The Fattorini-Hautus test

Transcription

1 The Fattorini-Hautus test Guillaume Olive Seminar, Shandong University Jinan, March

2 Plan Part 1: Background on controllability Part 2: Presentation of the Fattorini-Hautus test Part 3: Controllability of parabolic systems Part 4: Stabilization of integro-dierential equations Part 5: Perturbation theorems G. Olive The Fattorini-Hautus test 2 / 48

3 Part I Background on controllability G. Olive The Fattorini-Hautus test 3 / 48

4 System description Let H, U be two complex Hilbert spaces. Consider d y = Ay + Bu, t (, T ), dt y() = y, where T > is the time of control. (abst-ode) y is the state. y is the initial data. A : D(A) H H generates a C -semigroup (S(t)) t. u L 2 (, T ; U) is the control. B L(U, H) is a bounded linear operator. G. Olive The Fattorini-Hautus test 4 / 48

5 System description Let H, U be two complex Hilbert spaces. Consider d y = Ay + Bu, t (, T ), dt y() = y, where T > is the time of control. (abst-ode) y is the state. y is the initial data. A : D(A) H H generates a C -semigroup (S(t)) t. u L 2 (, T ; U) is the control. B L(U, H) is a bounded linear operator. Well-posedness : For every y H and u L 2 (, T ; U), there exists a unique solution t y(t) = S(t)y + S(t s)bu(s) ds, t [, T ]. Note that y C ([, T ]; H) with y(t) H C ( ) y H + u L 2 (,T ;U), t [, T ]. (cont-dep) G. Olive The Fattorini-Hautus test 4 / 48

6 Notions of controllability y(t ; ) y 1 y Figure Uncontrolled trajectory y : initial state, y 1 : target, y(t ; u) : value of the solution to (abst-ode) at time T with control u. G. Olive The Fattorini-Hautus test 5 / 48

7 Notions of controllability y(t ; ) y 1 = y(t ; u) y Figure Trajectory controlled exactly y : initial state, y 1 : target, y(t ; u) : value of the solution to (abst-ode) at time T with control u. Denition (abst-ode) is exactly controllable in time T if y, y 1 H, u L 2 (, T ; U), y(t ) = y 1. G. Olive The Fattorini-Hautus test 5 / 48

8 Notions of controllability y(t ; ) = y(t ; u) y Figure Trajectory controlled to y : initial state, y 1 : target, y(t ; u) : value of the solution to (abst-ode) at time T with control u. Denition (abst-ode) is null-controllable in time T if y H, u L 2 (, T ; U), y(t ) =. G. Olive The Fattorini-Hautus test 5 / 48

9 Notions of controllability y(t ; ) y 1 y(t ; u) ε y Figure Trajectory controlled approximately y : initial state, y 1 : target, y(t ; u) : value of the solution to (abst-ode) at time T with control u. Denition (abst-ode) is approximately controllable in time T if y, y 1 H, ε >, u L 2 (, T ; U), y(t ) y 1 H ε. G. Olive The Fattorini-Hautus test 5 / 48

10 Reformulation Let F T : H H y y(t ), and G T : L 2 (, T ; U) H u ŷ(t ), d y dt = Ay, t (, T ), y() = y, d ŷ dt = Aŷ + Bu, t (, T ), ŷ() =, so that y(t ) = F T y + G T u. G. Olive The Fattorini-Hautus test 6 / 48

11 Reformulation Let F T : H H y y(t ), and G T : L 2 (, T ; U) H u ŷ(t ), d y dt = Ay, t (, T ), y() = y, d ŷ dt = Aŷ + Bu, t (, T ), ŷ() =, so that y(t ) = F T y + G T u. Therefore, (abst-ode) is exactly controllable in time T if, and only if, Im G T = H. (1) (abst-ode) is null-controllable in time T if, and only if, Im F T Im G T. (2) (abst-ode) is approximately controllable in time T if, and only if, Im G T = H. (3) G. Olive The Fattorini-Hautus test 6 / 48

12 Reformulation Let F T : H H y y(t ), and G T : L 2 (, T ; U) H u ŷ(t ), d y dt = Ay, t (, T ), y() = y, d ŷ dt = Aŷ + Bu, t (, T ), ŷ() =, so that y(t ) = F T y + G T u. Therefore, (abst-ode) is exactly controllable in time T if, and only if, Im G T = H. (1) (abst-ode) is null-controllable in time T if, and only if, Im F T Im G T. (2) (abst-ode) is approximately controllable in time T if, and only if, Im G T = H. (3) Remark : If dim H < + (in particular A L(H)), then all these notions are equivalent : (1) (2) since Im F T = H. (1) (3) since dim Im G T < +. G. Olive The Fattorini-Hautus test 6 / 48

13 Duality By (cont-dep) we have F T L(H) and G T L(L 2 (, T ; U), H). Thus, (abst-ode) is exactly controllable in time T if, and only if, z 1 H C G T z1 H, z 1 H. (abst-ode) is null-controllable in time T if, and only if, F T z1 H C G T z1 H, z 1 H. (abst-ode) is approximately controllable in time T if, and only if, Let us compute G T. ker G T = {}. G. Olive The Fattorini-Hautus test 7 / 48

14 Duality By (cont-dep) we have F T L(H) and G T L(L 2 (, T ; U), H). Thus, (abst-ode) is exactly controllable in time T if, and only if, z 1 H C G T z1 H, z 1 H. (abst-ode) is null-controllable in time T if, and only if, F T z1 H C G T z1 H, z 1 H. (abst-ode) is approximately controllable in time T if, and only if, ker G T = {}. Let us compute GT. Multiplying (abst-ode) by z, solution to the adjoint system d dt z = A z, t (, T ), z(t ) = z 1, we obtain T y(t ) z 1 y z() = u(t) B z(t) dt. G. Olive The Fattorini-Hautus test 7 / 48

