Controllability of nonlinear systems

Transcription

1 Controllability of nonlinear systems Jean-Michel Coron Laboratory J.-L. Lions, University Pierre et Marie Curie (Paris 6) BCAM OPTPDE summer school, Bilbao, July

2 The linear test We consider the control system ẏ = f(y,u) where the state is y R n and the control is u R m. Let us assume that f(y e,u e ) = 0. The linearized control system at (y e,u e ) is the linear control system ẏ = Ay +Bu with (1) A := f y (y e,u e ), B := f u (y e,u e ). If the linearized control system ẏ = Ay + Bu is controllable, then ẏ = f(y,u) is small-time locally controllable at (y e,u e ).

3 Application to the baby stroller Let us recall that the baby stroller control system is (1) ẏ 1 = u 1 cosy 3, ẏ 2 = u 1 siny 3,ẏ 3 = u 2, n = 3, m = 2. The linearized control system at (0,0) R 3 R 2 is (2) ẏ 1 = u 1, ẏ 2 = 0, ẏ 3 = u 2, which is clearly not controllable. The linearized control system gives no information on the small-time local controllability at (0,0) R 3 R 2 of the baby stroller.

4 What to do if linearized control system is not controllable? Question: What to do if (1) ẏ = f y (y e,u e )y + f u (y e,u e )u is not controllable? In finite dimension: One uses iterated Lie brackets.

5 Lie brackets and iterated Lie brackets Definition (Lie brackets) (1) [X,Y](y) := Y (y)x(y) X (y)y(y). Iterated Lie brackets: [X,[X,Y]], [[Y,X],[X,[X,Y ]]] etc. Why Lie brackets are natural objects for controllability issues? For simplicity, from now on we assume that f(y,u) = f 0 (y)+ m u i f i (y). i=1 Drift: f 0. Driftless control systems: f 0 = 0.

6 Lie bracket for ẏ = u 1 f 1 (y)+u 2 f 2 (y) a

7 Lie bracket for ẏ = u 1 f 1 (y)+u 2 f 2 (y) y(ε) a (u 1,u 2 ) = (η 1,0)

8 Lie bracket for ẏ = u 1 f 1 (y)+u 2 f 2 (y) y(2ε) (u 1,u 2 ) = (0,η 2 ) y(ε) a (u 1,u 2 ) = (η 1,0)

9 Lie bracket for ẏ = u 1 f 1 (y)+u 2 f 2 (y) y(3ε) (u 1,u 2 ) = ( η 1,0) y(2ε) (u 1,u 2 ) = (0,η 2 ) y(ε) a (u 1,u 2 ) = (η 1,0)

10 Lie bracket for ẏ = u 1 f 1 (y)+u 2 f 2 (y) y(3ε) (u 1,u 2 ) = ( η 1,0) y(2ε) (u 1,u 2 ) = (0,η 2 ) (u 1,u 2 ) = (0, η 2 ) y(ε) y(4ε) a (u 1,u 2 ) = (η 1,0)

11 Lie bracket for ẏ = u 1 f 1 (y)+u 2 f 2 (y) y(3ε) (u 1,u 2 ) = ( η 1,0) y(2ε) (u 1,u 2 ) = (0,η 2 ) (u 1,u 2 ) = (0, η 2 ) y(ε) y(4ε) a+η 1 η 2 ε 2 [f 1,f 2 ](a) (ε 0 + ) a (u 1,u 2 ) = (η 1,0)

12 Controllability of driftless control systems: Local controllability Theorem (P. Rashevski (1938), W.-L. Chow (1939)) Let O be a nonempty open subset of R n and let y e O. Let us assume that, for some f 1,...,f m : O R n, (1) f(y,u) = m u i f i (y), (y,u) O R m. i=1 Let us also assume that { h(ye ); h Lie {f 1,...,f m } } = R n. (2) Then the control system ẏ = f(y, u) is small-time locally controllable at (y e,0) R n R m.

13 The baby stroller system: Controllability (1) ẏ 1 = u 1 cosy 3, ẏ 2 = u 1 siny 3,ẏ 3 = u 2, n = 3, m = 2. This system can be written as ẏ = u 1 f 1 (y)+u 2 f 2 (y), with (2) f 1 (y) = (cosy 3,siny 3,0) tr, f 2 (y) = (0,0,1) tr. One has (3) [f 1,f 2 ](y) = (siny 3, cosy 3,0) tr. Hence f 1 (0), f 2 (0) and [f 1,f 2 ](0) all together span all of R 3. This implies the small-time local controllability of the baby stroller at (0,0) R 3 R 2.

14 Controllability of driftless control systems: Global controllability Theorem (P. Rashevski (1938), W.-L. Chow (1939)) Let O be a connected nonempty open subset of R n. Let us assume that, for some f 1,...,f m : O R n, f(y,u) = m u i f i (y), (y,u) O R m. i=1 Let us also assume that { h(y); h Lie {f1,...,f m } } = R n, y O. Then, for every (y 0,y 1 ) O O and for every T > 0, there exists u belonging to L ((0,T);R m ) such that the solution of the Cauchy problem ẏ = f(y,u(t)), y(0) = y 0, satisfies y(t) O, t [0,T] and y(t) = y 1.

