Some results on the stabilization and on the controllability of hyperbolic systems

Transcription

1 Some results on the stabilization and on the controllability of hyperbolic systems Jean-Michel Coron Irrigation Channels and Related Problems Maiori (Salerno, Italy) October 02nd-04th 2008

2 Outline Stabilization: Dissipative boundary conditions The equations Main result Comparaison with prior results Proof of the exponential stability Application to the control of open channels

3 Outline Stabilization: Dissipative boundary conditions The equations Main result Comparaison with prior results Proof of the exponential stability Application to the control of open channels Controllability of hyperbolic systems The control problem Controllability theorem

4 Outline Stabilization: Dissipative boundary conditions The equations Main result Comparaison with prior results Proof of the exponential stability Application to the control of open channels Controllability of hyperbolic systems The control problem Controllability theorem A Water-tank control system Lie brackets The return method Controllability of a toy model

5 La Sambre

6 Commercial break

7 Commercial break JMC, Control and nonlinearity, Mathematical Surveys and Monographs, 136, 2007, 427 pp.

8 y t + A(y)y x = 0, y R n, x [0, 1], t [0, + ), (1)

9 y t + A(y)y x = 0, y R n, x [0, 1], t [0, + ), (1) Assumptions on A

10 y t + A(y)y x = 0, y R n, x [0, 1], t [0, + ), (1) Assumptions on A A(0) = diag (Λ 1, Λ 2,..., Λ n ),

11 y t + A(y)y x = 0, y R n, x [0, 1], t [0, + ), (1) Assumptions on A A(0) = diag (Λ 1, Λ 2,..., Λ n ), Λ i > 0, i {1,..., m}, Λ i < 0, i {m + 1,..., n},

12 y t + A(y)y x = 0, y R n, x [0, 1], t [0, + ), (1) Assumptions on A A(0) = diag (Λ 1, Λ 2,..., Λ n ), Λ i > 0, i {1,..., m}, Λ i < 0, i {m + 1,..., n}, Λ i Λ j, (i, j) {1,..., n} 2 such that i j.

13 Boundary conditions on y: ( ) ( ) y+ (t, 0) y+ (t, 1) = G, t [0, + ), (2) y (t, 1) y (t, 0) where

14 Boundary conditions on y: ( ) ( ) y+ (t, 0) y+ (t, 1) = G, t [0, + ), (2) y (t, 1) y (t, 0) where (i) y + R m and y R n m are defined by ( ) y+ y =, y

15 Boundary conditions on y: ( ) ( ) y+ (t, 0) y+ (t, 1) = G, t [0, + ), (2) y (t, 1) y (t, 0) where (i) y + R m and y R n m are defined by ( ) y+ y =, y (ii) the map G : R n R n vanishes at 0.

16 Notations For K M n,m (R), K := max{ Kx ; x R n, x = 1}.

17 Notations For K M n,m (R), If n = m, K := max{ Kx ; x R n, x = 1}. ρ 1 (K) := Inf { K 1 ; D n,+ }, where D n,+ denotes the set of n n real diagonal matrices with strictly positive diagonal elements.

18 Theorem 1.1 (JMC-G. Bastin-B. d Andréa-Novel (2008)). If ρ 1 (G (0)) < 1, then the equilibrium ȳ 0 of the quasi-linear hyperbolic system y t + A(y)y x = 0, with the above boundary conditions, is exponentially stable for the Sobolev H 2 -norm.

19 Estimate on the exponential decay rate For every ν (0, min{ Λ 1,..., Λ n } ln(ρ 1 (G (0)))), there exist ε > 0 and C > 0 such that, for every y 0 H 2 ((0, 1), R n ) satisfying y 0 H 2 ((0,1),R n ) < ε (and the usual compatibility conditions) the classical solution y to the Cauchy problem y t + A(y)y x = 0, y(0, x) = y 0 (x) + boundary conditions is defined on [0, + ) and satisfies y(t, ) H 2 ((0,1),R n ) Ce νt y 0 H 2 ((0,1),R n ), t [0, + ).

20 The Ta-tsien Li condition n R 2 (K) := Max { K ij ; i {1,..., n}}, j=1 ρ 2 (K) := Inf {R 2 ( K 1 ); D n,+ }. Theorem 1.2 (Ta-tsien Li (1994)). If ρ 2 (G (0)) < 1, then the equilibrium ȳ 0 of the quasi-linear hyperbolic system y t + A(y)y x = 0, with the above boundary conditions, is exponentially stable for the C 1 -norm.

21 The Ta-tsien Li approach

22 The Ta-tsien Li approach The Ta-tsien Li proof relies mainly on the use of direct estimates of the solutions and their derivatives along the characteristic curves.