15 Duality By (cont-dep) we have F T L(H) and G T L(L 2 (, T ; U), H). Thus, (abst-ode) is exactly controllable in time T if, and only if, z 1 H C G T z1 H, z 1 H. (abst-ode) is null-controllable in time T if, and only if, F T z1 H C G T z1 H, z 1 H. (abst-ode) is approximately controllable in time T if, and only if, ker G T = {}. Let us compute GT. Multiplying (abst-ode) by z, solution to the adjoint system d dt z = A z, t (, T ), z(t ) = z 1, we obtain This shows that T y(t ) z 1 y z() = u(t) B z(t) dt. F T : H H z 1 z(), G T : H L 2 (, T ; U) z 1 B z. G. Olive The Fattorini-Hautus test 7 / 48

16 Part II Presentation of the Fattorini-Hautus test G. Olive The Fattorini-Hautus test 8 / 48

17 Finite dimension Let A C n n and B C n m. d y dt = Ay + Bu, t (, T ), y() = y C n. (ODE) Theorem (Fattorini (1966), Hautus (1969)) (ODE) is controllable if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) G. Olive The Fattorini-Hautus test 9 / 48

18 Finite dimension Let A C n n and B C n m. d y dt = Ay + Bu, t (, T ), y() = y C n. (ODE) Theorem (Fattorini (1966), Hautus (1969)) (ODE) is controllable if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Proof : Let us denote S(t) = e ta. Let N = {z C n, B S(t) z =, t [, T ]}. We have to prove that N = {}. Taking t = we see that N ker B. Taking the derivative of the identity B S(t) z = we obtain A N N. Thus, if N {}, there exist eigen-elements λ C et ξ C n such that ξ, ξ ker(λ A ) ker B, a contradiction with (Fatt). G. Olive The Fattorini-Hautus test 9 / 48

19 Innite dimension Let H and U be two complex Hilbert spaces. We assume that A : D(A) H H generates C -semigroup on H. B L(U, H). d y dt = Ay + Bu, t (, T ), y() = y H. (4) Theorem (Fattorini (1966)) Assume that : (i) A generates an analytic C -semigroup. (ii) σ(a) = {λ k } has only isolated eigenvalues with nite (alg.) multiplicities. (iii) The family of generalized eigenvectors of A is complete in H. Then, (4) is approximately controllable if, and only if, Remarks : ker(λ A ) ker B = {}, λ C. (Fatt) (i) implies that the approximate controllability does not depend on the time T. (ii) is satised if the resolvent of A is compact. (iii) holds for perturbations of self-adjoint operators (Keldysh's theorem). G. Olive The Fattorini-Hautus test 1 / 48

20 Proof of the Fattorini-Hautus test Let N = {z H, B S(t) z = a.e. t (, + )}. The Laplace transform gives N = { z H, B (λ A ) 1 z =, λ ρ(a ) }. Let By (ii) we have dim Im P k P k = 1 (ξ A) 1 dξ. 2πi ξ λ k =ε k < + with Im Pk = ker(λ k A ) m k. Set N k,j = (λ k A ) j Pk N. Since N k,mk = {}, we have A N k,mk 1 N k,mk 1. Since dim N k,mk 1 < +, N k,mk 1 ker B, the Fattorini-Hautus test (Fatt) gives N k,mk 1 = {}. By iteration, N k, = Pk N = {}. Since this is true for every k, by (iii) we obtain N = {}. G. Olive The Fattorini-Hautus test 11 / 48

21 "Best" necessary condition If T > such that (A, B) is exactly controllable in time T, then ker(λ A ) ker B = {}, λ C. G. Olive The Fattorini-Hautus test 12 / 48

22 "Best" necessary condition If T > such that (A, B) is exactly controllable in time T, then ker(λ A ) ker B = {}, λ C. If T > such that (A, B) is null-controllable in time T, then ker(λ A ) ker B = {}, λ C. G. Olive The Fattorini-Hautus test 12 / 48

23 "Best" necessary condition If T > such that (A, B) is exactly controllable in time T, then ker(λ A ) ker B = {}, λ C. If T > such that (A, B) is null-controllable in time T, then ker(λ A ) ker B = {}, λ C. If T > such that (A, B) is approximatelly controllable in time T, then ker(λ A ) ker B = {}, λ C. G. Olive The Fattorini-Hautus test 12 / 48

24 "Best" necessary condition If T > such that (A, B) is exactly controllable in time T, then ker(λ A ) ker B = {}, λ C. If T > such that (A, B) is null-controllable in time T, then ker(λ A ) ker B = {}, λ C. If T > such that (A, B) is approximatelly controllable in time T, then ker(λ A ) ker B = {}, λ C. If (A, B) is rapidly stabilizable, then ker(λ A ) ker B = {}, λ C. G. Olive The Fattorini-Hautus test 12 / 48

25 Approximate controllability of the heat equation ty y = 1 ωu in (, T ) Ω, y = on (, T ) Ω, y() = y in Ω. (chal) tz z = in (, T ) Ω, z = on (, T ) Ω, z() = z in Ω. Theorem (chal) is approximately controllable (in time T for every T > ). G. Olive The Fattorini-Hautus test 13 / 48

26 Approximate controllability of the heat equation ty y = 1 ωu in (, T ) Ω, y = on (, T ) Ω, y() = y in Ω. (chal) tz z = in (, T ) Ω, z = on (, T ) Ω, z() = z in Ω. Theorem (chal) is approximately controllable (in time T for every T > ). "Classical" proof : We write + z(t) = α k e λ k t φ k, ( φ k = λ k φ k ). k=1 Using the analyticity in time, + 1 ωz(t) = α k e λ k t (1 ωφ k ) =, k=1 t [, + ). Multiplying by e λ 1t and letting t + : α 1 1 ωφ 1 = (rem : φ 1 ker( λ 1 ) ker 1 ω). Thus, α 1 =. Then we iterate, we multiply by e λ 2t to obtain α 2 =, etc. G. Olive The Fattorini-Hautus test 13 / 48

