Hypocoercivity without conﬁnement I: a mode-by-mode analysis

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a mode-by-mode analysis

Jean Dolbeault

http://www.ceremade.dauphine.fr/∼dolbeaul Ceremade, Université Paris-Dauphine

October 2, 2017

KI-Net conference onHypocoercivity and Sensitivity Analysis in Kinetic Equations and Uncertainty Quantification

Monday Oct 2, 2017 - Thursday Oct 5, 2017

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Outline

An abstract hypocoercivity result BAbstract statement

BA toy model

Mode-by-mode hypocoercivity

BFokker-Planck equation and scattering collision operators BA mode-by-mode hypocoercivity result

BEnlargement of the space by factorization BApplication to the torus

Application and computation of the constants BA more numerical point of view

The Kinetic equations without confinement

BHypocoercivity without confinement II:lecture by C. Schmeiser

J. Dolbeault Hypocoercivity without confinement I

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References

E. Bouin, J. Dolbeault, S. Mischler, C. Mouhot, C. Schmeiser.

Hypocoercivity without confinement. Preprint arXiv:1708.06180 J. Dolbeault, C. Mouhot, and C. Schmeiser. Hypocoercivity for kinetic equations with linear relaxation terms. Comptes Rendus Mathématique, 347(9-10):511 – 516, 2009.

J. Dolbeault, A. Klar, C. Mouhot, and C. Schmeiser. Exponential rate of convergence to equilibrium for a model describing fiber lay-down processes. Applied Mathematics Research eXpress, 2012.

J. Dolbeault, C. Mouhot, and C. Schmeiser. Hypocoercivity for linear kinetic equations conserving mass. Trans. Amer. Math. Soc., 367(6):3807–3828, 2015.

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An abstract hypocoercivity result

BAbstract statement BA toy model

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An abstract evolution equation

Let us consider the equation dF

dt +TF =LF (1)

In the framework of kinetic equations,TandLare respectively the transport and the collision operators

We assume thatTandL are respectively anti-Hermitian and Hermitian operators defined on the complex Hilbert space (H,h·,·i)

A:= 1 + (TΠ)^∗TΠ−1

(TΠ)^∗

∗denotes the adjoint with respect toh·,·i

Π is the orthogonal projection onto the null space ofL

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The assumptions

λm,λM, andCM are positive constants such that, for anyF∈H Bmicroscopic coercivity:

− hLF, Fi ≥λ_mk(1−Π)Fk² (H1) Bmacroscopic coercivity:

kTΠFk²≥λMkΠFk² (H2) Bparabolic macroscopic dynamics:

ΠTΠF = 0 (H3)

Bbounded auxiliary operators:

kAT(1−Π)Fk+kALFk ≤C_Mk(1−Π)Fk (H4) The estimate

1 2

d

dtkFk²=hLF, Fi ≤ −λmk(1−Π)Fk² is not enough to conclude thatkF(t,·)k² decays exponentially

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Equivalence and entropy decay

For someδ >0 to be determined later, the L² entropy / Lyapunov functional is defined by

H[F] := ¹₂kFk²+δRehAF, Fi

as in(Dolbeault-Mouhot-Schmeiser)so thathATΠF, Fi ∼ kΠFk² and

− d

dtH[F] = :D[F]

= − hLF, Fi+δhATΠF, Fi

− δRehTAF, Fi+δRehAT(1−Π)F, Fi −δRehALF, Fi Bfor anyδ >0 small enough andλ=λ(δ)

λH[F]≤D[F]

Bnorm equivalence ofH[F] andkFk² 2−δ

4 kFk²≤H[F]≤2 +δ 4 kFk²

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Exponential decay of the entropy