23 The Ta-tsien Li approach The Ta-tsien Li proof relies mainly on the use of direct estimates of the solutions and their derivatives along the characteristic curves. Robustness to small perturbations: C. Prieur, J. Winkin and G. Bastin (2008); V. Dos Santos and C. Prieur (2008).

24 C 1 /H 2 -exponential stability 1. Open problem : Does there exists K such that one has exponential stability for the C 1 -norm but not for the H 2 -norm?

25 C 1 /H 2 -exponential stability 1. Open problem : Does there exists K such that one has exponential stability for the C 1 -norm but not for the H 2 -norm? 2. Open problem : Does there exists K such that one has exponential stability for the H 2 -norm but not for the C 1 -norm?

26 Comparison of ρ2 and ρ1

27 Comparison of ρ 2 and ρ 1 Proposition 1.3. For every K M n,n (R), ρ 1 (K) ρ 2 (K). (3)

28 Comparison of ρ 2 and ρ 1 Proposition 1.3. For every K M n,n (R), ρ 1 (K) ρ 2 (K). (3) Example where (3) is strict: for a > 0, let ( ) a a K a := M a a 2,2 (R). Then ρ 1 (K a ) = 2a < 2a = ρ 2 (K a ).

29 Comparison of ρ 2 and ρ 1 Proposition 1.3. For every K M n,n (R), ρ 1 (K) ρ 2 (K). (3) Example where (3) is strict: for a > 0, let ( ) a a K a := M a a 2,2 (R). Then ρ 1 (K a ) = 2a < 2a = ρ 2 (K a ). Open problem: Does ρ 1 (K) < 1 implies the exponential stability for the C 1 -norm?

30 Comparison with stability conditions for linear hyperbolic systems For simplicity we assume that Λ i are all positive and consider we consider the special case of linear hyperbolic systems where y t + Λy x = 0, y(t, 0) = Ky(t, 1), Λ := diag (Λ 1,..., Λ n ), with Λ i > 0, i {1,..., n}. Theorem 1.4. Exponential stability for the C 1 -norm is equivalent to the exponential stability in the H 2 -norm.

31 A Necessary and sufficient condition for exponential stability Notation: r i = 1 Λ i, i {1,..., n}. Theorem 1.5. ȳ is exponentially stable for the system ẏ + Λy x = 0, y(t, 0) = Ky(t, 1) if and only if there exists δ > 0 such that ( ) det (Id n (diag (e r1z,..., e rnz ))K) = 0, z C (R(z) δ).

32 An example Let us choose λ 1 := 1, λ 2 := 2 (hence r 1 = 1 and r 2 = 1/2 and ( ) a a K a :=, a R. a a Then ρ 1 (K) = 2 a. Hence ρ 1 (K a ) < 1 is equivalent to a ( 1/2, 1/2).

33 An example Let us choose λ 1 := 1, λ 2 := 2 (hence r 1 = 1 and r 2 = 1/2 and ( ) a a K a :=, a R. a a Then ρ 1 (K) = 2 a. Hence ρ 1 (K a ) < 1 is equivalent to a ( 1/2, 1/2). However exponential stability is equivalent to a ( 1, 1/2).

34 Robustness issues For a positive integer n, let Then ( y1 y 2 Λ 1 := ) := 4n 4n + 1, Λ 2 = 4n 2n + 1. ( ( sin 4nπ(t (x/λ1 )) ) ) sin ( 4nπ(t (x/λ 2 )) ) is a solution of y t + Λy x, y(t, 0) = K 1/2 y(t, 1) which does not tends to 0 as t +. Hence one does not have exponential stability. However lim n + Λ 1 = 1 and lim n + Λ 2 = 2.

35 Robustness issues For a positive integer n, let Then ( y1 y 2 Λ 1 := ) := 4n 4n + 1, Λ 2 = 4n 2n + 1. ( ( sin 4nπ(t (x/λ1 )) ) ) sin ( 4nπ(t (x/λ 2 )) ) is a solution of y t + Λy x, y(t, 0) = K 1/2 y(t, 1) which does not tends to 0 as t +. Hence one does not have exponential stability. However lim n + Λ 1 = 1 and lim n + Λ 2 = 2.The exponential stability is not robust with respect to Λ: small perturbations of Λ can destroy the exponential stability.

36 Robust exponential stability Notations: ρ 0 (K) := max{ρ(diag (e ιθ 1,..., e ιθn )K); (θ 1,..., θ n ) tr R n }, Theorem 1.6 (R. Silkowski (1993)). If the (r 1,..., r n ) are rationally independent, the linear system y t + Λ x = 0, y(t, 0) = Ky(t, 1) is exponentially stable if and only if ρ 0 (K) < 1.