27 Part III Controllability of parabolic systems Joint work with Franck Boyer G. Olive The Fattorini-Hautus test 14 / 48

28 Toy model of parabolic systems We will focus on the distributed controllability of the following 2 2 system by 1 control : ty 1 y 1 = 1 ωu in (, T ) Ω, ty 2 y 2 = a 21 (x)y 1 in (, T ) Ω, y 1 = y 2 = y 1 () = y1, y 2() = y2, in Ω, on (, T ) Ω, (syst) where (y 1, y 2 ) is the state and (y 1, y 2 ) L2 (Ω) 2 the initial data, u L 2 (, T ; L 2 (Ω)) is the control, ω Ω localises in space the control, a 21 L (Ω) couples the second equation to the rst one. Remark : The controllability of (syst) by 2 controls is easy (apply Carleman estimates to both equations and add them up). Theorem (de Teresa (2)) Assume that there exist a nonempty open subset ω ω and ε > such that a 21 (x) ε, a.e. x ω. Then, (syst) is null-controllable in time T for every T >. This hypothesis thus requires that ω supp a 21. G. Olive The Fattorini-Hautus test 15 / 48

29 Spectral properties Let The adjoint of A is A =, D(A) = (H 2 (Ω) H 1 a 21 (Ω))2. A = a 21, D(A ) = D(A). G. Olive The Fattorini-Hautus test 16 / 48

30 Spectral properties Let The adjoint of A is The spectrums are A =, D(A) = (H 2 (Ω) H 1 a 21 (Ω))2. A = a 21, D(A ) = D(A). σ(a) = σ(a ) = { λ k } k N. G. Olive The Fattorini-Hautus test 16 / 48

31 Spectral properties Let The adjoint of A is The spectrums are A =, D(A) = (H 2 (Ω) H 1 a 21 (Ω))2. A = a 21, D(A ) = D(A). σ(a) = σ(a ) = { λ k } k N. Denoting P k the spectral projection of associated with λ k, the eigenspaces of A are ker( λ k A ) = V k W k, where V k = v v ker( λ k ), W k = S k(a 21 w) w w ker( λ k ) ker(p k a 21 ), where S k : f ker P k v ker P k with v the unique solution (in ker P k ) of ( λ k )v = f in Ω, v = on Ω. G. Olive The Fattorini-Hautus test 16 / 48

32 Sucient conditions Theorem (Kavian and de Teresa (21), Olive (214)) Assume that Then, (syst) is approximately controllable. ker( λ k ) ker(p k a 21 ) = {}, k N. (5) In general, (5) is not a necessary condition. (5) can be reformulated into ( ) det a 21 φ k,i φ k,j dx, k N, (6) Ω 1 i,j m k where φ k,1,..., φ k,mk is a basis of ker( λ k ). In the one-dimensional case Ω = (, 1) (denoting φ k,1 = φ k since m k = 1) I k = 1 a 21 (φ k ) 2 dx, k N. G. Olive The Fattorini-Hautus test 17 / 48

33 Proof of Olive (214) Let B = 1ω, D(B) = L 2 (Ω) 2. By the Fattorini-Hautus test, the approximate controllability is equivalent to By assumption so that As a result ker( λ k A ) ker B = {}, k N. W k = {}, k N, ker( λ k A ) = V k = v v ker( λ k ), k N. v ker( λ k A ) ker B (w = and v ker( λ k ) ker 1 ω). w The unique continuation for a single equation then gives v =. G. Olive The Fattorini-Hautus test 18 / 48

34 Reduction to a nonhomogeneous scalar elliptic problem In this part, we focus again on the approximate controllability. By the Fattorini-Hautus test, we have to study the property v λ k v = a 21 w in Ω w λ k w = in Ω = v = w = in Ω. u = in ω We treat this problem as a nonhomogeneous scalar equation : where F = a 21 w is known. v λ k v = F in Ω, G. Olive The Fattorini-Hautus test 19 / 48

35 Notations From now on, Ω = (, 1). ω Ω is still the control domain and ω is not necessarily connected. φ k denotes again the eigenfunctions of xx associated with λ k. G. Olive The Fattorini-Hautus test 2 / 48

36 Notations From now on, Ω = (, 1). ω Ω is still the control domain and ω is not necessarily connected. φ k denotes again the eigenfunctions of xx associated with λ k. ( ) Let C Ω\ω be the set of connected component of Ω\ω. G. Olive The Fattorini-Hautus test 2 / 48

37 Notations From now on, Ω = (, 1). ω Ω is still the control domain and ω is not necessarily connected. φ k denotes again the eigenfunctions of xx associated with λ k. ( ) Let C Ω\ω be the set of connected component of Ω\ω. ( ) For every C C Ω\ω and F L 2 (Ω), let M k (F, C) be the vector of R 2 dened by M k (F, C) = C F φ k dx if C Ω, M k (F, C) = C F φ k dx si C F φ k dx C Ω =, For instance, ω is connected = M k (F, C) = C F φ k dx, ( ) C C Ω\ω. G. Olive The Fattorini-Hautus test 2 / 48

38 Notations From now on, Ω = (, 1). ω Ω is still the control domain and ω is not necessarily connected. φ k denotes again the eigenfunctions of xx associated with λ k. ( ) Let C Ω\ω be the set of connected component of Ω\ω. ( ) For every C C Ω\ω and F L 2 (Ω), let M k (F, C) be the vector of R 2 dened by M k (F, C) = C F φ k dx if C Ω, M k (F, C) = C F φ k dx si C F φ k dx C Ω =, For instance, ω is connected = M k (F, C) = C F φ k dx, ( ) C C Ω\ω. Finally, for every F L 2 (Ω) we dene the following family of vectors of R 2 : ) M k (F, ω) = (M k (F, C)) ( (Ω\ω ) (R 2 ) C. C C Ω\ω G. Olive The Fattorini-Hautus test 2 / 48

