About the method of characteristics

Transcription

1 About the method of characteristics Francis Nier, IRMAR, Univ. Rennes 1 and INRIA project-team MICMAC. Joint work with Z. Ammari about bosonic mean-field dynamics June 3, 2013

2 Outline The problem An example Infinite dimensional measure transportation Why a probabilistic trajectory picture?

3 The problem Let E be a separable real Hilbert space. We want to prove that a Borel probability measure µ t on E, which depends continuously on the time t R and fulfills equals t µ + div (v(t, x)µ) = 0, µ t=0 = µ 0 µ t = Φ(t, 0) µ 0, when Φ(t, s) is a well-defined flow on E, associated with the ODE ẋ = v(t, x).

4 The problem Finite dimensional case: Assume that Φ(t, s) is a diffeomorphism on E for all t, s R. The transport equation is ( t ϕ+ v(t,.), x ϕ E ) dµ t (x)dt = 0, ϕ C0 (R E), or R E T 0 ( t ϕ+ v(t, x), x ϕ E ) dµ t (x)dt+ ϕ(0, x) dµ 0 (x) E E ϕ(t, x) dµ T (x) = 0, ϕ C0 ([0, T ] E). E

5 The problem Finite dimensional case: The method of characteristics consists in taking We get and E ϕ(x, t) = a(φ(t, t)x) a C 0 (E). ( t ϕ + v, ϕ )(Φ(t, 0)x, t) = d [a(φ(t, 0)x)] = 0 dt a(φ(t, 0)x) dµ 0 (x) = E a(x) dµ T (x), a C 0 (E), which is µ T = Φ(T, 0) µ 0.

6 The problem Infinite dimensional case: The weak formulation ( t ϕ + v(t,.), x ϕ E ) dµ t (x)dt = 0, R E is given for a class of test functions: the cylindrical functions ϕ C0,cyl (R E) or some polynomial functions on E.

7 The problem Infinite dimensional case: The weak formulation ( t ϕ + v(t,.), x ϕ E ) dµ t (x)dt = 0, R E is given for a class of test functions: the cylindrical functions ϕ C0,cyl (R E) or some polynomial functions on E. Problem: A nonlinear flow does not preserve these classes. We cannot take ϕ(x, t) = a(φ(t, t)x)!!!

8 An example, the Hartree flow The Hartree equation is given by i t z = z + (V z 2 )z, z t=0 = z 0. with V ( x) = V (x). Example : V (x) = ± 1 x in dimension 3.

9 An example, the Hartree flow The Hartree equation is given by i t z = z + (V z 2 )z, z t=0 = z 0. with V ( x) = V (x). Example : V (x) = ± 1 x in dimension 3. The equation can be written i t z = z E(z) with E(z) = z(x) 2 dx + 1 V (x y) z(x) 2 z(y) 2 dxdy R d 2 R 2d = z, z z 2, V (x y)z 2.

10 An example, the Hartree flow Set z t = e it z t and the equation becomes i t z t = z h( z, t) h(z, t) = z 2, V t z 2 V t = e it( x + y ) it( x + y ) (V (x y) )e

11 An example, the Hartree flow Set z t = e it z t and the equation becomes i t z t = z h( z, t) h(z, t) = z 2, V t z 2 V t = e it( x + y ) it( x + y ) (V (x y) )e When V = ± 1 x, d = 3, Hardy s inequality leads to ( ) z 1, V t z 2 z 3 L 2 C min zσ(1) H 1 z σ(2) L 2 z σ(3) L 2 σ S 3 z 1, V t z 2 z 3 H 1 C min σ S 3 ( zσ(1) H 1 z σ(2) H 1 z σ(3) L 2 )

12 An example, the Hartree flow Set z t = e it z t and the equation becomes i t z t = z h( z, t) h(z, t) = z 2, V t z 2 V t = e it( x + y ) it( x + y ) (V (x y) )e When V = ± 1 x, d = 3, Hardy s inequality leads to ( ) z 1, V t z 2 z 3 L 2 C min zσ(1) H 1 z σ(2) L 2 z σ(3) L 2 σ S 3 z 1, V t z 2 z 3 H 1 C min σ S 3 ( zσ(1) H 1 z σ(2) H 1 z σ(3) L 2 More generally it works if V satisfies V ( x) = V (x) and V (1 ) 1/2 L(L 2 (R d )). )

13 An example, the Hartree flow Theorem Assume V (x) = V ( x) and V (1 ) 1/2 L(L 2 (R d )). The equations i t z = z E(z) et i t z = z h( z, t) define flows Φ(t) and Φ(t, s) on H 1 (R d ; C). The norm L 2 and the energy E are invariant under Φ. The norm L 2 (R d ) is invariant under Φ and the velocity field v(z, t) = 1 i zh(z, t), satisfies v(z, t) H 1 C z 2 H 1 z L 2.

14 An example, the Hartree flow The mean field analysis for bosons interacting via a pair potential V (x y), leads to Borel probability measures on H 1 = H 1 (R d ; C) which verifies ( t ϕ + i( z h. z ϕ z ϕ. z h) dµ t (z)dt = 0 R H 1 for all ϕ C 0,cyl (R H1 ; R).