37 Robust exponential stability Notations: ρ 0 (K) := max{ρ(diag (e ιθ 1,..., e ιθn )K); (θ 1,..., θ n ) tr R n }, Theorem 1.6 (R. Silkowski (1993)). If the (r 1,..., r n ) are rationally independent, the linear system y t + Λ x = 0, y(t, 0) = Ky(t, 1) is exponentially stable if and only if ρ 0 (K) < 1. Note that ρ 0 (K) depends continuously on K and that (r 1,..., r n ) are rationally independent is a generic condition. Therefore, if one wants to have a natural robustness property with respect to the r i s, the condition for exponential stability is ρ 0 (K) < 1.

38 Robust exponential stability Notations: ρ 0 (K) := max{ρ(diag (e ιθ 1,..., e ιθn )K); (θ 1,..., θ n ) tr R n }, Theorem 1.6 (R. Silkowski (1993)). If the (r 1,..., r n ) are rationally independent, the linear system y t + Λ x = 0, y(t, 0) = Ky(t, 1) is exponentially stable if and only if ρ 0 (K) < 1. Note that ρ 0 (K) depends continuously on K and that (r 1,..., r n ) are rationally independent is a generic condition. Therefore, if one wants to have a natural robustness property with respect to the r i s, the condition for exponential stability is ρ 0 (K) < 1. This condition does not depend on the Λ i s!

39 Comparison of ρ0 and ρ1

40 Comparison of ρ 0 and ρ 1 Proposition 1.7 (JMC-G. Bastin-B. d Andréa-Novel (2008)). For every n N and for every K M n,n (R), ρ 0 (K) ρ 1 (K). For every n {1, 2, 3, 4, 5} and for every K M n,n (R), ρ 0 (K) = ρ 1 (K). For every n N \ {1, 2, 3, 4, 5}, there exists K M n,n (R) such that ρ 0 (K) < ρ 1 (K).

41 Comparison of ρ 0 and ρ 1 Proposition 1.7 (JMC-G. Bastin-B. d Andréa-Novel (2008)). For every n N and for every K M n,n (R), ρ 0 (K) ρ 1 (K). For every n {1, 2, 3, 4, 5} and for every K M n,n (R), ρ 0 (K) = ρ 1 (K). For every n N \ {1, 2, 3, 4, 5}, there exists K M n,n (R) such that ρ 0 (K) < ρ 1 (K). Open problem : Is ρ 0 (G (0)) < 1 a sufficient condition for exponential stability (for the H 2 -norm) in the nonlinear case?

42 Proof of the exponential stability if A is constant and G is linear Main tool: a Lyapunov approach. A(y) = Λ, G(y) = Ky. For simplicity, all the Λ i s are positive. Lyapunov function candidate: V (y) := 1 0 y tr Qye µx dx, Q is positive symmetric.

43 Proof of the exponential stability if A is constant and G is linear Main tool: a Lyapunov approach. A(y) = Λ, G(y) = Ky. For simplicity, all the Λ i s are positive. Lyapunov function candidate: V (y) := 1 0 y tr Qye µx dx, Q is positive symmetric. with V = = µ (y tr x ΛQy + y tr Qy x )Λe µx dx y tr ΛQy e µx dx B, B := [y tr ΛQye µx ] x=1 x=0 = y(1) tr (ΛQe µ K tr ΛQK)y(1)

44 Let D D n,+ be such that DKD 1 < 1 and let ξ := Dy(1). We take Q = D 2 Λ 1. Then B = e µ ξ 2 DKD 1 ξ 2. Therefore it suffices to take µ > 0 small enough. Remark 1.8. Introduction of µ:

45 Let D D n,+ be such that DKD 1 < 1 and let ξ := Dy(1). We take Q = D 2 Λ 1. Then B = e µ ξ 2 DKD 1 ξ 2. Therefore it suffices to take µ > 0 small enough. Remark 1.8. Introduction of µ: JMC (1998) for the stabilization of the Euler equations.

46 Let D D n,+ be such that DKD 1 < 1 and let ξ := Dy(1). We take Q = D 2 Λ 1. Then B = e µ ξ 2 DKD 1 ξ 2. Therefore it suffices to take µ > 0 small enough. Remark 1.8. Introduction of µ: JMC (1998) for the stabilization of the Euler equations. Cheng-Zhong Xu and Gauthier Sallet (2002) for symmetric linear hyperbolic systems.