39 Unique continuation for a 1D nonhomogeneous elliptic equation Theorem (Boyer and Olive (214)) Let k N and F L 2 (Ω). We have if, and only if, { v H 2 (Ω) H 1 (Ω), { xx v k 2 π 2 v = F in Ω, v = in ω, M k (F, ω) =. F = in ω, G. Olive The Fattorini-Hautus test 21 / 48

40 Application Theorem (Boyer and Olive (214)) Assume that ω supp a 21 =. Then, (syst) is approximately controllable if, and only if, M k (a 21 φ k, ω), k N. G. Olive The Fattorini-Hautus test 22 / 48

41 Simple conditions for the approximate controllability Corollary (Boyer and Olive (214)) Assume that ω supp a 21 =. 1 Sucient condition : (syst) is approximately controllable if a 21 satises I k = 1 a 21 (φ k ) 2 dx, k N. (7) 2 Necessary condition : if (syst) is approximately controllable and ω, supp a 21 are connected, then (7) has to hold. In general, (7) is not necessary. G. Olive The Fattorini-Hautus test 23 / 48

42 Role of the geometry of the control domain Let us take a look at the particular case ( a 21 (x) = x 1 2 ) 1 O (x), O = supp a 21 = Consider the two following geometric congurations for ω : ( 1 4, 3 ). 4 supp a 21 (a) ω is connected ω ω ω supp a 21 (b) ω is not connected (syst) is not approximately controllable in conguration (a). (syst) is approximately controllable in conguration (b). G. Olive The Fattorini-Hautus test 24 / 48

43 Part IV Stabilization of integro-dierential equations Joint work with Jean-Michel Coron and Long Hu G. Olive The Fattorini-Hautus test 25 / 48

44 The equation We consider u t(t, x) u x (t, x) = u(t, L) = u(, x) = g(x, y)u(t, y) dy U(t) u (x), t (, T ), x (, L), (transp-g) where : T > is the time of control and L > is the length of the domain. u is the initial data and u is the state. g L 2 ((, L) (, L)) is a given kernel. x L u (x) U(t) U L 2 (, T ) is a boundary control. T t G. Olive The Fattorini-Hautus test 26 / 48

45 An application : PDE-ODE systems Example borrowed from Smyshlyaev and Krstic (28) : u t(t, x) u x (t, x) = v(t, x), v xx (t, x) v(t, x) = u(t, x), u(t, L) = U(t), v x (t, ) =, u(, x) = u (x), v(t, L) = V (t). t (, T ), x (, L). Can we nd U, V as functions of u, v such that, for some T >, u(t, ) = v(t, ) =? (remark : u(t, ) = = v(t, ) = ). G. Olive The Fattorini-Hautus test 27 / 48

46 An application : PDE-ODE systems Example borrowed from Smyshlyaev and Krstic (28) : u t(t, x) u x (t, x) = v(t, x), v xx (t, x) v(t, x) = u(t, x), u(t, L) = U(t), v x (t, ) =, u(, x) = u (x), v(t, L) = V (t). t (, T ), x (, L). Can we nd U, V as functions of u, v such that, for some T >, u(t, ) = v(t, ) =? (remark : u(t, ) = = v(t, ) = ). First, we solve the ODE : ( v(t, x) = cosh(x) ) L x V (t) u(t, y) sinh(l y) dy + u(t, y) sinh(x y) dy. cosh(l) }{{}}{{} Fredholm Volterra G. Olive The Fattorini-Hautus test 27 / 48

47 An application : PDE-ODE systems Example borrowed from Smyshlyaev and Krstic (28) : u t(t, x) u x (t, x) = v(t, x), v xx (t, x) v(t, x) = u(t, x), u(t, L) = U(t), v x (t, ) =, u(, x) = u (x), v(t, L) = V (t). t (, T ), x (, L). Can we nd U, V as functions of u, v such that, for some T >, u(t, ) = v(t, ) =? (remark : u(t, ) = = v(t, ) = ). First, we solve the ODE : ( v(t, x) = cosh(x) ) L x V (t) u(t, y) sinh(l y) dy + u(t, y) sinh(x y) dy. cosh(l) }{{}}{{} Fredholm Volterra If we have 2 controls : take V such that v(t, ) = : Volterra integral. G. Olive The Fattorini-Hautus test 27 / 48

48 An application : PDE-ODE systems Example borrowed from Smyshlyaev and Krstic (28) : u t(t, x) u x (t, x) = v(t, x), v xx (t, x) v(t, x) = u(t, x), u(t, L) = U(t), v x (t, ) =, u(, x) = u (x), v(t, L) = V (t). t (, T ), x (, L). Can we nd U, V as functions of u, v such that, for some T >, u(t, ) = v(t, ) =? (remark : u(t, ) = = v(t, ) = ). First, we solve the ODE : ( v(t, x) = cosh(x) ) L x V (t) u(t, y) sinh(l y) dy + u(t, y) sinh(x y) dy. cosh(l) }{{}}{{} Fredholm Volterra If we have 2 controls : take V such that v(t, ) = : Volterra integral. If we have 1 control (V = ) : Fredholm integral. G. Olive The Fattorini-Hautus test 27 / 48

49 Notions of stabilization Stability (U(t) = ) : We say that (transp-g) is exp. stable if the solution u with U(t) = satises u(t) L 2 M ωe ωt u L 2, t, for some ω > and M ω >. stable in nite time T if the solution u with U(t) = satises u(t) =, t T. G. Olive The Fattorini-Hautus test 28 / 48

