Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

(hypo)coercive schemes for the Fokker-Planck equation F. Hérau (LMJL Nantes) with G. Dujardin (Inria Lille Europe) and P. Lafitte (Centrale Supélec) Workshop on kinetic and fluid PDEs Paris Descartes and Paris Diderot

Paris - 6 mars 2018

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Table of contents

1

Introduction

2

Continuous inhomogeneous case

3

Semi-discrete homogeneous case

4

Poincaré inequalities

5

All cases

6

Numerics

Poincaré inequalities

All cases

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Introduction We are interested in the study of the following inhomogeneous Fokker-Planck equation ∂t F + v ∂x F − ∂v (∂v + v )F = 0,

F |t=0 = F 0 ,

where 0 ≤ F = F (t, x , v ),

(t, x , v ) ∈ R+ × T × R

ZZ

Fdxdv = 1.

We focus in this talk on the case d = 1. The now rather standard hypocoercive methods give that F (t, x , v ) −→ M(x , v ), t→+∞

exponentially fast (for a large family of similar kinetic equations), where here the Maxwellian is given by 2 1 M(x , v ) = µ(v ) = √ e−v /2 . 2π

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

A much simpler equation (homogeneous kinetic equation) is ∂t F − ∂v (∂v + v )F = 0,

F |t=0 = F 0 ,

where 0 ≤ F = F (t, v ),

(t, v ) ∈ R+ × R,

Z

F dv = 1,

for which this is very easy to get (by "coercive" methods) that F (t, x , v ) −→ µ(v ). t→+∞

(This is just the heat equation for the harmonic oscillator.)

(1)

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Functional framework and proof for the homogeneous problem : set F = µ + µf , the equation is ∂t f + (−∂v + v )∂v f = 0 with f |t=0 = f 0 , consider f ∈ L2 (dµ) ⊂ L1 (dµ) (strictly smaller), R def R note that hf i = f dµ = f0 dµ = 0,

note that f ∈ L1 (dµ) ⇔ F ∈ L1 (dv ), compute 2 2 d dt kf kL2 (dµ) = −2 h(−∂v + v )∂v f , f iL2 (dµ) = −2 k∂v f kL2 (dµ) , 2

2

use Poincaré inequality kf kL2 (dµ) ≤ k∂v f kL2 (dµ) , so that d 2 2 kf kL2 (dµ) ≤ −2 kf kL2 (dµ) , dt

use Gronwall inequality kf kL2 (dµ) ≤ e−t f 0 L2 (dµ) ,

synthesis kF − Mk 1 ≤ kf k 2 ≤ e−t f 0 2 L (dv )

L (dµ)

L (dµ)

.

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

Many ingredients were involved in the short previous proof : Hilbertian framework, coercivity, Poincaré inequality, Gronwall lemma, existence of a Maxwellian ...

Aim of this talk : Explain how to adapt to the inhomogeneous case ✄ well understood and robust theory Explain how to discretize and numerically implement the problems ✄ new even in the homogeneous case

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

The continuous inhomogeneous case Perform the same change of variables F = µ + µf , and work in H 1 (dµdx ) ֒→ L2 (dµdx). The inhomogeneous equation reads ∂t f + v ∂x f + (−∂v + v )∂v f = 0, f |t=0 = f 0 , ZZ

 def hf i = f dµdx = f 0

very partial biblio (Guo, Villani-Desvillettes ... H. 06-07, Mouhot-Neumann 06, H.-Nier 04, Villani 07, Dolbeault-Mouhot-Schmeiser 15, etc...) robust proof (Boltzmann, Landau, fractional etc...) and methods (hypocoercivity, hypoellipticity) wide applications (hydro limits, perturbative NL solutions, VPFP, Landau damping, enlargement theory, statistical mechanics, low temperature, UQ etc

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

The continuous inhomogeneous case Perform the same change of variables F = µ + µf , and work in H 1 (dµdx ) ֒→ L2 (dµdx). The inhomogeneous equation reads ∂t f + v ∂x f + (−∂v + v )∂v f = 0, f |t=0 = f 0 , ZZ

 def hf i = f dµdx = f 0 commutator identity [∂v , v ∂x ] = ∂x (hypoellipticity results by Hörmander, Kohn, developped by Helffer, Nourrigat, mention subunit balls by Fefferman and subelliptic geometry etc... ).

how to discretize such an equality and equation ? fundamental point : have the simplest proofs and techniques in order to adapt them to the discretized cases.

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

The modified entropy

We define the entropy functional for C > D > E > 1, to be defined later on H : f 7→ C kf k2 + D k∂v f k2 + E h∂v f , ∂x f i + k∂x f k2 .

(2)

Then for C, D, E well chosen, we will prove that t 7→ H(f (t)) is nonincreasing when f solves the rescaled equation with initial datum f 0 ∈ H 1 (dµ). First note that if E 2 < D, H is equivalent to the H 1 (dµdx)-norm : 1 kf k2H 1 ≤ H(f ) ≤ 2C kf k2H 1 2 We have modified the norm in H 1 .