47 New difficulties if A(y) depends on y We try with the same V : with V = = µ ( y tr x A(y) tr Qy + y tr QA(y)y x ) e µx dx y tr A(y)Qye µx dx B + N 1 + N 2 N 1 := 1 0 N 2 := y tr (QA(y) A(y)Q)y x e µx dx, 1 0 y tr( A (y)y x ) trqye µx dx

48 Solution for N 1 Take Q depending on y such that A(y)Q(y) = Q(y)A(y), Q(0) = D 2 F (0) 1. (This is possible since the eigenvalues of F(0) are distinct.) Now with N 2 := V = µ y tr A(y)Q(y)ye µx dx B + N 2 y tr( A (y)y x Q(y) + A(y)Q (y)y x ) trye µx dx

49 Solution for N 1 Take Q depending on y such that A(y)Q(y) = Q(y)A(y), Q(0) = D 2 F (0) 1. (This is possible since the eigenvalues of F(0) are distinct.) Now with N 2 := V = µ 1 What to do with N 2? y tr A(y)Q(y)ye µx dx B + N 2 y tr( A (y)y x Q(y) + A(y)Q (y)y x ) trye µx dx

50 Solution for N 2 New Lyapunov function: with V (y) = V 1 (y) + V 2 (y) + V 3 (y) V 1 (y) = V 2 (y) = V 3 (y) = y tr Q(y)y e µx dx, y tr x R(y)y x e µx dx, y tr xxs(y)y xx e µx dx, where µ > 0, Q(y), R(y) and S(y) are symmetric positive definite matrices.

51 Choice of Q, R and S Commutations: A(y)Q(y) Q(y)A(y) = 0, A(y)R(y) R(y)A(y) = 0, A(y)S(y) S(y)A(y) = 0,

52 Choice of Q, R and S Commutations: A(y)Q(y) Q(y)A(y) = 0, A(y)R(y) R(y)A(y) = 0, A(y)S(y) S(y)A(y) = 0, Q(0) = D 2 A(0) 1, R(0) = D 2 A(0), S(0) = D 2 A(0) 3.

53 Estimates on V Lemma 1.9. If µ > 0 is small enough, there exist positive real constants α, β, δ such that, for every y : [0, 1] R n such that y C 0 ([0,1]) + y x C 0 ([0,1]) δ, we have 1 β 1 0 ( y 2 + y x 2 + y xx 2 )dx V (y) β V αv. 1 0 ( y 2 + y x 2 + y xx 2 )dx,...

54 La Sambre (G. Bastin, L. Moens,...)

55

56

57 Z 0 u 0 0 H(t,x) V(t,x) u L L x

58 The Saint-Venant equations The index i is for the i-th reach. Conservation of mass: H it + (H i V i ) x = 0,

59 The Saint-Venant equations The index i is for the i-th reach. Conservation of mass: H it + (H i V i ) x = 0, Conservation of momentum: V it + (gh i + V i 2 ) = 0. 2 x

60 The Saint-Venant equations The index i is for the i-th reach. Conservation of mass: H it + (H i V i ) x = 0, Conservation of momentum: V it + (gh i + V i 2 ) 2 Flow rate: Q i = H i V i. x = 0.

61 Barré de Saint-Venant (Adhémar-Jean-Claude) Théorie du mouvement non permanent des eaux, avec applications aux crues des rivières et à l introduction des marées dans leur lit, C. R. Acad. Sci. Paris Sér. I Math., vol. 53 (1871), pp

62 Barré de Saint-Venant (Adhémar-Jean-Claude) Théorie du mouvement non permanent des eaux, avec applications aux crues des rivières et à l introduction des marées dans leur lit, C. R. Acad. Sci. Paris Sér. I Math., vol. 53 (1871), pp

63 odelling Boundary conditions ydraulic gates Modelling Hydraulic gates ndeflow (sluice) Underflow (sluice) Undeflow (sluice) Overflow (spillway) Overflow (spillway) Overflow (spillway) u u u u u u u u 34

64 La Sambre: Hydraulic gates

65 Closed loop versus open loop 56

66 Work in progress: La Meuse

67 Complements Slope and friction: G. Bastin, JMC and B. d Andréa-Novel (2008; this workshop). Robust H -control design techniques: X. Litrico and D. Georges (2001), X. Litrico (2001). Spectral methods: X. Litrico and V. Fromion (2006). Observer for water level: G. Besançon, J.F. Dulhoste and D. Georges (2001). Integral action: V. Dos-Santos, G. Bastin, JMC and B. d andréa-novel (2007). Book: Automatique pour la gestion des ressources en eau D. Georges and X. Litrico (2002).