50 Notions of stabilization Stability (U(t) = ) : We say that (transp-g) is exp. stable if the solution u with U(t) = satises u(t) L 2 M ωe ωt u L 2, t, for some ω > and M ω >. stable in nite time T if the solution u with U(t) = satises u(t) =, t T. Stabilization (U(t) = Fu(t)) : We say that (transp-g) is exp. stabilizable if (transp-g) with U(t) = F ωu(t) is exp. stable. rap. stabilizable if this holds for every ω >. stabilizable in nite time T if (transp-g) with U(t) = Fu(t) is stable in nite time T. G. Olive The Fattorini-Hautus test 28 / 48

51 Relations between controllability and stabilization Stabilization : Finite time stabilization = rap. stabilization = exp. stabilization. Relations : Finite time stabilization = (NC). (NC) = rap. stabilization : Wonham (1967) in nite dimension Datko (1971) for bounded control operators G. Olive The Fattorini-Hautus test 29 / 48

52 Controllability of the transport equation Consider u t(t, x) u x (t, x) =, x L U(t) u(t, L) = U(t), u(, x) = u (x), (transp-) u (x) t (, T ), x (, L). t T Controllability : (transp-) is (exactly, null or approximately) controllable in time T if, and only if, T L. Stabilization : (transp-) is stable in nite time T = L. G. Olive The Fattorini-Hautus test 3 / 48

53 Abstract form of (transp-g ) Let us rewrite (transp-g) in the abstract form in L 2 (, L) : d u = dt Au + BU, t (, T ), u() = u, G. Olive The Fattorini-Hautus test 31 / 48

54 Abstract form of (transp-g ) Let us rewrite (transp-g) in the abstract form in L 2 (, L) : where the unbounded operator A is d u = dt Au + BU, t (, T ), u() = u, Au = u x + g(, y)u(y) dy, with domain D(A) = { u H 1 (, L) u(l) = }, and B L(C, D(A ) ) is BU, z D(A ),D(A ) = Uz(L). G. Olive The Fattorini-Hautus test 31 / 48

55 Abstract form of (transp-g ) Let us rewrite (transp-g) in the abstract form in L 2 (, L) : where the unbounded operator A is d u = dt Au + BU, t (, T ), u() = u, Au = u x + g(, y)u(y) dy, with domain D(A) = { u H 1 (, L) u(l) = }, and B L(C, D(A ) ) is BU, z D(A ),D(A ) = Uz(L). We can show that there exists a unique solution (by transposition) u C ([, T ]; L 2 (, L)). G. Olive The Fattorini-Hautus test 31 / 48

56 References on the stabilization of (transp-g ) We know that : In general, (transp-g) is not stable. G. Olive The Fattorini-Hautus test 32 / 48

57 References on the stabilization of (transp-g ) We know that : In general, (transp-g) is not stable. Indeed, g(x, y) = g large enough, λ >, ker(λ A) {}. G. Olive The Fattorini-Hautus test 32 / 48

58 References on the stabilization of (transp-g ) We know that : In general, (transp-g) is not stable. Indeed, g(x, y) = g large enough, λ >, ker(λ A) {}. (transp-g) is stabilizable in nite time T = L, if x g(x, y) =, x y (Volterra Integral dy). Smyshlyaev and Krstic (28). G. Olive The Fattorini-Hautus test 32 / 48

59 References on the stabilization of (transp-g ) We know that : In general, (transp-g) is not stable. Indeed, g(x, y) = g large enough, λ >, ker(λ A) {}. (transp-g) is stabilizable in nite time T = L, if x g(x, y) =, x y (Volterra Integral dy). Smyshlyaev and Krstic (28). (transp-g) is stabilizable in nite time T = L, if g is small enough. or g(x, y) = g 2(y) with 1 ( L g2(y) L y Argomedo-Bribiesca and Krstic (215). ) e λ (x y) dx dy, where λ = g2(y) dy. G. Olive The Fattorini-Hautus test 32 / 48

60 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) G. Olive The Fattorini-Hautus test 33 / 48

61 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Assumption (E) is satised in many cases : g small, g Volterra, g with separated variables,... G. Olive The Fattorini-Hautus test 33 / 48

62 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Assumption (E) is satised in many cases : g small, g Volterra, g with separated variables,... T = L is the optimal time of control : for g = (transp-g) is (transp-). G. Olive The Fattorini-Hautus test 33 / 48

63 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Assumption (E) is satised in many cases : g small, g Volterra, g with separated variables,... T = L is the optimal time of control : for g = (transp-g) is (transp-). (E) and (Fatt) are dierent. G. Olive The Fattorini-Hautus test 33 / 48

64 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Assumption (E) is satised in many cases : g small, g Volterra, g with separated variables,... T = L is the optimal time of control : for g = (transp-g) is (transp-). (E) and (Fatt) are dierent. In the nite dimensional case, (Fatt) characterizes the rap. stabilization. G. Olive The Fattorini-Hautus test 33 / 48

65 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Assumption (E) is satised in many cases : g small, g Volterra, g with separated variables,... T = L is the optimal time of control : for g = (transp-g) is (transp-). (E) and (Fatt) are dierent. In the nite dimensional case, (Fatt) characterizes the rap. stabilization. (Fatt) can fail for an arbitrary large number of λ. G. Olive The Fattorini-Hautus test 33 / 48