λ= _{3 (1+λ}^λ^M

M)minn

1, λm,_(1+λ^λ^m^λ^M

M)C_M²

o

,δ=¹₂ min

1, λm,_(1+λ^λ^m^λ^M

M)C_M²

h1(δ, λ) := (δ CM)²−4

λm− δ−2 +δ

4 λ δ λM

1 +λM

−2 +δ 4 λ

Theorem

LetL andTbe closed linear operators (respectively Hermitian and anti-Hermitian) onH. Under (H1)–(H4), for any t≥0

H[F(t,·)]≤H[F₀]e^−λ^?^t whereλ_? is characterized by

λ?:= sup

λ >0 : ∃δ >0s.t.h1(δ, λ) = 0, λm− δ−¹₄(2 +δ)λ >0

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Sketch of the proof

SinceATΠ = 1 + (TΠ)^∗TΠ−1

(TΠ)^∗TΠ, from (H1) and (H2)

− hLF, Fi+δhATΠF, Fi ≥λmk(1−Π)Fk²+ δ λM

1 +λM

kΠFk²

By (H4), we know that

|RehAT(1−Π)F, Fi+ RehALF, Fi| ≤CMkΠFk k(1−Π)Fk The equationG=AF is equivalent to (TΠ)^∗F =G+ (TΠ)^∗TΠG hTAF, Fi=hG,(TΠ)^∗Fi=kGk²+kTΠGk²=kAFk²+kTAFk² By the Cauchy-Schwarz inequality, for anyµ >0

hG,(TΠ)^∗Fi ≤ kTAFk k(1−Π)Fk ≤ 1

2µkTAFk²+µ

2k(1−Π)Fk² kAFk ≤ 1

2k(1−Π)Fk, kTAFk ≤ k(1−Π)Fk, |hTAF, Fi| ≤ k(1−Π)Fk² WithX :=k(1−Π)Fk andY :=kΠFk

D[F]−λH[F]≥(λm−δ)X²+ δ λM

1 +λM

Y²−δ CMX Y−2 +δ

4 λ(X²+Y²)

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Hypocoercivity

Corollary

For anyδ∈(0,2), ifλ(δ)is the largest positive root ofh₁(δ, λ) = 0for whichλm−δ−¹₄(2 +δ)λ >0, then for any solutionF of (1)

kF(t)k²≤ 2 +δ

2− δe^−λ(δ)^tkF(0)k² ∀t≥0 From the norm equivalence ofH[F] andkFk²

2−δ

4 kFk²≤H[F]≤2 +δ 4 kFk² We use ²⁻₄^δkF0k²≤H[F₀] so thatλ_?≥sup_δ∈(0,2)λ(δ)

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A toy problem

du

dt = (L−T)u , L=

0 0 0 −1

, T =

0 −k k 0

, k²≥Λ>0 Non-monotone decay, a well known picture:

see for instance(Filbet, Mouhot, Pareschi, 2006) H-theorem: _dt^d|u|²=−2u²₂

macroscopic limit: ^du_dt¹ =−k²u1

generalized entropy: H(u) =|u|²− _1+k^{δ k}₂ u₁u₂ dH

dt = −

2− δ k² 1 +k²

u²₂− δ k²

1 +k²u²₁+ δ k 1 +k²u1u2

≤ −(2−δ)u²₂− δΛ

1 + Λu²₁+δ 2u1u2

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Plots for the toy problem

1 2 3 4 5 6

0.2 0.4 0.6 0.8 1 u1²

1 2 3 4 5 6

0.1 0.2 0.3 0.4 0.5 0.6 0.7 u1²+u2²

1 2 3 4 5 6

0.2 0.4 0.6 0.8 1 u1²

1 2 3 4 5 6

0.2 0.4 0.6 0.8 1 H

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Mode-by-mode hypocoercivity

BFokker-Planck equation and scattering collision operators BA mode-by-mode hypocoercivity result

BEnlargement of the space by factorization BApplication to the torus

(Bouin, J.D., Mischler, Mouhot, Schmeiser)

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Fokker-Planck equation with general equilibria