68 Stabilization: Dissipative boundary conditions The equations Main result Comparaison with prior results Proof of the exponential stability Application to the control of open channels Controllability of hyperbolic systems The control problem Controllability theorem A Water-tank control system Lie brackets The return method Controllability of a toy model

69 The control problem Let T > 0. Given y 0 : [0, 1] R n and y 1 : [0, 1] R n. Does there exist u + : [0, T ] R m, u : [0, T ] R m such that the solution of the Cauchy problem y t +A(y)y x = 0, y + (t, 0) = u + (t), y (t, 0) = u (t), y(0, x) = y 0 (x), satisfies y(t, x) = y T (x)?

70 The control problem Let T > 0. Given y 0 : [0, 1] R n and y 1 : [0, 1] R n. Does there exist u + : [0, T ] R m, u : [0, T ] R m such that the solution of the Cauchy problem y t +A(y)y x = 0, y + (t, 0) = u + (t), y (t, 0) = u (t), y(0, x) = y 0 (x), satisfies y(t, x) = y T (x)? Local controllability: y 0 and y T are small. Remark 2.1. The control is on both sides. The case where the control is only on one side can also be considered.

71 Controllability theorem Theorem 2.2 (Ta-tsien Li and Bopeng Rao (2003)). (Local ( controllability { for the C 1 -norm) }) T > max 1 Λ,..., Λ m, Λ m+1,..., 1 Λ n

72 Controllability theorem Theorem 2.2 (Ta-tsien Li and Bopeng Rao (2003)). (Local ( controllability { for the C 1 -norm) }) T > max 1 Λ,..., Λ m, Λ m+1,..., 1 Λ n Remark 2.3. { } If T > max 1 Λ,..., Λ m, Λ m+1,..., 1 Λ n, the linearized control system at (ȳ, ū) = (0, 0) is controllable. One of the main ingredient of the proof: consider y t + A(y)y x = 0 as an evolution equation in x: A(y) 1 y t + y x = 0. Generalization: A(t, x, y): Zhiqiang Wang (2007). Applications to channels: M. Gugat and G. Leugering (2008). For the control on one side only, see Ta-tsien Li and Bopeng Rao (2002).

73 Stabilization: Dissipative boundary conditions The equations Main result Comparaison with prior results Proof of the exponential stability Application to the control of open channels Controllability of hyperbolic systems The control problem Controllability theorem A Water-tank control system Lie brackets The return method Controllability of a toy model

74 D partement de Math matique Orsay, France The physical system 26 April 2001 L H v D x

75 The model: Saint-Venant equations H t + (Hv) x = 0, t [0, T ], x [0, L], v t + (gh + v 2 ) = u (t), t [0, T ], x [0, L], 2 x v(t, 0) = v(t, L) = 0, t [0, T ], ds (t) = u (t), t [0, T ], dt dd (t) = s (t), t [0, T ]. dt u (t) is the horizontal acceleration of the tank in the absolute referential, g is the gravity constant, s is the horizontal velocity of the tank, D is the horizontal displacement of the tank.

76 State space d dt L H x (t, 0) = H x (t, L) 0 H (t, x) dx = 0, (= u(t)/g). Definition 3.1. The state space (denoted Y) is the set of Y = (H, v, s, D) C 1 ([0, L]) C 1 ([0, L]) R R satisfying v(0) = v(l) = 0, H x (0) = H x (L), L 0 H(x)dx = LH e.

77 Main result Theorem 3.2 (JMC (2002)). For T > 0 large enough the water-tank control system is locally controllable in time T around (Y e, u e ) := ((H e, 0, 0, 0), 0).

78 Main result Theorem 3.2 (JMC (2002)). For T > 0 large enough the water-tank control system is locally controllable in time T around (Y e, u e ) := ((H e, 0, 0, 0), 0). Prior work: F. Dubois, N. Petit and P. Rouchon (1999): Steady state controllability of the linearized control system. Linearized systems for other water tank problems: N. Petit and P. Rouchon (2002); see also Rouchon s talk.

79 The linearized control system Without loss of generality L = H e = g = 1. The linearized control system around (Y e, u e ) := ((1, 0, 0, 0), 0) is h t + v x = 0, t [0, T ], x [0, L], v t + h x = u (t), t [0, T ], x [0, L], v(t, 0) = v(t, L) = 0, t [0, T ], ds (t) = u (t), t [0, T ], dt dd (t) = s (t), t [0, T ]. dt

80 The linearized control system is not controllable For a function w : [0, 1] R, we denote by w ev the even part of w and by w od the odd part of w: w ev (x) := 1 2 (w(x) + w(1 x)), w od (x) := 1 (w(x) w(1 x)). 2 Σ 1 + vx ev = 0, vt ev + hx od = u (t), v ev (t, 0) = v ev (t, 1) = 0, ds dt h od t (t) = u (t), dd dt (t) = s (t),