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19 Drift and the Lie algebra rank condition We consider the control affine system ẏ = f 0 (y)+ m i=1 u if i (y) with f 0 (0) = 0. One says that this control system satisfies the Lie algebra rank condition at 0 R n if { (1) h(0); h Lie {f0,f 1,...,f m } } = R n. One has the following theorem. Theorem (R. Hermann (1963) and T. Nagano (1966)) If the f i s are analytic in a neighborhood of 0 R n and if the control system ẏ = f 0 (y)+ m i=1 f i(y) is small-time locally controllable at (0,0) R n R m, then this control system satisfies the Lie algebra rank condition at 0 R n.

20 Lie bracket for ẏ = f 0 (y)+uf 1 (y), with f 0 (a) = 0 a

21 Lie bracket for ẏ = f 0 (y)+uf 1 (y), with f 0 (a) = 0 y(ε) a u = η

22 Lie bracket for ẏ = f 0 (y)+uf 1 (y), with f 0 (a) = 0 u = η y(ε) y(2ε) a u = η

23 Lie bracket for ẏ = f 0 (y)+uf 1 (y), with f 0 (a) = 0 u = η y(ε) y(2ε) a+ηε 2 [f 0,f 1 ](a) ε 0 + a u = η

24 The Kalman rank condition and the Lie algebra rank condition Let us write the linear control system ẏ = Ay +Bu as ẏ = f 0 (y)+ m i=1 u if i (y) with (1) f 0 (y) = Ay, f i (y) = B i, B i R n, (B 1,...,B m ) = B. The Kalman rank condition is equivalent to the Lie algebra rank condition at 0 R n. Hence the Lie algebra rank condition is sufficient for two important cases, namely Linear systems and driftless control systems.

25 With a drift term: Not all the iterated Lie brackets are good We take n = 2 and m = 1 and consider the control system (1) ẏ 1 = y 2 2, ẏ 2 = u, where the state is y := (y 1,y 2 ) tr R 2 and the control is u R. This control system can be written as ẏ = f 0 (y)+uf 1 (y) with (2) f 0 (y) = (y 2 2,0) tr,f 1 (y) = (0,1) tr. One has [f 1,[f 1,f 0 ]] = (2,0) tr and therefore f 1 (0) and [f 1,[f 1,f 0 ]](0) span all of R 2. However the control system (1) is clearly not small-time locally controllable at (0,0) R 2 R.

26 References for sufficient or necessary conditions for small-time local controllability when there is a drift term A. Agrachev (1991), A. Agrachev and R. Gamkrelidze (1993), R. M. Bianchini and Stefani (1986), H. Frankowska (1987), M. Kawski (1990), H. Sussmann (1983, 1987), A. Tret yak (1990)....

27 Controllability of control systems modeled by linear PDE There are lot of powerful tools to study the controllability of linear control systems in infinite dimension. The most popular ones are based on the duality between observability and controllability (related to the J.-L. Lions Hilbert uniqueness method). This leads to try to prove observability inequalities. There are many methods to prove this observability inequalities. For example: Ingham s inequalities and harmonic analysis: D. Russell (1967), Multipliers method: Lop Fat Ho (1986), J.-L. Lions (1988), Microlocal analysis: C. Bardos-G. Lebeau-J. Rauch (1992), Carleman s inequalities: A. Fursikov, O. Imanuvilov, G. Lebeau and L. Robbiano ( ). However there are still plenty of open problems.

28 The linear test Of course when one wants to study the local controllability around an equilibrium of a control system in infinite dimension, the first step is to again study the controllability of the linearized control system. If this linearized control system is controllable, one can usually deduce the local controllability of the nonlinear control system. However this might be sometimes difficult due to some loss of derivatives issues. One needs to use suitable iterative schemes. Remark If the nonlinearity is not too big, one can get a global controllability result (E. Zuazua (1988), I. Lasiecka-R. Triggiani (1991) for semilinear wave equations, O. Imanuvilov (1995), A. Fursikov-O. Imanuvilov (1996), E. Fernández-Cara-E. Zuazua (2000) for semilinear heat equations).

29 The Euler control system Γ 0 Ω y(t, x) x Ω R n

30 Notations Let ϕ : Ω R n, x = (x 1,x 2,...,x n ) (ϕ 1 (x),ϕ 2 (x),...ϕ n (x)), we define div ϕ : Ω R and (ϕ )ϕ : Ω R n by (1) (2) div ϕ := [(ϕ )ϕ] k = If n = 2, one defines curl ϕ : Ω R by (3) n i=1 ϕ i x i, n i=1 ϕ i ϕ k x i curl ϕ := y 2 x 1 y 1 x 2. Finally, for π : Ω R, one defines π : Ω R n by ( π π :=, π,..., π ) (4). x 1 x 2 x n Similarly notations if ϕ or π depends also on time t (one fixes the time...).