66 Main result and consequences Let us consider the problem : Find θ H 1 (T +) H 1 (T ) such that : θ x (x, y) + θ y (x, y) + g(y, σ)θ(x, σ)dσ = g(y, x), θ(, y) =, θ(l, y) =, x, y (, L). (E) Theorem (Coron, Hu and Olive (216)) Assume that (E) has a solution. Then, (transp-g) is stabilizable in nite time T = L if, and only if, ker(λ A ) ker B = {}, λ C. (Fatt) Assumption (E) is satised in many cases : g small, g Volterra, g with separated variables,... T = L is the optimal time of control : for g = (transp-g) is (transp-). (E) and (Fatt) are dierent. In the nite dimensional case, (Fatt) characterizes the rap. stabilization. (Fatt) can fail for an arbitrary large number of λ. Important corollary : all the notions of controllability/stabilizability are equivalent, under assumption (E). G. Olive The Fattorini-Hautus test 33 / 48

67 Basic idea of Backstepping Find F and P such that d u = Au + B (Fu), dt u() = u. (initial system) transformation P d dt w = A w, w() = w. (target system) where : The target system is stable. P is invertible. Remark : Stability is preserved by change of variables. G. Olive The Fattorini-Hautus test 34 / 48

68 Basic idea of Backstepping Find F and P such that d u = Au + B (Fu), dt u() = u. (initial system) transformation P d dt w = A w, w() = w. (target system) where : The target system is stable. P is invertible. Remark : Stability is preserved by change of variables. In nite dimension, we can take A = A λ with λ > large enough, Coron (215). G. Olive The Fattorini-Hautus test 34 / 48

69 Choice of the target system For equation (transp-g), we choose as target system w t(t, x) w x (t, x) =, w(t, L) =, w(, x) = w (x), which is stable in nite time T = L : t (, + ), x (, L), (targ) w(t, ) =, t L. G. Olive The Fattorini-Hautus test 35 / 48

70 Choice of the transformation We look for P : L 2 (, L) L 2 (, L) in the form P = Id K, where, additionally, K is an integral operator with kernel k : u(t, x) = w(t, x) k(x, y)w(t, y)dy, (Fred-transfo) G. Olive The Fattorini-Hautus test 36 / 48

71 Choice of the transformation We look for P : L 2 (, L) L 2 (, L) in the form P = Id K, where, additionally, K is an integral operator with kernel k : u(t, x) = w(t, x) k(x, y)w(t, y)dy, (Fred-transfo) Goal : Find k such that : (Fred-transfo) maps (targ) into (transp-g). (Fred-transfo) is invertible. G. Olive The Fattorini-Hautus test 36 / 48

72 Choice of the transformation We look for P : L 2 (, L) L 2 (, L) in the form P = Id K, where, additionally, K is an integral operator with kernel k : u(t, x) = w(t, x) k(x, y)w(t, y)dy, (Fred-transfo) Goal : Find k such that : (Fred-transfo) maps (targ) into (transp-g). (Fred-transfo) is invertible. The feedback law F will then be given by the trace at x = L : Fu = k(l, y)(p 1 u)(y) dy. G. Olive The Fattorini-Hautus test 36 / 48

73 Choice of the transformation We look for P : L 2 (, L) L 2 (, L) in the form P = Id K, where, additionally, K is an integral operator with kernel k : u(t, x) = w(t, x) k(x, y)w(t, y)dy, (Fred-transfo) Goal : Find k such that : (Fred-transfo) maps (targ) into (transp-g). (Fred-transfo) is invertible. The feedback law F will then be given by the trace at x = L : Fu = k(l, y)(p 1 u)(y) dy. Fredholm transformations have been used in : Coron and L u (214) for the rap. stabilization of a Korteweg-de Vries equation. Coron and L u (215) for the rap. stabilization of a Kuramoto-Sivashinsky equ. Argomedo-Bribiesca and Krstic (215) for (transp-g). G. Olive The Fattorini-Hautus test 36 / 48

74 Formal derivation of the kernel equation Dierentiating (Fred-transfo) w.r.t t gives u t(t, x) =w t(t, x) k(x, y)w t(t, y)dy =w x (t, x) =w x (t, x) + k(x, y)w y (t, y) dy k y (x, y)w(t, y) dy k(x, L) w(t, L) + k(x, )w(t, ). }{{} = G. Olive The Fattorini-Hautus test 37 / 48

75 Formal derivation of the kernel equation Dierentiating (Fred-transfo) w.r.t t gives u t(t, x) =w t(t, x) k(x, y)w t(t, y)dy =w x (t, x) =w x (t, x) + k(x, y)w y (t, y) dy k y (x, y)w(t, y) dy k(x, L) w(t, L) + k(x, )w(t, ). }{{} = Dierentiating (Fred-transfo) w.r.t x gives u x (t, x) = w x (t, x) + k x (x, y)w(t, y)dy. G. Olive The Fattorini-Hautus test 37 / 48

76 Formal derivation of the kernel equation Dierentiating (Fred-transfo) w.r.t t gives u t(t, x) =w t(t, x) k(x, y)w t(t, y)dy =w x (t, x) =w x (t, x) + k(x, y)w y (t, y) dy k y (x, y)w(t, y) dy k(x, L) w(t, L) + k(x, )w(t, ). }{{} = Dierentiating (Fred-transfo) w.r.t x gives u x (t, x) = w x (t, x) + k x (x, y)w(t, y)dy. On the other hand, ( g(x, y)u(t, y) dy = g(x, y) + ) g(x, σ)k(σ, y) dσ w(t, y) dy. G. Olive The Fattorini-Hautus test 37 / 48

77 Formal derivation of the kernel equation Dierentiating (Fred-transfo) w.r.t t gives u t(t, x) =w t(t, x) k(x, y)w t(t, y)dy =w x (t, x) =w x (t, x) + k(x, y)w y (t, y) dy k y (x, y)w(t, y) dy k(x, L) w(t, L) + k(x, )w(t, ). }{{} = Dierentiating (Fred-transfo) w.r.t x gives u x (t, x) = w x (t, x) + k x (x, y)w(t, y)dy. On the other hand, ( g(x, y)u(t, y) dy = g(x, y) + ) g(x, σ)k(σ, y) dσ w(t, y) dy. As a result, k has to satisfy the following kernel equation : k y (x, y) + k x (x, y) + g(x, σ)k(σ, y)dσ = g(x, y), k(x, ) =. G. Olive The Fattorini-Hautus test 37 / 48