81 The linearized control system is not controllable For a function w : [0, 1] R, we denote by w ev the even part of w and by w od the odd part of w: w ev (x) := 1 2 (w(x) + w(1 x)), w od (x) := 1 (w(x) w(1 x)). 2 Σ 1 Σ 2 + vx ev = 0, vt ev + hx od = u (t), v ev (t, 0) = v ev (t, 1) = 0, ds dt h od t ht ev + vx od = 0, vt od + hx ev = 0, v od (t, 0) = v od (t, 1) = 0, (t) = u (t), dd dt (t) = s (t),

82 What to do if linearized control system is not controllable? We consider the control system ẋ = f (x, u) where the state is x R n and the control is u R m. Let us assume that f (x e, u e ) = 0. The linearized control system at (x e, u e ) is the linear control system ẋ = Ax + Bu with A = f x (x e, u e ), B = f u (x e, u e ). If the linearized control system ẋ = Ax + Bu is controllable, then ẋ = f (x, u) is locally controllable at (x e, u e ). Question: What to do if ẋ = Ax + Bu is not controllable?

83 What to do if linearized control system is not controllable? We consider the control system ẋ = f (x, u) where the state is x R n and the control is u R m. Let us assume that f (x e, u e ) = 0. The linearized control system at (x e, u e ) is the linear control system ẋ = Ax + Bu with A = f x (x e, u e ), B = f u (x e, u e ). If the linearized control system ẋ = Ax + Bu is controllable, then ẋ = f (x, u) is locally controllable at (x e, u e ). Question: What to do if ẋ = Ax + Bu is not controllable? In finite dimension: one uses iterated Lie brackets.

84 Lis brackets and iterated Lie brackets Definition 3.3 (Lie brackets). [X, Y ](x) := Y (x)x (x) X (x)y (x).

85 Lis brackets and iterated Lie brackets Definition 3.3 (Lie brackets). [X, Y ](x) := Y (x)x (x) X (x)y (x). In other words, the components of [X, Y ](x) are [X, Y ] j (x) = n k=1 X k (x) Y j x k (x) Y k (x) X j x k (x), j {1,..., n}.

86 Lis brackets and iterated Lie brackets Definition 3.3 (Lie brackets). [X, Y ](x) := Y (x)x (x) X (x)y (x). In other words, the components of [X, Y ](x) are [X, Y ] j (x) = n k=1 X k (x) Y j x k (x) Y k (x) X j x k (x), j {1,..., n}. Iterated Lie brackets: [X, [X, Y ]], [[Y, X ], [X, [X, Y ]]] etc.

87 Lie bracket for ẋ = u 1 f 1 (x) + u 2 f 2 (x) a

88 Lie bracket for ẋ = u 1 f 1 (x) + u 2 f 2 (x) x(ε) a (u 1, u 2 ) = (η 1, 0)

89 Lie bracket for ẋ = u 1 f 1 (x) + u 2 f 2 (x) x(2ε) (u 1, u 2 ) = (0, η 2 ) x(ε) a (u 1, u 2 ) = (η 1, 0)

90 Lie bracket for ẋ = u 1 f 1 (x) + u 2 f 2 (x) x(2ε) x(3ε) (u 1, u 2 ) = ( η 1, 0) (u 1, u 2 ) = (0, η 2 ) x(ε) a (u 1, u 2 ) = (η 1, 0)

91 Lie bracket for ẋ = u 1 f 1 (x) + u 2 f 2 (x) x(2ε) x(3ε) (u 1, u 2 ) = ( η 1, 0) (u 1, u 2 ) = (0, η 2 ) (u 1, u 2 ) = (0, η 2 ) x(ε) x(4ε) a (u 1, u 2 ) = (η 1, 0)

92 Lie bracket for ẋ = u 1 f 1 (x) + u 2 f 2 (x) x(2ε) x(3ε) (u 1, u 2 ) = ( η 1, 0) (u 1, u 2 ) = (0, η 2 ) (u 1, u 2 ) = (0, η 2 ) x(ε) x(4ε) a + η 1 η 2 ε 2 [f 1, f 2 ](a) a (u 1, u 2 ) = (η 1, 0)

