find the two frequencies ω ±
⎧ ⎨ ⎩ 1 2 V 11 α 1 + V 22 α 2 ± 1 2 V 11 α 1 − V 22 α 2 2 + 4 V 2 12 α 1 α 2 1/2 ⎫ ⎬ ⎭ 1/2 . (4.153) Obviously if V 12 = 0 (hence if the original system is in diagonal form) we find ω + = V 11 /α 1 , ω
− = V 22 /α 2 . 158 The dynamics of discrete systems. Lagrangian formalism 4.10 Example 4.19 A cylindrical container of height h is closed at the boundary and is divided into three sections by two pistons of mass m, which can slide without friction. Each section contains the same amount of gas, for which we suppose the law P v = constant
is applicable. Write the Lagrange equations describing the motion of the two pistons, find the stable equilibrium configuration and study the small oscillations of the system around it. Let x
1 , x
2 indicate the distance of the pistons from one of the two bases. Then (x 1
2 ), on the first piston there acts the force F 1
c x 1 − c x 2 − x
1 , c > 0 constant, and on the second piston the force F 2 = c x 2 − x
1 − c h − x
2 . Use the dimensionless variables f i = hF
i /c, ξ
i = x
i /h, i = 1, 2, and write f 1
1 ξ 1 − 1 ξ 2 − ξ
1 , f 2 = 1 ξ 2 − ξ 1 − 1 1 − ξ
2 . This is a conservative system of forces, with potential V (ξ 1 , ξ
2 ) =
− log[ξ 1 (ξ 2 − ξ 1 )(1
− ξ 2 )]. Recall that V is expressed in dimensionless variables while the corresponding physical quantity is V = cV . The Lagrangian in the original variables is L = 1 2
x 2 1 + ˙ x 2 2 ) − cV and can be replaced by the dimensionless Lagrangian L =
1 2 dξ 1 dτ 2 + dξ 2 dτ 2 − V (ξ 1 , ξ
2 ), by introducing the change of time-scale τ = t/t 0 , with t
2 0 = mh 2 /c. The equations of motion become d 2 ξ 1 dτ 2 = 1 ξ 1 − 1 ξ 2 − ξ 1 , d 2 ξ 2 dτ 2 = 1 ξ 2 − ξ
1 − 1 1 − ξ
2 . It is easily verified that the only equilibrium configuration is given by ξ 1 = 1 3 , ξ 2 = 2 3 . The Hessian matrix of V (ξ 1 , ξ
2 ) is
⎛ ⎜ ⎜ ⎝ 1 ξ 2 1 + 1 (ξ 2 − ξ 1 ) 2 − 1 (ξ 2 − ξ 1 ) 2 − 1 (ξ 2 − ξ
1 ) 2 1 (1 − ξ 2 ) 2 ⎞ ⎟ ⎟ ⎠ . 4.11 The dynamics of discrete systems. Lagrangian formalism 159 At the equilibrium, this becomes V = 9 2 −1 −1 1 , which is positive definite, with eigenvalues given as solutions of λ 2 − 27λ + 81 = 0, namely λ 1 = 9 2 (3 − √ 5), λ 2 = 9 2 (3 +
√ 5). Hence the equilibrium is stable. The Hessian matrix of the kinetic energy is the identity matrix. Therefore the equations describing small oscillations are d 2
2 Q + V Q = 0, with Q =
ξ 1 ξ 2 , and √ λ 1 , √ λ 2 give the dimensionless frequencies directly (we obtain ω i =
i /t 0 , i = 1, 2). The normal modes are obtained by setting X = W T Q, where W is such that W T V W = λ 1 0 0 λ 2 . We easily find that W = 1 5 1/4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 2 √ 5 − 1
1/2 2 √ 5 + 1 1/2
− √ 5 − 1 2 1/2 √ 5 + 1
2 1/2
⎞ ⎟ ⎟ ⎟ ⎟ ⎠ . By writing Q = W X we can describe the small motions of the pistons as combinations of the harmonic motions X 1 , X 2 . 4.11 Lyapunov functions In the previous section we have introduced the concept of stability of equilibrium points, for the system of differential equations (4.131). In particular, we have analysed the stability of the equilibrium of holonomic systems, with smooth fixed constraints, and subject to conservative forces. We now discuss some extensions and one additional criterion for stability. We start by observing that the conditions guaranteeing the stability of the equilibrium in the case of conservative forces must still hold if we introduce dissipative forces. T heorem 4.7 Theorem 4.4 is still valid if in addition to forces with potential energy V (q) there exist dissipative forces. 160 The dynamics of discrete systems. Lagrangian formalism 4.11 Proof