31 Controllability problem We denote by ν : Ω R n the outward unit normal vector field to Ω. Let T > 0. Let y 0, y 1 : Ω R n be such that (1) div y 0 = div y 1 = 0, y 0 ν = y 1 ν = 0 on Ω\Γ 0. Does there exist y : [0,T] Ω R n and p : [0,T] Ω R such that (2) (3) (4) y t +(y )y + p = 0, div y = 0, y ν = 0 on [0,T] ( Ω\Γ 0 ), y(0, ) = y 0, y(t, ) = y 1? For the control, many choices are in fact possible. For example, for n = 2, one can take 1 y ν on Γ 0 with Γ 0 y ν = 0, 2 curl y at the points of [0,T] Γ 0 where y ν < 0.

32 A case without controllability Γ 0 Ω Γ 1 R 2 Ω Ω

33 Proof of the noncontrollability Let us give it only for n = 2. Let γ 0 be a Jordan curve in Ω. Let, for t [0, T], γ(t) be the Jordan curve obtained, at time t [0, T], from the points of the fluids which, at time 0, were on γ 0. The Kelvin law tells us that, if γ(t) does not intersect Γ 0, y(t, ) (1) ds = y(0, ) ds, t [0,T], γ(t) γ 0 We take γ 0 := Γ 1. Then γ(t) = Γ 1 for every t [0,T]. Hence, if y 1 (2) ds y 0 ds, Γ 1 Γ 1 one cannot steer the control system from y 0 to y 1. More generally, for every n {2,3}, if Γ 0 does not intersect every connected component of the boundary Ω of Ω, the Euler control system is not controllable. This is the only obstruction to the controllability of the Euler control system.

34 Controllability of the Euler control system Theorem (JMC for n = 2 (1996), O. Glass for n = 3 (2000)) Assume that Γ 0 intersects every connected component of Ω. Then the Euler control system is globally controllable in every time: For every T > 0, for every y 0, y 1 : Ω R n such that (1) div y 0 = div y 1 = 0, y 0 ν = y 1 ν = 0 on Ω\Γ 0, there exist y : [0,T] Ω R n and p : [0,T] Ω R such that (2) (3) (4) y t +(y )y + p = 0, div y = 0, y ν = 0 on [0,T] ( Ω\Γ 0 ) y(0, ) = y 0, y(t, ) = y 1.

35 Comments on the proof First try: One studies (as usual) the controllability of the linearized control system around 0. This linearized control system is the control system (1) y t + p = 0,div y = 0, y ν = 0 on [0,T] ( Ω\Γ 0 ). For simplicity we assume that n = 2. Taking the curl of the first equation, on gets, (curl y) t = 0. Hence curl y remains constant along the trajectories of the linearized control system. Hence the linearized control system is not controllable.

36 Iterated Lie brackets and PDE control systems Iterative Lie brackets have been used successfully for some control PDE systems: Euler and Navier Stokes control systems (different from the one considered here): A. Agrachev and A. Sarychev (2005); A. Shirikyan (2006, 2007), H. Nersisyan (2010), Schrödinger control system: T. Chambrion, P. Mason, M. Sigalotti and U. Boscain (2009). However they do not seem to work here.

37 A problem with Lie brackets for PDE control systems Consider the simplest PDE control system (1) y t +y x = 0, x [0,L],y(t,0) = u(t). It is a control system where, at time t, the state is y(t, ) : (0,L) R and the control is u(t) R. Formally it can be written in the form ẏ = f 0 (y)+uf 1 (y). Here f 0 is linear and f 1 is constant.

38 Lie bracket for ẏ = f 0 (y)+uf 1 (y), with f 0 (a) = 0 u = η y(ε) y(2ε) a+ηε 2 [f 0,f 1 ](a) ε 0 + a u = η

39 Problems of the Lie brackets for PDE control systems (continued) Let us consider, for ε > 0, the control defined on [0,2ε] by u(t) := η for t (0,ε), u(t) := η for t (ε,2ε). Let y : (0,2ε) (0,L) R be the solution of the Cauchy problem (1) (2) 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, (3) (4) y(2ε,x) = η, x (0,ε), y(2ε,x) = η, x (ε,2ε), y(2ε,x) = 0, x (2ε,L).

40 Problems of the Lie brackets for PDE control systems (continued) (1) y(2ε, ) y(0, ) ε 2 + as ε 0 +. L 2 (0,L) For every φ H 2 (0,L), one gets after suitable computations (2) lim ε 0 + L 0 ( ) y(2ε,x) y(0,x) φ(x) dx = ηφ (0). ε 2 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. How to avoid the use of iterated Lie brackets?