93 Lie bracket for ẋ = f 0 (x) + uf 1 (x), f 0 (a) = 0 a

94 Lie bracket for ẋ = f 0 (x) + uf 1 (x), f 0 (a) = 0 x(ε) a u = η

95 Lie bracket for ẋ = f 0 (x) + uf 1 (x), f 0 (a) = 0 u = η x(ε) x(2ε) a u = η

96 Lie bracket for ẋ = f 0 (x) + uf 1 (x), f 0 (a) = 0 u = η x(ε) x(2ε) a + ηε 2 [f 0, f 1 ](a) a u = η

97 The baby stroller system ẋ 1 = u 1, ẋ 2 = u 2, ẋ 3 = u 1 x 2 u 2 x 1, n = 3, m = 2. The linearized control system at (0, 0) R 3 R 2 is ẋ 1 = u 1, ẋ 2 = u 2, ẋ 3 = 0, which is not controllable. Our system can be written as ẋ = u 1 f 1 (x) + u 2 f 2 (x), with f 1 (x) = (1, 0, x 2 ) tr, f 2 (x) = (0, 1, x 1 ) tr. One has [f 1, f 2 ](x) = (0, 0, 2) tr. Hence f 1 (0), f 2 (0) and [f 1, f 2 ](0) span all of R 3. This implies the local controllability of the baby stroller at (0, 0) R 3 R 2.

98 Problems of the Lie brackets for PDE control systems Consider the simplest PDE control system Σ : y t + y x = 0, x [0, L], y(t, 0) = u(t)

99 Let us consider, for ε > 0, the control defined on [0, 2ε] by u(t) = 1 for t (0, ε), u(t) = 1 for t (ε, 2ε). Let y : (0, 2ε) (0, L) R be the solution of the Cauchy problem y t + y x = 0, t (0, 2ε), x (0, L), y(t, 0) = u(t), t (0, 2ε), y(0, x) = 0, x (0, L). Then one readily gets, if 2ε L, y(2ε, x) = 1, x (0, ε), y(2ε, x) = 1, x (ε, 2ε), y(2ε, x) = 0, x (2ε, L).

100 Problems of the Lie brackets for PDE control systems (continued) For every φ H 2 (0, L), one gets after suitable computations 1 lim ε 0 + ε 2 L 0 φ(x)(y(2ε, x) y(0, x))dx = φ (0). So, in some sense, we could say that [f 0, f 1 ] = δ 0. Unfortunately it is not clear how to use this derivative of a Dirac mass at 0.

101 The return method x T t

102 The return method x x(t) T t

103 The return method x x(t) x 1 x 0 T t B 0 B 1

104 The return method x x(t) x(t) ε η x 1 x 0 T t B 0 B 1

105 The return method: An example in finite dimension We go back to the baby stroller system BS : ẋ 1 = u 1, ẋ 2 = u 2, ẋ 3 = u 1 x 2 u 2 x 1. For every ū : [0, T ] R 2 such that, for every t in [0, T ], ū(t t) = ū(t), every solution x : [0, T ] R 3 of x 1 = ū 1, x 2 = ū 2, x 3 = ū 1 x 2 ū 2 x 1 satisfies x(0) = x(t ). The linearized control system around ( x, ū) is L : ẋ 1 = u 1, ẋ 2 = u 2, ẋ 3 = ū 1 x 2 ū 2 x 1 + u 1 x 2 u 2 x 1, which is controllable if (and only if) ū 0....

106 The return method: An example in finite dimension We go back to the baby stroller system BS : ẋ 1 = u 1, ẋ 2 = u 2, ẋ 3 = u 1 x 2 u 2 x 1. For every ū : [0, T ] R 2 such that, for every t in [0, T ], ū(t t) = ū(t), every solution x : [0, T ] R 3 of x 1 = ū 1, x 2 = ū 2, x 3 = ū 1 x 2 ū 2 x 1 satisfies x(0) = x(t ). The linearized control system around ( x, ū) is L : ẋ 1 = u 1, ẋ 2 = u 2, ẋ 3 = ū 1 x 2 ū 2 x 1 + u 1 x 2 u 2 x 1, which is controllable if (and only if) ū We have got the controllability of the baby stroller system without using Lie brackets. We have only used controllability results for linear control systems.

107 Return method: references Stabilization of driftless systems in finite dimension: JMC (1992). Euler equations of incompressible fluids: JMC (1993,1996), O. Glass (1997,2000). Control of driftless systems in finite dimension: E.D. Sontag (1995). Navier-Stokes equations JMC (1996), JMC and A. Fursikov (1996), A. Fursikov and O. Imanuvilov (1999), S. Guerrero, O. Imanuvilov and J.-P. Puel (2006). Burgers equation Th. Horsin (1998), M. Chapouly (2006). Saint-Venant equations: JMC (2002) Vlasov Poisson: O. Glass (2003). Isentropic Euler equations: O. Glass (2006). Schrödinger equation: K. Beauchard (2005), K. Beauchard and JMC (2006). Korteweg de Vries equation: M. Chapouly (2008).