(3)

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

✄ First term d kf k2 = 2 h∂t f , f i = −2 hv ∂x f , f i − 2 h(−∂v + v )∂v f , f i = −2 k∂v f k2 dt ✄ Second term d 2 k∂v f k = 2 h∂v (∂t f ), ∂v f i dt = −2 h∂v (v ∂x f + (−∂v + v )∂v f ), ∂v f i

= −2 hv ∂x ∂v f , ∂v f i − 2 h[∂v , v ∂x ] f , ∂v f i − 2 h∂v (−∂v + v )∂v f , ∂v f i . = −2 h∂x f , ∂v f i − 2 k(−∂v + v )∂v f k

✄ Last term

2

d 2 2 k∂x f k = −2 k∂v ∂x f k dt

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

✄ Third important term d h∂x f , ∂v f i dt = − h∂x (v ∂x f + (−∂v + v )∂v f ), ∂v f i − h∂x f , ∂v (v ∂x f + (−∂v + v )∂v f )i = − hv ∂x (∂x f ), ∂v f i − h(−∂v + v )∂v f , ∂x ∂v f i − h∂x f , [∂v , v ∂x ] f i − h∂x f , v ∂x ∂v f i − h∂x f , [∂v , (−∂v + v )] ∂v f i − h(−∂v + v )∂v f , ∂x ∂v f i .

we have hv ∂x ∂x f , ∂v f i + h∂x f , v ∂x ∂v f i = 0. and [∂v , (−∂v + v )] = 1 so that d 2 h∂x f , ∂v f i = − k∂x f k + 2 h(−∂v + v )∂v f , ∂x ∂v f i − h∂x f , ∂v f i . dt

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

✄ Entropy dissipation inequality d 2 2 2 2 H(f ) = −2C k∂v f k − 2D k(−∂v + v )∂v f k − E k∂x f k − 2 k∂x ∂v f k dt − 2(D + E) h∂x f , ∂v f i − 2E h(−∂v + v )∂v f , ∂x ∂v f i . Therefore, using Cauchy-Schwartz : for 1 < E < D < C well chosen, E d H(f ) ≤ −C k∂v f k2 − (E − 1/2) k∂x f k2 ≤ − (k∂v f k2 + k∂x f k2 ). dt 2 Using the Poincaré inequality in space and velocity E E E cp d 2 2 2 H(f ) ≤ − (k∂v f k + k∂x f k ) − cp kf k ≤ − H(f ). dt 4 4 4 2C ✷

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

✄ Synthesis We pose 2κ =

E cp and we get by Gronwall lemma 4 2C

Theorem For all f 0 ∈ H 1 , the solution f to the rescaled inhomogeneous Fokker-Planck equation satisifies for all t ≥ 0,

2 1 1 2 2 kf kL2 ≤ kf kH 1 ≤ H(f (t)) ≤ e−2κt H(0) ≤ 2Ce−2κt f 0 H 1 . 2 2

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

We want to discretize the equation in space, velocity and time, with preservation of the long time behavior, (hypo)coercivity, the notion of Maxwellian. Keywords and the discrete case Equation ? derivative ? Hilbert space ? Maxwellian ? Gronwall ? Poincaré ? commutators ? local ? Huge literature : (Herda, Bessemoulin, AP, Jin, etc etc...). In the spirit of hypocoercivity for short time : Poretta-Zuazua

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

The semi-discrete homogeneous case

We want to discretize (and implement) the equation ∂t F − ∂v (∂v + v )F = 0. We look for a discretization only in velocity. ✄ The velocity derivative : for F ∈ ℓ1 (Z), define Dv F ∈ ℓ1 (Z∗ ) by (Dv F )i =

Fi − Fi−1 for i > 0, h

(Dv F )i =

Fi+1 − Fi for i < 0. h

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

✄ The Maxwellian : solving equation (Dv + v )µh = 0 yields µhi = Q|i|

ch

l=0 (1

+ hvi )

,

i ∈ Z.

Then µh is even, positive, in ℓ1 . Proof by direct computation : (Dv + v )µh = 0 writes  h µ − µhi−1   i + vi µhi = 0 h h h   µi+1 − µi + v µh = 0 i i h

which gives the expression of µh ∈ ℓ1 .

for i > 0 for i < 0,

All cases

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

✄ The "adjoint" : for G ∈ ℓ1 (Z∗ ), define Dv♯ F ∈ ℓ1 (Z) (Dv♯ G)i =

Gi+1 − Gi for i > 0, h (Dv♯ G)0 =

(Dv♯ G)i =

Gi − Gi−1 for i < 0, h

G1 − G−1 . h

✄ The Hilbert spaces : we pose F = µh + µh f and consider X f ∈ ℓ2 (µh ) ⇔ fi2 µhi < +∞ i

then denoting µ♯i = µhi−1 for i > 0 and µ♯i = µhi+1 for i < 0 −Dv♯ ((Dv + v )µh f ) = Dv♯ (µ♯ (Dv f )i ) = µhi (−Dv♯ + v )Dv f .

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

The discrete equation is then ∂t F − Dv♯ (Dv + v )F = 0,

F |t=0 = F 0 .