41 The return method (JMC (1992)) y 0 T t

42 The return method (JMC (1992)) y ȳ(t) 0 T t

43 The return method (JMC (1992)) y ȳ(t) 0 y 0 y 1 T t B 0 B 1

44 The return method (JMC (1992)) y ȳ(t) y(t) 0 y 0 y 1 T t B 0 B 1

45 The return method (JMC (1992)) y ȳ(t) y(t) ε η 0 y 0 T B 0 B 1 y 1 t

46 The return method: An example in finite dimension We go back to the baby stroller control system (1) ẏ 1 = u 1 cosy 3, ẏ 2 = u 1 siny 3, ẏ 3 = u 2. For every ū : [0,T] R 2 such that, for every t in [0,T], ū(t t) = ū(t), every solution ȳ : [0,T] R 3 of (2) ȳ 1 = ū 1 cosȳ 3, ȳ 2 = ū 1 sinȳ 3, ȳ 3 = ū 2, satisfies ȳ(0) = ȳ(t). The linearized control system around (ȳ, ū) is { ẏ1 = ū 1 y 3 sinȳ 3 +u 1 cosȳ 3, ẏ 2 = ū 1 y 3 cosȳ 3 +u 1 sinȳ 3, (3) ẏ 3 = u 2, which is controllable if (and only if) ū 0. We have got the controllability of the baby stroller system without using Lie brackets. We have only used controllability results for linear control systems.

47 No loss with the return method We consider the control system Σ : ẏ = f 0 (y)+ m u i f i (y), where the state is y R n and the control is u R m. We assume that f 0 (0) = 0 and that the f i s, i {0,1,...,m} are of class C in a neighborhood of 0 R n. One has the following proposition. i=1 Proposition (E. Sontag (1988), JMC (1994)) Let us assume that Σ satisfies the Lie algebra rank condition at 0 R n and is STLC at (0,0) R n R m. Then, for every ε > 0, there exists ū L ((0,ε);R m ) satisfying u(t) ε, t [0,T], such that, if ȳ : [0,ε] R n is the solution of ȳ = f(ȳ,ū(t)), ȳ(0) = 0, then ȳ(t) = 0, the linearized control system around (ȳ,ū) is controllable.

48 The return method and the controllability of the Euler equations One looks for (ȳ, p) : [0,T] Ω R n R such that (1) (2) (3) (4) ȳ t +(ȳ ȳ)+ p = 0, div ȳ = 0, ȳ ν = 0 on [0,T] ( Ω\Γ 0 ), ȳ(t, ) = ȳ(0, ) = 0, the linearized control system around (ȳ, p) is controllable.

49 Construction of (ȳ, p) Take θ : Ω R such that θ = 0 in Ω, θ ν = 0 on Ω\Γ 0. Take α : [0,T] R such that α(0) = α(t) = 0. Finally, define (ȳ, p) : [0,T] Ω R 2 R by (1) ȳ(t,x) := α(t) θ(x), p(t,x) := α(t)θ(x) α(t)2 2 θ(x) 2. Then (ȳ, p) is a trajectory of the Euler control system which goes from 0 to 0.

50 Controllability of the linearized control system around (ȳ, p) if n = 2 The linearized control system around (ȳ, p) is { yt +(ȳ )y +(y )ȳ + p = 0, div y = 0 in [0,T] Ω, (1) y ν = 0 on [0,T] ( Ω\Γ 0 ). Again we assume that n = 2. Taking once more the curl of the first equation, one gets (2) (curl y) t +(ȳ )(curl y) = 0. This is a simple transport equation on curl y. If there exists a Ω such that θ(a) = 0, then ȳ(t,a) = 0 and (curl y) t (t,a) = 0 showing that (2) is not controllable. This is the only obstruction: If θ does not vanish in Ω, one can prove that (2) (and then (1)) is controllable if T 0 α(t)dt is large enough.

51 A good θ for n = 2 and Ω simply connected Γ 0 Ω Ω R 2

52 A good θ for n = 2 and Ω simply connected Γ + Γ 0 Ω Γ Ω R 2

53 A good θ for n = 2 and Ω simply connected Γ + Γ 0 Ω Γ R 2 Ω g : Ω R Ωgds = 0, {g > 0} = Γ +, {g < 0} = Γ

54 A good θ for n = 2 and Ω simply connected θ = 0, θ ν = g on Ω Γ + Γ 0 Ω Γ R 2 Ω g : Ω R Ωgds = 0, {g > 0} = Γ +, {g < 0} = Γ

55 A good θ for n = 2 and Ω simply connected θ = 0, θ ν = g on Ω Γ + Γ 0 Ω Γ θ R 2 Ω g : Ω R Ωgds = 0, {g > 0} = Γ +, {g < 0} = Γ

56 From local controllability to global controllability A simple scaling argument: if (y,p) : [0,1] Ω : R n R is a trajectory of our control system, then, for every ε > 0, (y ε,p ε ) : [0,ε] Ω R n R defined by y ε (t,x) := 1 ( ) t ε y ε,x, p ε (t,x) := 1 ( ) t (1) ε 2p ε,x is also a trajectory of our control system.

57 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), JMC and S. Guerrero (2009), M. Chapouly (2009). Burgers equation: Th. Horsin (1998), M. Chapouly (2006), O. Imanuvilov and J.-P. Puel (2009). Saint-Venant equations: JMC (2002). Vlasov Poisson: O. Glass (2003).