The proof of Theorem 4.4 is based only on the fact that the trajectories originating within the set Ω remain there for all subsequent times. This is true if energy is conserved, but also if energy is dissipated. Dissipation helps stability, and in addition it may have the effect of bringing the system back to the equilibrium configuration, starting from a small enough perturbation, either in finite time or asymptotically for t → +∞. This is the case of asymptotic stability (see Definition 3.5). D efinition 4.7 A point x 0 of stable equilibrium for the system (4.131) is asymp- totically stable if there exists a δ > 0 such that for every x(0) in the neighbourhood |x(0) − x 0 | < δ one has |x(t) − x 0 | → 0 for t → +∞. Example 4.20 For the harmonic damped motion (3.35) the point x = 0 is a point of equilibrium, and it is asymptotically stable (see (3.38)). Recall the case of the linear system ˙x = Ax (Example 4.14); in this case we can deduce that x = 0 is an equilibrium point which is asymptotically stable if all eigenvalues λ j of the matrix A have negative real part: Re λ j < 0, j = 1, . . . , n. The Dirichlet stability criterion (Theorem 4.4) is a special case of a well-known method for analysing stability, based on the so-called Lyapunov function. We consider again the system (4.131) and an equilibrium point x 0 ; with reference to these we give the following definition. D efinition 4.8 Let Ω be a neighbourhood of x 0 , and let Λ ∈ C
1 ( Ω ) be a function with an isolated minimum at x 0 (assume
Λ (x 0 ) = 0). If for the field w(x) of system (10.1) we have that w(x) · ∇
Λ (x)
≤ 0, ∀ x ∈ Ω , (4.154) then
Λ is a Lyapunov function for the system. Note that the meaning of (4.154) is that d dt Λ (x(t))
≤ 0 along the solutions x(t) of the system (4.131). Clearly for any holonomic system the total energy is a Lyapunov function in the phase space, in a neighbourhood of a local isolated minimum of the potential energy. The following theorem has a proof analogous to the proof of Theorem 4.4. T heorem 4.8 If x 0 is such that there exists a Lyapunov function for the system (4.131) then it is a stable equilibrium point. A more specific case is the following. 4.11 The dynamics of discrete systems. Lagrangian formalism 161 T
w(x) · Λ (x) < 0, x =
/ x 0 , x ∈ Ω , (4.155)
then x 0 is asymptotically stable. Proof Consider the sets A = {x ∈
Ω | Λ (x) ≤ }.
Then A ⊂ A if
< and moreover diam A → 0 for → 0. Since along the trajectories of (4.131) ˙ Λ
Ω must cross the boundary ∂A with decreasing. If the point tends to ∂A ∗ for some
∗ > 0,
we would have ˙ Λ ≤ −α for some α > 0 and ∀ t > 0, which cannot hold; indeed, this would yield Λ → −∞, contradicting the hypothesis that Λ (x 0 ) = 0 is a minimum.
Example 4.21 For the damped harmonic oscillator (3.35), or equivalently for the system ˙x = w, ˙
−(2βw + ω 2 x), β > 0, (4.156)
Λ (x, w) =
1 2 (w 2 +ω 2 x 2 ) has an isolated minimum at the equilibrium point and ˙ Λ = −2βw 2 < 0 away from the origin. We can therefore apply Theorem 4.8. Lyapunov’s method can be invoked to establish instability. T heorem 4.10 Let x 0 be an equilibrium point for the system (4.131). Suppose that there exist a neighbourhood Ω of x 0 and a function Λ ∗
1 ( Ω ) ( Λ ∗ (x 0 ) = 0) such that w(x)
· ∇ Λ ∗ (x) > 0, x =
/ x 0 , x ∈ Ω , (4.157)
and that x 0 is an accumulation point for the positivity set of Λ ∗ . Then x 0 is unstable. Proof Consider a ball B δ (x
) of centre x 0 and radius δ such that B δ ⊂ Ω and let x(0)
∈ B δ (x 0 ) be such that Λ ∗
in the set M 0 , where Λ ∗ (x) > Λ ∗ (x(0)). In the intersection of this set with B δ (x 0 ) the scalar product w · ∇ Λ
has a positive infimum, while in this set Λ ∗ is bounded. It follows that x(t) must leave B δ in a finite time. Example 4.22 Consider the system ˙ x = w,
˙ w = ω
2 x (4.158) 162 The dynamics of discrete systems. Lagrangian formalism 4.12 for which (0, 0) is the (only) equilibrium point. Consider the function Λ ∗ = xw. In the plane (x, ω) this function is positive in the first and third quadrant and w ∂ Λ ∗ ∂x + ω 2 x ∂ Λ ∗ ∂w = w
2 + ω
2 x 2 > 0 away from the origin. Instability follows. Another useful result on instability, whose assumptions are less restrictive than those of Theorem 4.9 is the following. T heorem 4.11 (ˇCetaev) Suppose that there exists an open connected set Ω 1 (possibly unbounded) with x 0 ∈ ∂
Ω 1 , and a function Λ ∗ ∈ C 1 ( Ω 1 ), such that Λ ∗
Ω 1 and Λ ∗ (x 0 ) = 0, for which (4.157) holds inside Ω 1
0 is unstable. Proof This is just an extension of the previous theorem. With x(0) ∈ Ω
, the trajectory cannot reach the boundary of Ω 1
Λ ∗ is increasing) and cannot stay indefinitely inside B δ (x 0 ) ∩ M 0 . Example 4.23 The origin is the only point of equilibrium for the system ˙ x = w, ˙ w = ω
2 |x|.