78 The equation of the adjoint kernel Let us introduce the adjoint kernel k (x, y) = k(y, x). Then, k has to verify kx (x, y) + ky (x, y) + g(y, σ)k (x, σ)dσ = g(y, x), k (, y) =, x, y (, L). y L k (, y) L x G. Olive The Fattorini-Hautus test 38 / 48

79 The equation of the adjoint kernel Let us introduce the adjoint kernel k (x, y) = k(y, x). Then, k has to verify kx (x, y) + ky (x, y) + g(y, σ)k (x, σ)dσ = g(y, x), k (x, ) = U(x), k (, y) =, (well posed for every U), x, y (, L). There is an innite number of choices for the kernel. y L k (, y) / / / / / / / / / / U(x) L x G. Olive The Fattorini-Hautus test 38 / 48

80 The equation of the adjoint kernel Let us introduce the adjoint kernel k (x, y) = k(y, x). Then, k has to verify kx (x, y) + ky (x, y) + g(y, σ)k (x, σ)dσ = g(y, x), k (x, ) = U(x), k (, y) =, (well posed for every U), x, y (, L). There is an innite number of choices for the kernel. y L k (, y) / / / / / / / / / / U(x) L x PROBLEM : not every corresponding (Fred-transfo) is invertible. G. Olive The Fattorini-Hautus test 38 / 48

81 Assumption (E) With the assumption (E), we assume that there exists U such that the solution to kx (x, y) + ky (x, y) + g(y, σ)k (x, σ)dσ = g(y, x), k x, y (, L), (x, ) = U(x), k (, y) =, satises the nal condition k (L, ) =. We will prove that (Fred-transfo) is then invertible, if (Fatt) holds. G. Olive The Fattorini-Hautus test 39 / 48

82 Invertibility of the transformation We want to prove that P = Id K is invertible. Clearly, Id K is invertible Id K is invertible. G. Olive The Fattorini-Hautus test 4 / 48

83 Invertibility of the transformation We want to prove that P = Id K is invertible. Clearly, Id K is invertible Id K is invertible. Since K is compact, by the Fredholm alternative Id K is invertible N = ker(id K ) = {}, and dim N < +. G. Olive The Fattorini-Hautus test 4 / 48

84 Invertibility of the transformation We want to prove that P = Id K is invertible. Clearly, Id K is invertible Id K is invertible. Since K is compact, by the Fredholm alternative Id K is invertible N = ker(id K ) = {}, and dim N < +. We can establish that : N ker B, thanks to the nal condition k (L, ) =. N is stable by A, thanks to the kernel equation and N ker B. G. Olive The Fattorini-Hautus test 4 / 48

85 Invertibility of the transformation We want to prove that P = Id K is invertible. Clearly, Id K is invertible Id K is invertible. Since K is compact, by the Fredholm alternative Id K is invertible N = ker(id K ) = {}, and dim N < +. We can establish that : N ker B, thanks to the nal condition k (L, ) =. N is stable by A, thanks to the kernel equation and N ker B. Since N is nite dimensional, A N has at least one eigenfunction : A ξ = λξ, ξ N, ξ. Thus, ξ ker(λ A ) ker B, but a contradiction with (Fatt). ξ, G. Olive The Fattorini-Hautus test 4 / 48

86 Controllability implies stabilization Proposition (Coron, Hu and Olive (216)) Assume that (transp-g) is null-controllable in time T = L. Then, (E) holds and (Fatt) is satised. G. Olive The Fattorini-Hautus test 41 / 48

87 Controllability implies stabilization Proposition (Coron, Hu and Olive (216)) Assume that (transp-g) is null-controllable in time T = L. Then, (E) holds and (Fatt) is satised. Proof : Firstly, we solve the free nonhomogeneous equation : p x (x, y) + p y (x, y) + g(y, σ)p(x, σ)dσ = g(y, x), p(x, ) =, p(, y) =, x, y (, L). G. Olive The Fattorini-Hautus test 41 / 48

88 Controllability implies stabilization Proposition (Coron, Hu and Olive (216)) Assume that (transp-g) is null-controllable in time T = L. Then, (E) holds and (Fatt) is satised. Proof : Firstly, we solve the free nonhomogeneous equation : p x (x, y) + p y (x, y) + g(y, σ)p(x, σ)dσ = g(y, x), p(x, ) =, p(, y) =, Secondly, we pick a control U such that q x (x, y) + q y (x, y) + g(y, σ)q(x, σ)dσ =, q(x, ) = U(x), q(, y) =, q(l, y) = p(l, y), x, y (, L). x, y (, L). G. Olive The Fattorini-Hautus test 41 / 48

89 Controllability implies stabilization Proposition (Coron, Hu and Olive (216)) Assume that (transp-g) is null-controllable in time T = L. Then, (E) holds and (Fatt) is satised. Proof : Firstly, we solve the free nonhomogeneous equation : p x (x, y) + p y (x, y) + g(y, σ)p(x, σ)dσ = g(y, x), p(x, ) =, p(, y) =, Secondly, we pick a control U such that q x (x, y) + q y (x, y) + g(y, σ)q(x, σ)dσ =, q(x, ) = U(x), q(, y) =, q(l, y) = p(l, y), x, y (, L). x, y (, L). Then, θ = p + q. G. Olive The Fattorini-Hautus test 41 / 48