58 Return method: references (continued) Isentropic Euler equations: O. Glass (2006). Schrödinger equation: K. Beauchard (2005), K. Beauchard and JMC (2006). Camassa-Holm equation: O. Glass (2008), V. Perrollaz (2010). Korteweg de Vries equation: M. Chapouly (2008). Hyperbolic equations: JMC, O. Glass and Z. Wang (2009). Ensemble controllability of Bloch equations: K. Beauchard, JMC and P. Rouchon (2010). Parabolic systems: JMC, S. Guerrero, L. Rosier (2010). Uniform controllability of scalar conservation laws in the vanishing viscosity limit: M. Léautaud (2010).

59 The Navier-Stokes control system The Navier-Stokes control system is deduced from the Euler equations by adding the linear term y: the equation is now (1) y t y +(y )y + p = 0, div y = 0. For the boundary condition, one requires now that (2) y = 0 on [0,T] ( Ω\Γ 0 ). For the control, one can take, for example, y on [0,T] Γ 0.

60 Smoothing effects and a new notion of (global) controllability: A. Fursikov and O. Imanuvilov (1995) y ŷ y 0 T t

61 Smoothing effects and a new notion of (global) controllability: A. Fursikov and O. Imanuvilov (1995) y ŷ y 0 T t

62 Smoothing effects and a new notion of local controllability A. Fursikov and O. Imanuvilov (1995) y ŷ T t

63 Smoothing effects and a new notion of local controllability A. Fursikov and O. Imanuvilov (1995) y ŷ T t

64 Smoothing effects and a new notion of local controllability A. Fursikov and O. Imanuvilov (1995) y ŷ y 0 T t

65 Smoothing effects and a new notion of local controllability A. Fursikov and O. Imanuvilov (1995) y ŷ y 0 T t

66 Local controllability Theorem (A. Fursikov and O. Imanuvilov (1994), O. Imanuvilov (1998, 2001), E. Fernandez-Cara, S. Guerrero, O. Imanuvilov and J.-P. Puel (2004)) The Navier-Stokes control system is locally controllable. The proof relies on the controllability of the linearized control system around ŷ (which is obtained by Carleman inequalities).

67 Global controllability Theorem (JMC (1996), JMC and A. Fursikov (1996)) The Navier-Stokes control system is globally controllable if Γ 0 = Ω. Open problem Does the above global controllability result hold even if Γ 0 Ω?

68 Sketch of the proof of the global result y T t

69 Sketch of the proof of the global result y T t

70 Sketch of the proof of the global result y T B 1 B 2 t

71 Sketch of the proof of the global result y ȳ(t) B 1 B 2 T t

72 Sketch of the proof of the global result y ȳ(t) B 1 B 2 T t B 3 B 4

73 The strategy explained on a finite dimensional case We consider the following control system (1) ẏ = F(y)+Bu(t), where the state is y R n, the control is u R m, B is a n m matrix and F C 1 (R n ;R n ) is quadratic: F(λy) = λ 2 F(y), λ [0,+ ), y R n. We assume that there exists a trajectory (ȳ,ū) C 0 ([0,T 0 ];R n ) L ((0,T 0 );R m ) of the control system (1) such that the linearized control system around (ȳ,ū) is controllable and such that ȳ(0) = ȳ(t 0 ) = 0. Remark One has F(0) = 0. Hence (0,0) is an equilibrium of the control system (1). The linearized control system around this equilibrium is ẏ = Bu, which is not controllable if (and only if) B is not onto.

74 Let A be a n n matrix and let us consider the following control system (1) ẏ = Ay +F(y)+Bu(t), where the state is y R n, the control is u R m. For the application to incompressible fluids, (1) is the Euler control system and (1) is the Navier-Stokes control system. One has the following theorem. Theorem Under the above assumptions, the control system (1) is globally controllable in arbitrary time: For every T > 0, for every y 0 R n and for every y 1 R n, there exists u L ((0,T);R m ) such that (ẏ = f(y,u(t)), y(0) = y 0 ) ( y(t) = y 1).

75 Proof of the controllability theorem Let y 0 R n and y 1 R n. Let R n G : R L ((0,T 0 );R m ) (ε,ũ) ỹ(t 0 ) εy 1 where ỹ : [0,T 0 ] R n is the solution of ỹ = F(ỹ)+εAỹ +Bũ(t), ỹ(0) = εy 0. The map G is of class C 1 in a neighborhood of (0,ū). One has G(0,ū) = 0. Moreover G ũ (0,ū)v = y(t 0) where y : [0,T 0 ] R n is the solution of ẏ = F (ȳ)y +Bv, y(0) = 0. Hence G ũ (0,ū) is onto. Therefore there exist ε 0 > 0 and a C 1 -map ε ( ε 0,ε 0 ) ũ ε L ((0,T 0 );R m ) such that G(ε,ũ ε ) = 0, ε ( ε 0,ε 0 ), ũ 0 = ū.