We consider the Cauchy problem

∂tf+v· ∇xf =Lf , f(0, x, v) =f0(x, v) (2) for a distribution functionf(t, x, v), withpositionvariablex∈R^d or x∈T^d the flat d-dimensional torus

Fokker-Planckcollision operator with a general equilibriumM Lf =∇v·h

M∇v M⁻¹fi

Notation and assumptions: anadmissible local equilibriumM is positive, radially symmetric and

Z

R^d

M(v)dv= 1, dγ=γ(v)dv:= dv M(v) γis anexponential weightif

lim

|v|→∞

|v|^k

γ(v) = lim

|v|→∞M(v)|v|^k= 0 ∀k∈(d,∞)

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Definitions

Θ = 1 d

Z

R^d

|v|²M(v)dv= Z

R^d

(v·e)²M(v)dv for an arbitrarye∈S^d−1

Z

R^d

v⊗v M(v)dv= Θ Id Then

θ= 1

d k∇vMk²_L2(dγ)=4 d

Z

R^d

|∇v

√ M

2dv <∞

IfM(v) = ^e⁻

1 2|v|2

(2π)^d/2, then Θ = 1 andθ= 1 σ:= 1

2 pθ/Θ

Microscopic coercivity property(Poincaré inequality): for all u=M⁻¹F ∈H¹(M dv)

Z

R^d

|∇u|²M dv≥λm

Z

R^d

u−

Z

R^d

u M dv 2

M dv

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Scattering collision operators

Scatteringcollision operator Lf =

Z

R^d

σ(·, v⁰) f(v⁰)M(·)−f(·)M(v⁰) dv⁰

Main assumption on thescattering rateσ: for some positive, finiteσ 1≤σ(v, v⁰)≤σ ∀v, v⁰∈R^d

Example: linear BGK operator

Lf =M ρ_f−f , ρ_f(t, x) = Z

R^d

f(t, x, v)dv Local mass conservation

Z

R^d

Lf dv= 0 and we have

Z

R^d

|Lf|²dγ≤4σ² Z

R^d

|M ρf−f|²dγ

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The symmetry condition Z

R^d

σ(v, v⁰)−σ(v⁰, v)

M(v⁰)dv⁰ = 0 ∀v∈R^d implies thelocal mass conservationR

R^dLf dv= 0

Micro-reversibility,i.e., the symmetry ofσ, is not required

The null space ofLis spanned by the local equilibriumM andLonly acts on the velocity variable

Microscopic coercivity property: for someλm>0 1

2 Z Z

R^d×R^d

σ(v, v⁰)M(v)M(v⁰) (u(v)−u(v⁰))²dv⁰dv

≥λm

Z

R^d

(u−ρu M)²M dv holds according to Proposition 2.2 of(Degond, Goudon, Poupaud, 2000)for allu=M⁻¹F ∈L²(M dv). Ifσ≡1, thenλ_m= 1

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Fourier modes

In order to perform amode-by-mode hypocoercivityanalysis, we introduce the Fourier representation with respect tox,

f(t, x, v) = Z

R^d

fˆ(t, ξ, v)e^{−i x·ξ}dµ(ξ)

dµ(ξ) = (2π)^−ddξ anddξ is the Lesbesgue measure ifx∈R^d dµ(ξ) = (2π)^−d P

z∈Z^dδ(ξ−z) is discrete forx∈T^d

Parseval’s identity ifξ∈Z^d and Plancherel’s formula ifx∈R^d read kf(t,·, v)k_L2(dx)=

fˆ(t,·, v)

_L₂_(dµ(ξ))

The Cauchy problem is now decoupled in theξ-direction

∂_tfˆ+Tfˆ=Lf ,ˆ fˆ(0, ξ, v) = ˆf₀(ξ, v) Tfˆ=i(v·ξ) ˆf

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For any fixedξ∈R^d, let us apply the abstract result with H= L²(dγ), kFk²=