15 An example, the Hartree flow The mean field analysis for bosons interacting via a pair potential V (x y), leads to Borel probability measures on H 1 = H 1 (R d ; C) which verifies ( t ϕ + i( z h. z ϕ z ϕ. z h) dµ t (z)dt = 0 R H 1 for all ϕ C 0,cyl (R H1 ; R). For a cylindrical function f C 0,cyl (H1 ; R) on H 1, we define z f by u H 1 (R d ), u, z f H 1 = u, z f. Its gradient for the real structure on H 1 = H 1 (R d ; C) with the scalar product u 1, u 2 H 1 = Re u 1, u 2 R H 1 is given by = 2 z

16 An example, the Hartree flow The mean field analysis for bosons interacting via a pair potential V (x y), leads to Borel probability measures on H 1 = H 1 (R d ; C) which verifies ( t ϕ + i( z h. z ϕ z ϕ. z h) dµ t (z)dt = 0 R H 1 for all ϕ C 0,cyl (R H1 ; R). The above equation is the weak version of t µ + div (v t µ) = 0, v t = 1 i zh(z, t) with cylindrical test functions on R H 1 R, H1 R = H1 (R d ; C) being the real Hilbert space with the scalar product u, v H 1 R and z H 1 R = z H 1.

17 An example, the Hartree flow Remark : With good assumptions (on the initial mean field data), one verifies that the measure µ t is continuous w.r.t the Wasserstein distance W 2 (µ 1, µ 2 ) = ( inf { H 1 R H1 R z 2 z 1 2 H 1 dµ(z 1, z 2 ), Π j, µ = µ j }) 1/2. Also, for all t R H 1 R z 4 H 1 z 2 L 2 dµ t (z) C and T 0 H 1 R v(t, z) 2 H 1 dµ t (z) C T. This strong continuity property is requires intermediate steps

18 Solving a transport equation in Prob 2 (E) E real separable Hilbert space. Prob 2 (E) is the space of Borel probability measures on E, µ, such that x 2 dµ(x) < + E The Wasserstein distance W 2 on Prob 2 (E) is given by { } W2 2 (µ 1, µ 2 ) = inf x 2 x 1 2 dµ(x 1, x 2 ), (Π j ) µ = µ j. E 2

19 Solving a transport equation in Prob 2 (E) For T > 0, set Γ T = C 0 ([ T, T ]; E) endowed with the norm γ = max t [ T,T ] γ(t) or the distance d(γ, γ ) = max t [ T,T ] ( n N γ(t) γ (t), e n 2 2 n) 1/2 with (e n ) n N ONB of E For a Borel probability measure η on E Γ T define the evaluation map at time t [ T, T ] by ϕ dµ η t = ϕ(γ(t)) dη(x, γ), ϕ C0,cyl (E). E Γ T

20 Solving a transport equation in Prob 2 (E) For T > 0, set Γ T = C 0 ([ T, T ]; E) endowed with the norm γ = max t [ T,T ] γ(t) or the distance d(γ, γ ) = max t [ T,T ] ( n N γ(t) γ (t), e n 2 2 n) 1/2 with (e n ) n N ONB of E For a Borel probability measure η on E Γ T define the evaluation map at time t [ T, T ] by ϕ dµ η t = ϕ(γ(t)) dη(x, γ), ϕ C0,cyl (E). E Γ T Actually, µ η t = (e t ) η with e t : (x, γ) E Γ T γ(t) E.

21 Solving a transport equation in Prob 2 (E) The following result is the infinite dimensional version of a result by Ambrosio-Gigli-Savaré. Proposition If µ t : [ T, T ] Prob 2 (E) is a W 2 -continuous solution to the equation t µ + div (v(t, x))µ) = 0 on ( T, T ), for a Borel velocity field v(t, x) = v t (x) on E such that v t L 2 (E,µ t) L 1 ([ T, T ]) ; then there exists a Borel probability measure η on E Γ T which satisfies η is carried by the set of pairs (x, γ) such that γ AC 2 ([ T, T ]; E) solves γ(t) = v t (γ(t)) for almost all t ( T, T ) and γ(0) = x. µ t = µ η t for all t [ T, T ].

22 Solving a transport equation in Prob 2 (E) Corollary If for all x E, the Cauchy problem γ(t) = v t (γ(t)), γ(0) = x is well posed and defines a flow Φ(t, s) on E, then µ t = Φ(t, 0) µ 0.