76 Let ỹ ε : [0,T 0 ] R n be the solution of the Cauchy problem ỹ ε = F(ỹ ε )+εaỹ ε +Bũ ε (t), ỹ ε (0) = εy 0. Then ỹ ε (T 0 ) = εy 1. Let y : [0,εT 0 ] R n and u : [0,εT 0 ] R m be defined by y(t) := 1 ( ) t εỹε, u(t) := 1 ( ) t ε ε 2ũε. ε Then ẏ = F(y)+Ay +Bu, y(0) = y 0 and y(εt 0 ) = y 1. This concludes the proof of the controllability theorem if T is small enough. If T is not small, it suffices, with ε > 0 small enough, to go from y 0 to 0 during the interval of time [0,ε], stay at 0 during the interval of time [ε,t ε] and finally go from 0 to y 1 during the interval of time [T ε,t].

77 Morality The morality behind Theorem 4.4 is that the quadratic term F(y) wins against the linear term Ay. Note, however, that for Euler/Navier-Stokes equations the linear term µ y contains more derivatives than the quadratic term (y )y. This creates many new difficulties and the proof requires important modifications. In particular, one first gets a global approximate controllability result and then concludes with a local controllability result. Of course, as one can see by looking at the proof of the controllability theorem, this method works only if we have a (good) convergence of the solution of the Navier-Stokes equations to the solution of the Euler equations when the viscosity tends to 0.

78 Let us recall that this is not known even in dimension n = 2 if there is no control. More precisely, let us assume that Ω is of class C, that n = 2 and that ϕ C0 (Ω;R2 ) is such that div ϕ = 0. Let T > 0. Let y C ([0,T] Ω;R 2 ) and p C ([0,T] Ω) be the solution to the Euler equations (E) y t +(y )y + p = 0, div y = 0, in (0,T) Ω, y ν = 0 on [0,T] Ω, y(0, ) = ϕ on Ω. Let ε (0,1]. Let y ε C ([0,T] Ω;R 2 ) and p ε C ([0,T] Ω) be the solution to the Navier-Stokes equations yt ε ε y ε +(y ε )y ε + p ε = 0, div y ε = 0, in (0,T) Ω, (NS) y ε = 0 on [0,T] Ω, y(0, ) = ϕ on Ω. One knows that there exists C > 0 such that y ε C 0 ([0,T];L 2 (Ω;R 2 )) C, for every ε (0,1].

79 One has the following challenging open problems. Open problem (Convergence of Navier-Stokes to Euler) (i) Does y ε converge weakly to y in L 2 ((0,T) Ω;R 2 ) as ε 0 +? (ii) Let K be a compact subset of Ω and m be a positive integer. Does y ε [0,T] K converge to y [0,T] K in C m ([0,T] K;R 2 ) as ε 0 +? (Of course, due to the difference of boundary conditions between the Euler equations and the Navier-Stokes equations, one does not have a positive answer to this last question if K = Ω.)

80 Main difficulty for the return method: Return to the initial state

81 Main difficulty for the return method: Return to the initial state Initial state

82 Main difficulty for the return method: Return to the initial state Initial state Intermediate state

83 Main difficulty for the return method: Return to the initial state Initial state Intermediate state It is difficult to return to the initial state.

84 A water-tank control system

85 Saint-Venant equations: Notations H D v x The horizontal velocity v is taken with respect to the one of the tank.

86 The model: Saint-Venant equations (1) (2) (3) (4) (5) H t +(Hv) x = 0, t [0,T], x [0,L], ) v t + (gh + v2 = u(t), t [0,T], x [0,L], 2 x v(t,0) = v(t,l) = 0, t [0,T], ṡ(t) = u(t), t [0,T], Ḋ(t) = s(t), t [0,T]. 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.

87 State space (1) (2) d dt L 0 H(t,x)dx = 0, H x (t,0) = H x (t,l) (= u(t)/g). Definition (State space) 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 (3) v(0) = v(l) = 0, H x (0) = H x (L), L 0 H(x)dx = LH e.

88 Main result Theorem (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): 1 The linearized control system around (Y e,u e ) is not controllable, 2 Steady state controllability of the linearized control system around (Y e,u e ).

89 Corollary 1: Destroy waves u

90 Corollary 2: Steady-state controllability D 1 u

91 The linearized control system Without loss of generality L = H e = g = 1. For simplicity, we forget s and D. The linearized control system around (Y e,u e ) := ((1,0),0) is (1) (2) (3) 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],

92 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)), wod (x) := 1 2 (w(x) w(1 x)). Σ 1 h od t +vx ev = 0, vt ev +h od x = u(t), v ev (t,0) = v ev (t,1) = 0, Σ 2 h ev t +vx od = 0, vt od +h ev x = 0, v od (t,0) = v od (t,1) = 0, The Σ 2 part is clearly not controllable.

93 Toy model T { T1 { ẏ1 = y 2, ẏ 2 = y 1 +u, T 2 { ẏ3 = y 4, ẏ 4 = y 3 +2y 1 y 2, where the state is y = (y 1,y 2,y 3,y 4 ) tr R 4 and the control is u R. The linearized control system of T around (y e,u e ) := (0,0) is (1) ẏ 1 = y 2, ẏ 2 = y 1 +u, ẏ 3 = y 4, ẏ 4 = y 3, which is not controllable.