Z

R^d

|F|²dγ , ΠF =M Z

R^d

F dv=M ρF

andTfˆ=i(v·ξ) ˆf,TΠF=i(v·ξ)ρFM, kTΠFk²=|ρF|²

Z

R^d

|v·ξ|²M(v)dv = Θ|ξ|²|ρF|²= Θ|ξ|²kΠFk² (H2)Macroscopic coercivitykTΠFk²≥λMkΠFk²: λM = Θ|ξ|² (H3)R

R^dv M(v)dv= 0 The operatorAis given by

AF = −i ξ·R

R^dv⁰F(v⁰)dv⁰ 1 + Θ|ξ|² M

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A mode-by-mode hypocoercivity result

kAFk=kA(1−Π)Fk ≤ 1 1 + Θ|ξ|²

Z

R^d

|(1−Π)F|

√M |v·ξ|√ M dv

≤ 1

1 + Θ|ξ|²k(1−Π)Fk Z

R^d

(v·ξ)²M dv 1/2

=

√ Θ|ξ|

1 + Θ|ξ|²k(1−Π)Fk Scattering operatorkLFk²≤4σ²k(1−Π)Fk²

Fokker-Planck (FP) operator kALFk ≤ 2

1 + Θ|ξ|² Z

R^d

|(1−Π)F|

√

M |ξ·∇v

√

M|dv≤

√ θ|ξ|

1 + Θ|ξ|²k(1−Π)Fk In both cases withκ=√

θ(FP) orκ= 2σ√

Θ we obtain kALFk ≤ κ|ξ|

1 + Θ|ξ|²k(1−Π)Fk

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TAF(v) =−(v·ξ)M 1 + Θ|ξ|²

Z

R^d

(v⁰·ξ) (1−Π)F(v⁰)dv⁰ is estimated by

kTAFk ≤ Θ|ξ|²

1 + Θ|ξ|²k(1−Π)Fk (H4) holds withCM = ^κ_1+Θ^|ξ|+Θ_|ξ|^|ξ|₂²

Two elementary estimates Θ|ξ|²

1 + Θ|ξ|² ≥ Θ max{1,Θ}

|ξ|² 1 +|ξ|² λM

(1 +λ_M)C_M² =Θ 1 + Θ|ξ|²

(κ+ Θ|ξ|)² ≥ Θ κ²+ Θ

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Theorem

Let us consider an admissibleM and a collision operatorL satisfying Assumption(H), and takeξ∈R^d. Iffˆis a solution such that fˆ₀(ξ,·)∈L²(dγ), then for any t≥0, we have

fˆ(t, ξ,·)

2

L²(dγ)≤3e⁻^µ^ξ^t

fˆ0(ξ,·)

2 L²(dγ)

where

µ_ξ := Λ|ξ|²

1 +|ξ|² and Λ = Θ

3 max{1,Θ} min

1, λ_mΘ κ²+ Θ

withκ= 2σ√

Θfor scattering operators andκ=√

θfor (FP) operators

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Enlargement of the space by factorization

A simple case (factorization of order 1) of thefactorization method of(Gualdani, Mischler, Mouhot)

Theorem

LetB1,B2 be Banach spaces and letB2 be continuously imbedded in B1, i.e.,k · k1≤c₁k · k2. LetB andA+Bbe the generators of the strongly continuous semigroupse^Bt ande^(A+B)^ton B1. If for all t≥0,

e^(A+B)t

2→2≤c2e^−λ²^t, e^Bt

_1→1≤c3e^−λ¹^t, kAk_1→2≤c4

wherek · k_i→j denotes the operator norm for linear mappings fromBi

toBj. Then there exists a positive constantC=C(c1, c2, c3, c4)such that, for allt≥0,

e^(A+B)^t _1→1≤

( C 1 +|λ1−λ₂|⁻¹

e⁻^min{λ¹^,λ²^}^t forλ₁6=λ₂ C(1 +t)e^−λ¹^t forλ₁=λ₂

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Integrating the identity _ds^d e^(A+B)se^B^(t−s)