(4.159) The function Λ ∗
Λ ∗ = ω 2 x |x| + w 2 and it satisfies the hypotheses of Theorem 4.10, with Ω 1 taken equal to the first quadrant (note that the hypotheses of Theorem 4.9 are not satisfied). 4.12 Problems
1. Two point particles with mass, (P 1 , m 1 ), (P
2 , m
2 ), are constrained on two vertical lines r 1 , r 2 , at a distance d. The two points attract each other with an elastic force of constant k and both are attracted by a fixed point O, placed at an equal distance from the two lines, with an elastic force of equal constant. Write down Lagrange’s equations and show that the motion can be decomposed into two harmonic oscillations around the equilibrium configuration. Determine also the constraint reactions. 2. In a horizontal plane, two point particles (P 1 , m
1 ), (P
2 , m
2 ) attract each other with an elastic force of constant k and are constrained on a smooth circle of centre O and radius R. They are also attracted by two points O 1 , O
2 , respectively, with an elastic force of equal constant. The latter points are at a distance 2R from O and such that the radii O 1 − O and O 2 − O form a right angle. Find the equilibrium configurations of the system and study the small oscillations around the stable equilibrium configuration. 4.12 The dynamics of discrete systems. Lagrangian formalism 163 3. Find the normal modes when the number of degrees of freedom of the system is equal to two, and the matrix V is diagonal. 4. In a horizontal plane two point particles (P 1 , m
1 ), (P
2 , m
2 ) are attracted respectively by two fixed points O 1 , O 2 in the plane with elastic forces of equal constant. The two particles are subject to the rigidity constraint |P 1 − P 2 | = |O 1 − O 2 |. Find the normal modes of the system. 5. Determine the fundamental frequencies and the normal modes of oscillation of a system of equal point particles constrained to move on a line and sequen- tially linked by springs with an elastic constant equal to k. The first particle is elastically attracted by the origin with a constant k and the last particle is elastically attracted by a fixed point at a distance a > 0 from the origin with a constant k. Solution
Let q i be the coordinate of the ith particle. Then the equilibrium positions are q i = ai/( + 1), i = 1, . . . , , the fundamental frequencies are ω i = 2 k m sin π 2 i + 1 and the normal modes are q i
ai + 1
+ j =1 2 + 1
sin jiπ
+ 1 X i . 6. Consider l equal point particles P 1 , P
2 , . . . , P l (l > 2) on a circle of radius R and centre O. All particles move without friction and the point P i is attracted by its neighbouring points P i −1 , P i +1 with an elastic force (set P 0 = P l ). Write
down the potential of the system and prove that the configurations in which neighbouring rays form equal angles are equilibrium configurations. Study its stability (up to rotations). Compute the fundamental frequencies for l = 3. What is the general procedure? 7. A point particle of mass m is constrained to move along a curve of equation ζ = Aξ
2n , where A > 0 and n ≥ 1 is an integer. The curve rotates in three- dimensional Euclidean space with angular velocity ω around the z-axis and at time t = 0 belongs to the vertical (x, z) plane. Prove that, if ξ is chosen as the generalised coordinate, the Lagrangian of the system is equal to L = m
1 + 4n 2 A 2 ξ 4n−2 ˙ ξ 2 − mgAξ 2n − m 2 ω 2 ξ 2 . Prove that if n = 1 the only equilibrium position of the system is ξ = 0; the equilibrium is stable if ω 2
ξ = ±
2 2ngA
1/(2n−2) are positions of stable equilibrium, while ξ = 0 is unstable. Compute the frequencies of the small oscillations around the stable equilibrium positions.