94 Controllability of the toy model If y(0) = 0, (1) (2) y 3 (T) = T 0 T y 4 (T) = y 2 1(T) y 2 1(t)cos(T t)dt, 0 y 2 1(t)sin(T t)dt. Hence T is not controllable in time T π. Using explicit computations one can show that T is (locally) controllable in time T > π. Remark For linear systems 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.

95 How to recover the large time local controllability of T The first point is at least to find a trajectory such that the linearized control system around it is controllable. We try the simplest possible trajectories, namely equilibrium points. Let γ R and define (1) ((y γ 1,yγ 2,yγ 3,yγ 4 )tr,u γ ) := ((γ,0,0,0) tr,γ). Then ((y γ 1,yγ 2,yγ 3,yγ 4 )tr,u γ ) is an equilibrium of T. The linearized control system of T at this equilibrium is (2) ẏ 1 = y 2, ẏ 2 = y 1 +u, ẏ 3 = y 4, ẏ 4 = y 3 +2γy 2, which is controllable if (and only if) γ 0.

96 How to recover the large time local controllability of T y t

97 How to recover the large time local controllability of T y y γ (t) t

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

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

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

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

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

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

104 Construction of the blue trajectory One uses quasi-static deformations. Let g C 2 ([0,1];R) be such that g(0) = 0, g(1) = 1. Let ũ : [0,1/ε] R be defined by ũ(t) := γg(εt), t [0,1/ε]. Let ỹ := (ỹ 1,ỹ 2,ỹ 3,ỹ 4 ) tr : [0,1/ε] R 4 be defined by requiring (1) (2) ỹ 1 = ỹ 2, ỹ 2 = ỹ 1 +ũ, ỹ 3 = ỹ 4, ỹ 4 = ỹ 3 +2ỹ 1 ỹ 2, ỹ(0) = 0. One easily checks that ỹ(1/ε) (γ,0,0,0) tr as ε 0.

105 (y γ,u γ ) for the water-tank ( ) 1 u(t) = γ, h = γ 2 y, v = 0.

106 Difficulties Loss of derivatives. Solution: one uses the iterative scheme inspired by the usual one to prove the existence to y t +A(y)y x = 0, y(0,x) = ϕ(x), namely Σ nlinear : yt n+1 +A(y n )yx n+1 = 0, y n+1 (0,x) = ϕ(x). However, I have only been able to prove that the control system corresponding to Σ nlinear is controllable for (h n,v n ) satisfying some resonance conditions. Hence one has also to insure that (h n+1,v n+1 ) satisfies these resonance conditions. This turns out to be possible. (For control system, resonance is good: when there is a resonance, with a small action we get a strong effect).

107 An open problem What is the minimal time for the local controllability of the water tank control system? 1 A simple observation on the speed of propagation shows that the time for local controllability is at least 1. 2 For the linearized control system around h = γ((1/2) x), v = 0 the minimal time for controllability tends to 2 as γ 0. 3 For our toy model, there is no minimal time for the controllability around ((γ,0,0,0) tr,γ). However for the local the controllability of the nonlinear system the minimal time is π > 0. 4 For a related problem (a quantum particle in a moving box; P. Rouchon (2003), K. Beauchard (2005)), there is again no minimal time for the controllability of the linearized control system around the analogue of ((γ,0,0,0) tr,γ) and there is a minimal time for the local controllability of the nonlinear system (JMC (2006)). Again, for this related problem, the optimal time for the local controllability is not known.

108 Power series expansions: The KdV control system (1) (2) y t +y x +y xxx +yy x = 0, t [0,T], x [0,L], y(t,0) = y(t,l) = 0, y x (t,l) = u(t), t [0,T]. where, at time t [0,T], the control is u R and the state is y(t, ) L 2 (0,L). Remark Prior pioneer work on the controllability of the Korteweg-de Vries equation (with periodic boundary conditions and internal controls): D. Russell and B.-Y. Zhang (1996).

109 Controllability of the linearized control system The linearized control system (around 0) is (1) (2) y t +y x +y xxx = 0, t [0,T], x [0,L], y(t,0) = y(t,l) = 0, y x (t,l) = u(t), t [0,T]. where, at time t [0,T], the control is u R and the state is y(t, ) L 2 (0,L). Theorem (L. Rosier (1997)) For every T > 0, the linearized control system is controllable in time T if and only { } k L N := 2π 2 +kl+l 2, k N, l N. 3

110 Application to the nonlinear system Theorem (L. Rosier (1997)) For every T > 0, the KdV control system is locally controllable (around 0) in time T if L N. Question: Does one have controllability if L N?

111 Controllability when L N Theorem (JMC and E. Crépeau (2004)) If L = 2π (which is in N: take k = l = 1), for every T > 0 the KdV control system is locally controllable (around 0) in time T. Theorem (E. Cerpa (2007), E. Cerpa and E. Crépeau (2008)) For every L N, there exists T > 0 such that the KdV control system is locally controllable (around 0) in time T.