=e^(A+B)^sAe^B^(t−s) with respect tos∈[0, t] gives

e^(A+B)t=e^B^t+ Z t

0

e^(A+B)sAe^B^(t−s)ds

The proof is completed by the straightforward computation e^(A+B)^t

1→1≤c₃e^−λ¹^t+c₁ Z t

0

e^(A+B)^sAe^B(t−s) 1→2ds

≤c₃e^−λ¹^t+c₁c₂c₃c₄e^−λ¹^t Z t

0

e^(λ¹^−λ²⁾^sds

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Weights with polynomial growth

Let us consider the measure

dγ_k :=γ_k(v)dv where γ_k(v) =πd/2 Γ((k−d)/2)

Γ(k/2) 1 +|v|²^k/2 for an arbitraryk∈(d,+∞)

We chooseB1= L²(dγk) andB2= L²(dγ) Theorem

LetΛ = _{3 max{1,Θ}}^Θ minn 1,_κ^λ2^m+Θ^Θ

o

andk∈(d,∞]. For anyξ∈R^d if fˆis a solution with initial datum fˆ₀(ξ,·)∈L²(dγ_k), then there exists a constantC=C(k, d, σ)such that

fˆ(t, ξ,·)

2

L²(dγ_k)≤C e⁻^µ^ξ^t

fˆ0(ξ,·)

2

L²(dγ_k) ∀t≥0

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Fokker-Planck: AF =N χRF andBF=−i(v·ξ)F+LF−AF N andR are two positive constants,χ is a smooth cut-off function andχR:=χ(·/R)

For anyRandN large enough, according to Lemma 3.8 of(Mischler, Mouhot, 2016)

Z

R^d

(L−A)(F)F dγk ≤ −λ1

Z

R^d

F²dγk

for someλ₁>0 ifk > d, andλ₂=µ_ξ/2≤1/4 Scattering operator:

AF(v) = M(v) Z

R^d

σ(v, v⁰)F(v⁰)dv⁰ BF(v) = −

i(v·ξ) + Z

R^d

σ(v, v⁰)M(v⁰)dv⁰

F(v) Boundedness: kAFk_L2(dγ)≤σ R

R^dγ_k⁻¹dv^1/2

kFk_L2(dγk)

λ1= 1 andλ2=µξ/2≤1/4

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Exponential convergence to equilibrium in T

^d

The unique global equilibrium in the casex∈T^d is given by f∞(x, v) =ρ∞M(v) with ρ∞= 1

|T^d| Z Z

T^d×R^d

f0dx dv

Theorem

Assume thatk∈(d,∞]andγ has an exponential growth ifk <∞.

We consider an admissibleM, a collision operatorLsatisfying Assumption(H), andΛgiven by (3). There exists a positive constant Ck such that the solutionf of (2) onT^d×R^d with initial datum f0∈L²(dx dγk)satisfies

kf(t,·,·)−f_∞k_L2(dx dγk)≤Ck kf0−f_∞k_L2(dx dγk) e⁻¹⁴^Λ^t ∀t≥0

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If we represent the flat torusT^d by the box [0,2π)^d with periodic boundary conditions, the Fourier variable satisfiesξ∈Z^d. Forξ= 0, the microscopic coercivity implies

fˆ(t,0,·)−fˆ_∞(0,·)

_L₂_(dγ)≤

fˆ₀(0,·)−fˆ_∞(0,·)

_L₂_(dγ) e^−t

Otherwiseµ_ξ ≥Λ/2 for anyξ6= 0

Parseval’s identity applies, with measuredγ(v) andC_∞=√ 3 The result with weightγk follows from the factorization result for someCk>0

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Computation of the constants

BA more numerical point of view

Two simple examples: Ldenotes either theFokker-Planck operator L1f := ∆vf+∇v·(v f)

or thelinear BGK operator

L₂f :=Πf −f

Πf =ρ_fM is the projection operator on the normalized Gaussian function

M(v) = e⁻¹²^|v|² (2π)^d/2 andρf :=R

R^df dv is the spatial density

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Where do we have space for improvements ?