23 Solving a transport equation in Prob 2 (E) Corollary If for all x E, the Cauchy problem γ(t) = v t (γ(t)), γ(0) = x E is well posed and defines a flow Φ(t, s) on E, then µ t = Φ(t, 0) µ 0. Proof: ϕ dµ t = ϕ(γ(t)) dη(x, γ) = [ϕ Φ(t, s)](γ(s)) dη(x, γ) E Γ T E Γ T = ϕ Φ(t, s) dµ s. E

24 Why a probabilistic trajectory picture? Notations: (e n ) n N Hilbert basis of E.

25 Why a probabilistic trajectory picture? Notations: (e n ) n N Hilbert basis of E. π d, π d,t, ˆπ d = π d,t π d are defined by E x π d (x) = ( e n, x, n d) R d, d R d (y 1,..., y d ) π d,t (y) = y n e n E, n=1 d E x ˆπ d (x) = e n, x e n E. n=1

26 Why a probabilistic trajectory picture? Notations: (e n ) n N Hilbert basis of E. π d, π d,t, ˆπ d = π d,t π d are defined by E x π d (x) = ( e n, x, n d) R d, d R d (y 1,..., y d ) π d,t (y) = y n e n E, E x ˆπ d (x) = n=1 d e n, x e n E. n=1 µ d t = π d µ t ˆµ d t = ˆπ d µ t = µ d t δ 0 when E = F d F d, F d R d. { µt,y, y R d} is the disintegration of µ t w.r.t µ d t.

27 Why a probabilistic trajectory picture? Notations: (e n ) n N Hilbert basis of E. π d, π d,t, ˆπ d = π d,t π d are defined by E x π d (x) = ( e n, x, n d) R d, d R d (y 1,..., y d ) π d,t (y) = y n e n E, E x ˆπ d (x) = n=1 d e n, x e n E. n=1 µ d t = π d µ t ˆµ d t = ˆπ d µ t = µ d t δ 0 when E = F d F d, F d R d. { µt,y, y R d} is the disintegration of µ t w.r.t µ d t. v d t (y) = ˆv d t (y) = (π d ) 1 (y) (ˆπ d ) 1 (ˆπ d y) π d (v t (x)) dµ t,y (x) for a.e.y R d, ˆπ d (v t (x)) dµ t,π d y (x) for a.e.y E.

28 Why a probabilistic trajectory picture? t,y F d x v t (x ) 1 x 3 1 v t (x ) 3 y = (x) d F d x 2 v t (x ) 2 t µ t + div (v t µ t ) = 0.

29 Why a probabilistic trajectory picture? t,y F d x 3 x 1 v t (x ) 3 v t (x ) 1 x 2 d v t (y) y d = (x) v t (x ) 2 F d t µ d t + div (v d t µ d t ) = 0.

30 Why a probabilistic trajectory picture? t,y F d x 3 v t (x ) 3 x x 1 x v t (x ) 1 2 d v t (x) d v t (y) y d = (x) v t (x ) 2 F d t ˆµ d t + div (ˆv d t ˆµ d t ) = 0.

31 Why a probabilistic trajectory picture? Properties: ˆv d t L 2 (E,ˆµ d t ) = v d t L 2 (R d,µ d t ) v t L 2 (E,µ t).

32 Why a probabilistic trajectory picture? Properties: ˆv d t L 2 (E,ˆµ d t ) = v d t L 2 (R d,µ d t ) v t L 2 (E,µ t). W 2 (µ d t 1, µ d t 2 ) t 2 t 1 v d t L 2 (R d,µ d t )dt t 2 t 1 v t L 2 (E,µ t)dt.

33 Why a probabilistic trajectory picture? Properties: ˆv d t L 2 (E,ˆµ d t ) = v d t L 2 (R d,µ d t ) v t L 2 (E,µ t). W 2 (µ d t 1, µ d t 2 ) t 2 t 1 vt d L 2 (R d,µ d )dt t 2 t t 1 v t L 2 (E,µ t)dt. The sequence (ˆµ d t ) d N converges weakly narrowly to µ t with the estimate W 2 (µ t2, µ t1 ) lim inf d W 2(ˆµ d t 2, ˆµ d t 1 ).

34 Why a probabilistic trajectory picture? Properties: ˆv d t L 2 (E,ˆµ d t ) = v d t L 2 (R d,µ d t ) v t L 2 (E,µ t). W 2 (µ d t 1, µ d t 2 ) t 2 t 1 vt d L 2 (R d,µ d )dt t 2 t t 1 v t L 2 (E,µ t)dt. The sequence (ˆµ d t ) d N converges weakly narrowly to µ t with the estimate W 2 (µ t2, µ t1 ) lim inf d W 2(ˆµ d t 2, ˆµ d t 1 ). Similar properties hold for the families η d, ˆη d after projecting the trajectories.

35 Why a probabilistic trajectory picture? Properties: ˆv d t L 2 (E,ˆµ d t ) = v d t L 2 (R d,µ d t ) v t L 2 (E,µ t). W 2 (µ d t 1, µ d t 2 ) t 2 t 1 vt d L 2 (R d,µ d )dt t 2 t t 1 v t L 2 (E,µ t)dt. The sequence (ˆµ d t ) d N converges weakly narrowly to µ t with the estimate W 2 (µ t2, µ t1 ) lim inf d W 2(ˆµ d t 2, ˆµ d t 1 ). Similar properties hold for the families η d, ˆη d after projecting the trajectories. In this approximation process, the finite dimensional vector field v d t may be (is) singular no uniqueness of trajectories.