108 Return method: references (continued) See also JMC, Control and nonlinearity, Mathematical Surveys and Monographs, 136, 2007, 427 pp.

109 A toy model for the water tank control system If h := H 1, ht od + vx ev = (h ev v ev + h od v od ) x, Σ 1 vt ev + hx od = u (t) (v ev v od ) x, v ev (t, 0) = v ev (t, 1) = 0, ds dd dt (t) = u (t), dt (t) = s (t),

110 A toy model for the water tank control system If h := H 1, ht od + vx ev = (h ev v ev + h od v od ) x, Σ 1 vt ev + hx od = u (t) (v ev v od ) x, v ev (t, 0) = v ev (t, 1) = 0, ds dd dt (t) = u (t), dt Σ 2 ht ev + vx od = (h ev v od + h od v ev ) x, vt od (t, x) + hx ev = 1 2 ((v ev ) 2 + (v od ) 2 ) x, v od (t, 0) = v od (t, 1) = 0, (t) = s (t),

111 Toy model (continued) T T 1 { ẋ1 = x 2, ẋ 2 = x 1 + u, ṡ = u, Ḋ = s T 2 { ẋ3 = x 4, ẋ 4 = x 3 + 2x 1 x 2. (4) where the state is x = (x 1, x 2, x 3, x 4, s, D) tr R 6 and the control is u R.

112 Toy model (continued) T T 1 { ẋ1 = x 2, ẋ 2 = x 1 + u, ṡ = u, Ḋ = s T 2 { ẋ3 = x 4, ẋ 4 = x 3 + 2x 1 x 2. (4) where the state is x = (x 1, x 2, x 3, x 4, s, D) tr R 6 and the control is u R. The linearized control system of T around (x e, u e ) := (0, 0) is ẋ 1 = ẋ 1 = x 2, ẋ 2 = x 1 + u, ṡ = u, Ḋ = s, ẋ 3 = x 4, ẋ 4 = x 3, which is not controllable.

113 Controllability of the toy model If x(0) = 0, x 3 (T ) = T 0 x 4 (T ) = x 2 1 (T ) x 2 1 (t) cos(t t)dt, T 0 x 2 1 (t) sin(t t)dt.

114 Controllability of the toy model If x(0) = 0, x 3 (T ) = T 0 x 4 (T ) = x 2 1 (T ) x 2 1 (t) cos(t t)dt, T Hence T is not controllable in time T π. 0 x 2 1 (t) sin(t t)dt.

115 Controllability of the toy model If x(0) = 0, x 3 (T ) = T 0 x 4 (T ) = x 2 1 (T ) x 2 1 (t) cos(t t)dt, T 0 x 2 1 (t) sin(t t)dt. Hence T is not controllable in time T π. Using explicit explicit computations one can show that T is (locally) controllable in time T > π. Remark 3.4. For linear system in finite dimension, the controllability in large time implies the controllability in small time. This is no longer for linear PDE. This is also no longer true for nonlinear systems in finite dimension.

116 How to recover the large time local controllability of T We forget about s and D for simplicity. Let γ R and define ((x γ 1, x γ 2, x γ 3, x γ 4 )tr, u γ ) := ((γ, 0, 0, 0) tr, γ). Then ((x γ 1, x γ 2, x γ 3, x γ 4 )tr, u γ ) is an equilibrium of T.

117 How to recover the large time local controllability of T We forget about s and D for simplicity. Let γ R and define ((x γ 1, x γ 2, x γ 3, x γ 4 )tr, u γ ) := ((γ, 0, 0, 0) tr, γ). Then ((x γ 1, x γ 2, x γ 3, x γ 4 )tr, u γ ) is an equilibrium of T. The linearized control system of T at this equilibrium is ẋ 1 = ẋ 1 = x 2, ẋ 2 = x 1 + u, ẋ 3 = x 4, ẋ 4 = x 3 + 2γx 2, which is controllable if (and only if) γ 0.

118 How to recover the large time local controllability of T x t

119 How to recover the large time local controllability of T x x γ (t) t

120 How to recover the large time local controllability of T x x γ (t) B 1 B 2 t

121 How to recover the large time local controllability of T x x γ (t) B 1 B 2 t

122 How to recover the large time local controllability of T x x γ (t) B 1 B 2 t

123 How to recover the large time local controllability of T x x γ (t) B 1 B 2 B 0 a t

124 How to recover the large time local controllability of T x x γ (t) B 1 B 2 B 0 a t

125 How to recover the large time local controllability of T x x γ (t) B 1 B 2 B 0 a b B 3 T t

126 Open problems

127 Open problems Optimal time for local controllability.

128 Open problems Optimal time for local controllability. Stabilizing feedbacks for the water-tank control system.