WithX :=k(1−Π)Fk andY :=kΠFk, we wrote D[F]−λH[F]

≥(λm− δ)X²+ δ λM

1 +λ_M Y²−δ CMX Y −λ

2 X²+Y²+δ X Y

≥(λm−δ)X²+ δ λM

1 +λ_M Y²−δ CMX Y −2 +δ

4 λ(X²+Y²) We can directly study the positivity condition for the quadratic form

(λm− δ)X²+ δ λM

2 X²+Y²+δ X Y λm= 1,λM =|ξ|²andCM =|ξ|(1 +|ξ|)/(1 +|ξ|²)

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2 4 6 8 10 0.1

0.2 0.3 0.4

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Withλ_m= 1, λ_M =|ξ|² andC_M =|ξ|(1 +|ξ|)/(1 +|ξ|²), we optimizeλunder the condition that the quadratic form

(λm− δ)X²+ δ λM

2 X²+Y²+δ X Y is positive, thus getting aλ(ξ)

By taking alsoδ=δ(ξ) whereξis seen as a parameter, we get a better estimate ofλ(ξ)

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2 4 6 8 10 0.1

0.2 0.3 0.4 0.5

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By takingδ=δ(ξ), for each value ofξ we build a different Lyapunov function, namely

Hξ[F] := ¹₂kFk²+δ(ξ) RehAF, Fi where the operatorAis given by

AF = −i ξ·R

R^dv⁰F(v⁰)dv⁰ 1 +|ξ|² M

We can consider

A_εF = −i ξ·R

R^dv⁰F(v⁰)dv⁰ ε+|ξ|² M and look for the optimal value ofε...

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2 4 6 8 10 0.1

0.2 0.3 0.4 0.5

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The dependence ofλin εis monotone, and the limit asε→0+ gives the optimal estimate ofλ. The operator

A₀F= −i ξ·R

R^dv⁰F(v⁰)dv⁰

|ξ|² M

is not bounded anymore, but estimates still make sense and lim_ξ→0δ(ξ) = 0 (see below)

2 4 6 8 10

0.05 0.10 0.15 0.20 0.25 0.30 0.35

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2 4 6 8 10 0.1

0.2 0.3 0.4

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Theorem (Hypocoercivity onT^d with exponential weight) Assume thatL=L1 or L=L2. If f is a solution, then

kf(t,·,·)−f_∞k²_L2(dx dγ)≤C?kf0k²_L2(dx dγ)e⁻^λ^?^t ∀t≥0 withf∞(x, v) =M(v)RR

T^d×R^df0(x, v)dx dv C?≈1.75863 andλ?= ₁₃²(5−2√

3)≈0.236292.

Warning: work in progress

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These slides can be found at

http://www.ceremade.dauphine.fr/∼dolbeaul/Lectures/

BLectures The papers can be found at

http://www.ceremade.dauphine.fr/∼dolbeaul/Preprints/

BPreprints / papers

For final versions, useDolbeaultas login andJeanas password

Hypocoercivity without conﬁnement I: a mode-by-mode analysis

a mode-by-mode analysis

Outline

References

An abstract hypocoercivity result

An abstract evolution equation

The assumptions

Equivalence and entropy decay

Exponential decay of the entropy

Sketch of the proof

Hypocoercivity

A toy problem

Plots for the toy problem

Mode-by-mode hypocoercivity

Fokker-Planck equation with general equilibria

Scattering collision operators

Fourier modes

A mode-by-mode hypocoercivity result

Enlargement of the space by factorization

Weights with polynomial growth

Exponential convergence to equilibrium in T

Computation of the constants

Thank you for your attention !