3.3 Hamiltonian case
3.3.3 Convergence to equilibrium
= −{Φ, H} , (3.37)
i.e. iff Ψ(q, t) := Φ −Rt
0hΦsids satisfies the Liouville equation Ψt+ {Ψ, H} = 0. But this is not possible, since then Ψ(q, t) must depend on p. Indeed, the Liouville equation for Ψ(q, t) implies
Ψ(q, t + ∆t) = Ψ(q, t) − ∆t ∇qΨ(q, t) · ∇pH(q, p) + o(∆t) .
For physical Hamiltonians the right hand side depends on p. Then, the only possibility is that Ψ, and thus Φ, does not depend on q, so that ρ = ρeq.
The conclusion of the above argument is that F (ρ) → min F = F (ρeq) as t → +∞. However, one cannot simply and straightforwardly conclude that ρ → ρeq, which is meaningful only when the norm ruling the distances is specified. Notice that, due to some technical reasons, in infinite-dimensional problems the Lyapunov function method does not ensure stability, in general.
3.3.3 Convergence to equilibrium
In the sequel it is proven that the quantity η := (ρ − ρeq)/ρeq converges exponentially fast to zero in the L2(µeq) norm. More precisely, one proves that there exists a constant C > 0 such that dkηk/dt ≤ −Ckηk, where kηk2 := R
Γη2dµeq = hη2ic. As a consequence, for any initial relative error η0 such that R
Γη0dµeq = 0 and kη0k < +∞, one has kη(t)k ≤ e−Ctkη0k, which implies kη(t)k → 0 in a characteristic time of the order 1/C.
First of all, the Fokker-Planck equation is rewritten for the variable η. Upon substituting ρ = ρeq(1 + η) into (3.21)-(3.22), one easily gets (do it)
ηt= −{η, H} + γQ(η) , (3.38)
where
Q(η) := −∇pH · ∇pη + T ∆pη . (3.39) Exercise 3.3. Show that (3.38)-(3.39) imply that R ηdµeq is independent of time. Hint: show that R {η, H}dµeq = 0, and that R Q(η)dµeq = 0.
Exercise 3.4. Prove that R 2η{η, H}ρeqdV = 0 and that R 2ηQ(η)ρeqdV = −2TR |∇pη|2dµeq. On multiplying (3.38) by 2ηρeq and integrating on the whole phase space, one gets
dkηk2 dt = −
Z
2η{η, H}ρeqdV + γ Z
2ηQ(η)ρeqdV = −2γT Z
|∇pη|2dµeq . (3.40) Now, one makes use of the following technical inequality, the so-called Chernoff-Poincar´e in-equality.
Theorem 3.1 (CP inequality [12]). Let dµG(x) := e−x2/2dx/√
2π be the normal measure (Gaus-sian with zero mean and unit variance) on R. Then, for any function f ∈ L2(µG) one has
[f0(x)]2
G≥[f − hf iG]2
G , (3.41)
where h·iG :=R · dµG.
Now, by a (nontrivial) extension of the above inequality to dimension n, observing that the equilibrium density on the momenta is Gaussian, and that R ηdµeq = 0, one proves that there exists a constant C > 0 such that
Z
|∇pη|2dµeq≥ C Z
η2dµeq . (3.42)
As a consequence, (3.40) implies dkηk2
dt =≤ −2 τ
Z
η2dµeq = −2
τkηk2 ⇔ dkηk
dt ≤ −1
τkηk , (3.43)
where 1/τ := γT C. Integrating the above differential inequality one gets kηk(t) ≤ e−t/τkηk(0) ,
which implies kηk(t) → 0 as t → +∞ for any η(0) ∈ L2(µeq) with zero average. For what concerns the expectation of any observable F ∈ L2(µeq), the convergence in norm to zero of η impliesR F ρdV → R F dµeqin the sup-norm sense (prove it by means of the Schwartz inequality in L2(µeq)).
Chapter 4
Kinetic theory of gasses and fluids
The treatment of this topics makes reference to the book [24]; see also [41] and [48].
45
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