The Liouville Equation as a Hamiltonian System
- PDF / 489,071 Bytes
- 5 Pages / 612 x 792 pts (letter) Page_size
- 96 Downloads / 178 Views
Liouville Equation as a Hamiltonian System V. V. Kozlov1* 1
Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, 119991 Russia
Received March 24, 2020; in final form, March 24, 2020; accepted Aril 20, 2020
Abstract—Smooth dynamical systems on closed manifolds with invariant measure are considered. The evolution of the density of a nonstationary invariant measure is described by the well-known Liouville equation. For ergodic dynamical systems, the Liouville equation is expressed in Hamiltonian form. An infinite collection of quadratic invariants that are pairwise in involution with respect to the Poisson bracket generated by the Hamiltonian structure is indicated. DOI: 10.1134/S0001434620090035 Keywords: Liouville equation, Hamiltonian systems, integrability.
1. Let v be a smooth vector field on a closed smooth n-dimensional manifold M = {x} whose phase flow preserves the measure dμ = λ(x) dn x with positive density λ: div λv = 0. Let there be another (this time, nonstationary) invariant measure ρ dμ. The fundamental (for statistical mechanics) Liouville equation ∂ρλ + div(ρλv) = 0 ∂t leads to the following linear evolution equation for the function ρ(t, x): ∂ρ + Lv ρ = 0, (1) ∂t where Lv is the differential operator along the field v. In other words, ρ is a first integral of the dynamical system x ∈ M.
x˙ = v(x),
(2)
In ergodic theory, an essential role is played by the one-parameter group of Koopman operators ϕ → U t ϕ = ϕ(gt x),
t ∈ R,
where {g t } is the phase flow of the dynamical system (2). These operators are unitary in the real Hilbert space L2 (M, μ). Obviously, d and U t = exp(tLv ). Lv = U t dt t=0 2. The domain of the operator Lv includes C 1 (M ), the space of continuously differentiable functions, which is, obviously, dense in L2 (M, μ). Lemma 1. If f ∈ C 1 (M ), then
ˆ Lv f dμ = 0. M
*
E-mail: [email protected]
339
340
KOZLOV
Indeed, in view of the invariance condition for the measure μ, we have λLv f = div(f λv) = 0. It remains to apply the Gauss–Ostrogradskii theorem. Lemma 2. For any functions f and g from C 1 (M ), ˆ ˆ (Lv f )g dμ = − M
(3)
f (Lv g) dμ. M
This formula follows from Lemma 1 applied to the product f g of the functions. Corollary 1. The following equality holds: ˆ f (Lv f ) dμ = 0. M
Let R be the image of C 1 (M ) under the linear mapping Lv . Obviously, the mapping Lv : C 1 (M ) → R is not invertible, because it takes all the constant functions to zero. To invert the operator Lv , (1) we narrow the space C 1 (M ) to a subspace of functions with zero mean (we denote this vector space by S); (2) we shall only consider vector fields v for which the system (2) does not admit nonconstant continuously differentiable first integrals. In particular, property (2) necessarily holds for ergodic systems. However, it should be noted that the dynamical system (2) may be nonergodic but admit no nonconstant first integrals from C 1 (M ) (for more general problems of this kind, see [1] and [2]). We note that R lies in the space of continuous functions with zero mean. Thu
Data Loading...