With F = µh + µh f , we have Proposition The equation satisfied by f is the following ∂t f + (−Dv♯ + v )Dv f = 0,

f |t=0 = f 0 .

The operator (−Dv♯ + v )Dv is selfadjoint non-negative in ℓ2 (µh ) : D E D E (−Dv♯ + v )Dv f , g = hDv f , Dv gi♯ = f , (−Dv♯ + v )Dv g , where

2

ϕ ∈ ℓ2 (µ♯ ) ⇔ kϕk♯ =

X i6=0

ϕ2i µ♯i < ∞.

Constant sequences are the equilibrium states of the equation.

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

We can do the same proof as in the continuous case P def P note that hf i = fi µhi = fi0 µhi = 0, D E 2 2 compute dtd kf k = −2 (−Dv♯ + v )Dv f , f = −2 kDv f k♯ , use Poincaré inequality kf k2 ≤ kDv f k2♯ ,

use Gronwall inequality kf k ≤ e−t f 0 ,

synthesis F − µh ℓ1 ≤ kf k ≤ e−t f 0 .

? ? Poincare inequality ? ?

All cases

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

We can do the same proof as in the continuous case P def P note that hf i = fi µhi = fi0 µhi = 0, D E 2 2 compute dtd kf k = −2 (−Dv♯ + v )Dv f , f = −2 kDv f k♯ , use Poincaré inequality kf k2 ≤ kDv f k2♯ ,

use Gronwall inequality kf k ≤ e−t f 0 ,

synthesis F − µh ℓ1 ≤ kf k ≤ e−t f 0 .

? ? Poincare inequality ? ?

All cases

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Poincaré inequality Lemma (adapted proof of that by H. Poincare (1912)) 2

2

For all f ∈ H 1 (µ) with hf i = 0, we have kf kL2 (dµ) ≤ k∂v f kL2 (dµ) Proof. Denote f (v ) = f , f (v ′ ) = f ′ , dµ = µ(v )dv and dµ′ = µ(v ′ )dv ′ .

Z

1 f dµ = 2 R 2

ZZ

1 (f − f ) dµdµ = 2 R2 ′

′

2

ZZ

R2

Z

v′

v

∂v f (w)dw

!2

From Cauchy Schwartz 1 f dµ ≤ 2 R

Z

2

ZZ

R2

Z

v

v′

2

|∂v f (w)| dw

!

(v ′ − v )dµdµ′

dµdµ′

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

Rv 2 Denote F (v ) = a |∂v f (w)| dw. Then ZZ Z 1 (F ′ − F ) (v ′ − v )dµdµ′ f 2 dµ ≤ 2 2 R R  ZZ ZZ ZZ ZZ 1 ′ ′ F ′ v dµdµ′ Fv ′ dµdµ′ − Fv dµdµ′ − F v dµdµ′ + = 2 R2 R2 R2 R2 Z Fv dµ, = R

Note that ∂v µ = −v µ and perform an integration by parts Z Z Z Z Z 2 f dµ ≤ Fv µdv = − F ∂v µdv = ∂v F µdv = |∂v f |2 dµ. R

R

R

R

R

✷

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Discrete Poincaré inequality Proposition (Discrete Poincaré inequality) Let f be a sequence in H 1 . Then, 2

2

kf − hf ikℓ2 (µh ) ≤ kDv f kℓ2 (µ♯ ) . Proof. Assume hf i = 0 and write X X 1X fi2 µi = (fj − fi )2 µi µj = (fj − fi )2 µi µj 2 i

i,j

i0

i D > E > 1 well  chosen. There exists k , t0 , h0 > 0 such that for all f 0 ∈ H 1 with f 0 = 0, all δt ∈ (0, t0 ), and all h ∈ (0, h0 ) the sequence defined by the implicit Euler scheme satisfies for all n ∈ N,

2 1 n 2 kf kH 1 ≤ H(f n ) ≤ H(f 0 )e−knδt ≤ 2C f 0 H 1 e−knδt . 2

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Numerics

Elements of proof Consider 2

2

2

H(f ) = C kf k + D kDv f k♯ + E hDv f , SDx f i♯ + kDx f k .

(4)

with S = [Dv , v ] and therefore SDx = [Dv , vDx ] : (Sg)i = gi−1 for i ≥ 1

(Sg)i = gi+1 for i ≤ −1.

We have for example Dv (−Dv♯ + v )S − S(−Dv♯ + v )Dv = S + δ, where δ is the singular operator from ℓ2 to ℓ2♯ defined for f ∈ ℓ2 by (δf )j = 0 if |j| ≥ 2,

(δf )−1 =

f1 − f0 , h2

(δf )1 = −

f0 − f−1 . h2

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Poincaré inequalities

All cases

Especially for the singular term involving δ, we have for all ε > 0,

2 1

2 ♯ hδDx f , Dv f i♯ ≤ (−Dv + v )Dv f + ε kDv Dx f k♯ . ε

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Numerics

Poincaré inequalities

All cases

Numerics

Introduction

Continuous inhomogeneous case

Semi-discrete homogeneous case

Thank you !

Poincaré inequalities

All cases

Numerics