90 Controllability implies stabilization Proposition (Coron, Hu and Olive (216)) Assume that (transp-g) is null-controllable in time T = L. Then, (E) holds and (Fatt) is satised. Proof : Firstly, we solve the free nonhomogeneous equation : p x (x, y) + p y (x, y) + g(y, σ)p(x, σ)dσ = g(y, x), p(x, ) =, p(, y) =, Secondly, we pick a control U such that q x (x, y) + q y (x, y) + g(y, σ)q(x, σ)dσ =, q(x, ) = U(x), q(, y) =, q(l, y) = p(l, y), x, y (, L). x, y (, L). Then, θ = p + q. Remark : The null-controllability assumption is stronger than (Fatt). G. Olive The Fattorini-Hautus test 41 / 48

91 Example of g with separated variables Proposition (Coron, Hu and Olive (216)) Assume that g(x, y) = g 1 (x)g 2 (y). Then, (E) has a solution. G. Olive The Fattorini-Hautus test 42 / 48

92 Example of g with separated variables Proposition (Coron, Hu and Olive (216)) Assume that Then, (E) has a solution. Moreover, (Fatt) is equivalent to g(x, y) = g 1 (x)g 2 (y). ( x ) e λx g 1 (x) e λy g 2 (y) dy dx 1, λ Z(g 2 ), { where Z(g 2 ) = λ C : } L eλy g 2 (y) dy =. G. Olive The Fattorini-Hautus test 42 / 48

93 Example of g with separated variables In particular, if we assume g(x, y) = g 1 (x), then (Fatt) is equivalent to 1 λ k ( λ where λ k = 2kπ L i for k and λ = g 1(x) dx. ) e λ k x g 1 (x) dx 1, k (k Z), (8) G. Olive The Fattorini-Hautus test 43 / 48

94 Example of g with separated variables In particular, if we assume g(x, y) = g 1 (x), then (Fatt) is equivalent to 1 λ k ( λ where λ k = 2kπ L i for k and λ = g 1(x) dx. ) e λ k x g 1 (x) dx 1, k (k Z), (8) Moreover, (8) has to be checked only for a nite number of k since 1 λ k ( λ ) e λ k x g 1 (x) dx. k ± G. Olive The Fattorini-Hautus test 43 / 48

95 Example of g with separated variables In particular, if we assume g(x, y) = g 1 (x), then (Fatt) is equivalent to 1 λ k ( λ where λ k = 2kπ L i for k and λ = g 1(x) dx. ) e λ k x g 1 (x) dx 1, k (k Z), (8) Moreover, (8) has to be checked only for a nite number of k since 1 λ k ( λ ) e λ k x g 1 (x) dx. k ± On the other hand, (8) can fail for an arbitrary large number N of k. For instance : g(x, y) = g 1 (x) = 2 L N k=1 2kπ L ( ) 2kπ sin L x. G. Olive The Fattorini-Hautus test 43 / 48

96 Example of g with separated variables Finally, if then (Fatt) is equivalent to where λ = g 2(y) dy. g(x, y) = g 2 (y), e λy g 2 (y) dy si λ, y g 2 (y) dy 1 si λ =, G. Olive The Fattorini-Hautus test 44 / 48

97 Example of g with separated variables Finally, if then (Fatt) is equivalent to g(x, y) = g 2 (y), e λy g 2 (y) dy si λ, y g 2 (y) dy 1 si λ =, where λ = g 2(y) dy. Equivalent to the condition of Argomedo-Bribiesca and Krstic (215) But the kernels are dierent : x g 2 (y) dy, si (x, y) T +, θ(x, y) = θ(x, y) = g 2 (y) dy, si (x, y) T, (unless λ = ). x x e λ(x y) g 2 (y) dy, G. Olive The Fattorini-Hautus test 44 / 48

98 Part V Perturbation theorems Joint work with Michel Duprez G. Olive The Fattorini-Hautus test 45 / 48

99 First perturbation theorem Let H and U be two Hilbert spaces. Assume that A : D(A ) H H generates a C -semigroup on H. B L(U, H) is bounded. K L(H). Let us form A K = A + K, D(A K ) = D(A ). Theorem (Duprez and Olive, 216) We assume that : T > such that (A, B) is exactly controllable in time T. K is compact. (A K, B) is approximatively controllable in time T. Then, (A K, B) is exactly controllable in time T. This is known as the uniqueness-compactness argument. Introduced in control theory by E. Zuazua (1987). G. Olive The Fattorini-Hautus test 46 / 48

100 Second perturbation theorem Theorem (Duprez and Olive, 216) We assume that : T > such that (A, B) is exactly controllable in time T. K is compact. (A K, B) satises the Fattorini-Hautus test ker(λ A K ) ker B = {}, λ C. Then, (A K, B) is exactly controllable in time T for every T > T. G. Olive The Fattorini-Hautus test 47 / 48

101 Second perturbation theorem Theorem (Duprez and Olive, 216) We assume that : T > such that (A, B) is exactly controllable in time T. K is compact. (A K, B) satises the Fattorini-Hautus test ker(λ A K ) ker B = {}, λ C. Then, (A K, B) is exactly controllable in time T for every T > T. Applications : Controllability of integro-dierential equations. Controllability of systems of wave equations. Controllability of parabolic systems (by transmutation). etc. G. Olive The Fattorini-Hautus test 47 / 48

102 Some references Boundary approximate controllability of some linear parabolic systems, G. Olive, Evol. Equ. Control Theory 3 (214), no. 1, Approximate controllability conditions for some linear 1D parabolic systems with space-dependent coecients, F. Boyer and G. Olive, Math. Control Relat. Fields 4 (214), no. 3, Stabilization and controllability of rst-order integro-dierential hyperbolic equations, J.-M. Coron, L. Hu and G. Olive, J. Funct. Anal. 271 (216), Perturbations of controlled systems, M. Duprez and G. Olive, submitted (216). Thank you for your attention! articles available at G. Olive The Fattorini-Hautus test 48 / 48