112 Strategy of the proof: power series expansion. Example with L = 2π. For every trajectory (y,u) of the linearized control system around 0 d dt 2π 0 (1 cos(x))y(t,x)dx = 0. This is is the only obstacle to the controllability of the linearized control system: Proposition (L. Rosier (1997)) Let H := {y L 2 (0,L); L 0 (1 cos(x))y(x)dx = 0}. For every (y 0,y 1 ) H H, there exists u L 2 (0,T) such that the solution to the Cauchy problem y t +y x +y xxx = 0, y(t,0) = y(t,l) = 0, y x (t,l) = u(t), t [0,T], satisfies y(t,x) = y 1 (x), x [0,L]. y(0,x) = y 0 (x), x [0,L],

113 We explain the method on the control system of finite dimension ẏ = f(y,u), where the state is y R n and the control is u R m. We assume that (0,0) R n R m is an equilibrium of the control system ẏ = f(y,u), i.e. that f(0,0) = 0. Let with H := Span{A i Bu; u R m, i {0,...,n 1}} A := f f (0,0), B := y u (0,0). If H = R n, the linearized control system around (0,0) is controllable and therefore the nonlinear control system ẏ = f(y,u) is small-time locally controllable at (0,0) R n R m.

114 Let us look at the case where the dimension of H is n 1. Let us make a (formal) power series expansion of the control system ẏ = f(y,u) in (y, u) around 0. We write y = y 1 +y , u = v 1 +v The order 1 is given by (y 1,v 1 ); the order 2 is given by (y 2,v 2 ) and so on. The dynamics of these different orders are given by (1) ẏ 1 = f y (0,0)y1 + f u (0,0)v1, (2) ẏ 2 = f y (0,0)y2 + f u (0,0)v f 2 y 2(0,0)(y1,y 1 ) + 2 f y u (0,0)(y1,v 1 )+ 1 2 f 2 u 2(0,0)(v1,v 1 ), and so on.

115 Let e 1 H. Let T > 0. Let us assume that there are controls v 1 ± and v 2 ±, both in L ((0,T);R m ), such that, if y 1 ± and y 2 ± are solutions of (1) (2) ẏ 1 ± = f y (0,0)y1 ± + f u (0,0)v1 ±, y± 1 (0) = 0, (3) ẏ± 2 = f y (0,0)y2 ± + f u (0,0)v2 ± f 2 y 2(0,0)(y1 ±,y1 ± ) + 2 f y u (0,0)(y1 ±,u 1 ±)+ 1 2 f 2 u 2(0,0)(u1 ±,u 1 ±), (4) y± 2 (0) = 0, then (5) y± 1 (T) = 0, (6) y±(t) 2 = ±e 1.

116 Let (e i ) i {2,...n} be a basis of H. By the definition of H, there are (u i ) i=2,...,n, all in L (0,T) m, such that, if (y i ) i=2,...,n are the solutions of (1) (2) ẏ i = f y (0,0)y i + f u (0,0)u i, y i (0) = 0, then, for every i {2,...,n}, (3) y i (T) = e i. Now let (4) b = n b i e i i=1 be a point in R n. Let v 1 and v 2, both in L ((0,T);R m ), be defined by the following (5) (6) If b 1 0, then v 1 := v 1 + and v 2 := v 2 +, If b 1 < 0, then v 1 := v 1 and v 2 := v 2.

117 Let u : (0,T) R m be defined by (1) u(t) := b 1 1/2 v 1 (t)+ b 1 v 2 (t)+ Let y : [0,T] R n be the solution of n b i u i (t). i=2 (2) ẏ = f(y,u(t)), y(0) = 0. Then one has, as b 0, (3) y(t) = b+o(b). Hence, using the Brouwer fixed-point theorem and standard estimates on ordinary differential equations, one gets the local controllability of ẏ = f(y,u) (around (0,0) R n R m ) in time T, that is, for every ε > 0, there exists η > 0 such that, for every (a,b) R n R n with a < η and b < η, there exists a trajectory (y,u) : [0,T] R n R m of the control system ẏ = f(y,u) such that (4) (5) y(0) = a, y(t) = b, u(t) ε, t (0,T).

118 Bad and good news for L = 2π Bad news: The order 2 is not sufficient. One needs to go to the order 3 Good news: the fact that the order is odd allows to get the local controllability in arbitrary small time. The reason: If one can move in the direction ξ H one can move in the direction ξ. Hence it suffices to argue by contradiction (assume that it is impossible to enter in H in small time etc.)

119 Open problems 1 Is there a minimal time for local controllability for some values of L? 2 Do we have global controllability? This is open even with three boundary controls: (1) (2) y t +y x +y xxx +yy x = 0, y(t,0) = u 1 (t), y(t,l) = u 2 (t), y x (t,l) = u 3 (t). Note that one has global controllability for (3) (4) y t +y x +y xxx +yy x = u 4 (t), y(t,0) = u 1 (t), y(t,l) = u 2 (t), y x (t,l) = u 3 (t). (M. Chapouly (2009)). The proof uses the return method as for the Navier-Stokes control system.