164 The dynamics of discrete systems. Lagrangian formalism 4.12 8. A point particle of mass m is constrained to move on an ellipsoid of equation ξ 2 a 2 + η 2 + ζ 2 b 2 = 1, where a > b > 0. The ellipsoid rotates in space around the y-axis with angular velocity ω. At the instant t = 0 the principal axes ξ, η and ζ coincide with the axes x, y and z. Prove that, after setting ξ = a cos θ, η = b sin θ sin ϕ, ζ = b sin θ cos ϕ, the kinetic energy of the point is T = T 2 + T
1 + T
0 , where
T 2 = m 2 (a 2 sin
2 θ + b
2 cos
2 θ) ˙
θ 2 + b 2 sin
2 θ ˙
ϕ 2 , T 1 = abmω[cos ϕ ˙ θ − sin θ cos θ sin ϕ ˙ϕ], T 0
m 2 a 2 ω 2 cos 2 θ + b 2 ω 2 sin 2 θ cos 2 ϕ .
9. Two point particles of mass m constrained to the vertical axis mutually interact with an elastic force of constant k. The first point is also elastically attracted to the point z = 0 by a spring of constant k. Let z 1 and z 2 be the
coordinates of the two points. Prove that the Lagrangian of the system is L =
m 2 ˙ z 2 1 + ˙ z 2 2 − k 2 z 2 1 + (z
2 1 − z 2 2 ) − mgz 1 − mgz 2 . Determine the equilibrium positions, discuss their stability and compute the fundamental frequencies of the small oscillations around the equilibrium positions, and the normal modes. 10. A point particle of mass m and electric charge e is in motion in space under the action of a central field with potential energy V and of a magnetic field B = (0, 0, B). Prove that if the initial velocity is v = (v 1 , v 2 , 0) the motion takes place in the (x, y) plane. Write the Lagrangian in the plane polar coordinates (r, ϕ), and prove that the coordinate ϕ is cyclic. Use this fact to reduce the problem to one-dimensional motion and find the trajectories in the case V (r) = 1 2 ω 2 r 2 . 11. A point particle P of mass m is constrained to move along the parabola y = a + bx 2 , with a, b being given positive constants. A point Q of mass m is constrained to move along the line y = (tan α)x. P and Q interact with an attractive elastic force of constant k. Write the expression for the Lagrangian and find the equilibrium positions depending on the parameter α. Study the stability and compute the frequency of the small oscillations around the stable equilibrium position. 12. A point particle of mass m moves on a torus of equation x 2
2 + z
2 − 2a y
2 + z
2 + a
2 − b
2 = 0,
4.14 The dynamics of discrete systems. Lagrangian formalism 165 where 0 < b < a, under the action of the force due to its weight F = (0, 0, −mg). Write down the Lagrangian, find the equilibrium positions and study their sta- bility. Compute the principal curvature of the torus at the points (0, 0, −a − b),
(0, 0, −a + b), (0, 0, a − b), (0, 0, a + b). 13. A point particle of unit mass is constrained to move on the sphere x 2 + y 2 + z 2 = 1 under the action of the force field F = ( −ax, ay, −bz), where a, b are given constants. Write down the Lagrangian and reduce the problem to one-dimensional motion. 4.13
Additional remarks and bibliographical notes The theory of stability is much more extensive than that presented in Section 4.10. The concept of stability is very important when studying all phenomena modelled by systems of differential equations of the same kind as system (4.131). It is not surprising then that the literature on the subject is very extensive, and that research in this field is still very active. The beginning of the theory is in a memoir, published in 1892, by A. Lyapunov (in Russian). The book of La Salle and Lefschetz (1961) is a particularly simple and concise read. In addition, we note a recent book of Amann (1990), containing a vast bibliography. Finally, we recall that Definition 4.1 of a holonomic system with smooth constraints is traditionally given by introducing the so-called virtual (infinitesimal) displacements instead of the virtual velocities, and hence the definition is known as the virtual work principle. 4.14 Additional solved problems Problem 1 Consider a rigid plane plate, bounded and with a smooth boundary, lying in a vertical plane. The boundary γ (or a part of it) of the plate rolls without sliding on a horizontal line, with respect to which the plate lies in the upper half-plane (the ascending orientation on the vertical is assumed as the positive orientation). In an equilibrium configuration the centre of mass G is on the vertical of the contact point O (Fig. 4.5). (i) Prove that the stability condition for the equilibrium is that the height h of the centre of mass is less than the curvature radius k −1 0 of γ at O. (ii) Compute the period of small oscillations under the above hypotheses. Solution (i) With reference to Fig. 4.5, let us compute the height of the centre of mass in the configuration when the contact point on the supporting line is moved from O to C. Equivalently we can compute it in the frame of reference t, n, the tangent and principal normal unit vectors to γ at C.
166 The dynamics of discrete systems. Lagrangian formalism 4.14
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