Control and Observation for Stochastic Partial Differential Equations

Transcription

1 Control and Observation for Stochastic Partial Differential Equations Qi Lü LJLL, UPMC October 5, 2013

2 Advertisement Do the controllability problems for stochastic ordinary differential equations(sdes for short) and stochastic partial differential equations deserve to be studied?

3 Advertisement Do the controllability problems for stochastic ordinary differential equations(sdes for short) and stochastic partial differential equations deserve to be studied? Yes. I can find many evidences to support my conclusion.

4 Advertisement Do the controllability problems for stochastic ordinary differential equations(sdes for short) and stochastic partial differential equations deserve to be studied? Yes. I can find many evidences to support my conclusion. I only point out that, in the control theory of deterministic ODEs and PDEs, both the optimal control problems and the controllability problems are studied extensively. However, for the stochastic counterpart, the situation is quite different.

5 Advertisement Many excellent mathematicians have left their footprints in the stochastic optimal control theory.

6 Advertisement Many excellent mathematicians have left their footprints in the stochastic optimal control theory. A very brief list includes J.-L. Lions, A. Bensoussan, W. H. Fleming, J.-M. Bismut, P.-L. Lions, etc.

7 Advertisement Many excellent mathematicians have left their footprints in the stochastic optimal control theory. A very brief list includes J.-L. Lions, A. Bensoussan, W. H. Fleming, J.-M. Bismut, P.-L. Lions, etc. The stochastic controllability problems attract very few mathematicians. It seems that this area is almost an uncultivated land.

8 Outline 1. A toy model 2. Exact controllability of Stochastic transport equations 3. Some recent controllability results for SPDEs and open problems

9 1. A toy model Let us recall some basic result for deterministic ODE.

10 1. A toy model Let us recall some basic result for deterministic ODE. Consider the following controlled system: dy = Ay + Bu, dt t y(0) = y 0. (0,T), (1) where A R n n, B R n m, T > 0. System (1) is said to be exactly controllable on (0,T) if for any y 0,y 1 R n, there exists a u L 2 (0,T; R m ) such that y(t) = y 1.

11 1. A toy model Let us recall some basic result for deterministic ODE. Consider the following controlled system: dy = Ay + Bu, dt t y(0) = y 0. (0,T), (1) where A R n n, B R n m, T > 0. System (1) is said to be exactly controllable on (0,T) if for any y 0,y 1 R n, there exists a u L 2 (0,T; R m ) such that y(t) = y 1. Theorem 1ñSystem (1) is exactly controllable on (0,T) rank(b,ab,a 2 B,,A n 1 B) = n.

12 Consider a linear stochastic differential equation: { dy = ( Ay + Bu ) dt + CydW (t), t 0, y(0) = y 0 R n, where A,C R n n and B R n m, {W (t)} t 0 is a standard one dimensional Brownian motion. (2)

13 Consider a linear stochastic differential equation: { dy = ( Ay + Bu ) dt + CydW (t), t 0, y(0) = y 0 R n, where A,C R n n and B R n m, {W (t)} t 0 is a standard one dimensional Brownian motion. System (2) is said to be exactly controllable if for any y 0 R n and y T L 2 F T (Ω; R n ), there exists a control u( ) L 2 F (Ω;L2 (0,T; R m )) such that the corresponding solution y( ) of (2) satisfies y(t) = y T. (2)

14 u( ) L 2 F (Ω;L2 (0,T; R m )) means that one can only utilize the information of the noise {W (s)} 0 s t which can be observed at time t. This is a key point in the study of stochastic control problems.

15 u( ) L 2 F (Ω;L2 (0,T; R m )) means that one can only utilize the information of the noise {W (s)} 0 s t which can be observed at time t. This is a key point in the study of stochastic control problems. L 2 F T (Ω; R n ) is the natural state space at time T when there is noise.

16 u( ) L 2 F (Ω;L2 (0,T; R m )) means that one can only utilize the information of the noise {W (s)} 0 s t which can be observed at time t. This is a key point in the study of stochastic control problems. L 2 F T (Ω; R n ) is the natural state space at time T when there is noise. In virtue of a result by S. Peng (1994), system (2) is NOT exactly controllable for any B and any m!

17 Thus, we consider the following alternative system: { dy = ( Ay + Bu ) dt + (Cy + Dv)dW (t), t 0, y(0) = y 0 R n, where A,C,D R n n, B R n m, u L 2 F (0,T; Rm ) and v L 2 F (0,T; Rn ). (3)

18 Thus, we consider the following alternative system: { dy = ( Ay + Bu ) dt + (Cy + Dv)dW (t), t 0, y(0) = y 0 R n, where A,C,D R n n, B R n m, u L 2 F (0,T; Rm ) and v L 2 F (0,T; Rn ). Theorem 2ñSystem (4) is exactly controllable on (0,T) rank(b,ab,a 2 B,,A n 1 B) = n and rank(d) = n. (3)

19 Is Theorem 2 a trivial corollary of Theorem 1?

20 Is Theorem 2 a trivial corollary of Theorem 1? Taking v(t) = D 1 Cy(t), we get { dy = ( Ay + Bu ) dt, t 0, y(0) = y 0 R n. Is this system exactly controllable when (A,B) satisfies the Kalman rank condition? NO. (4)

21 Is Theorem 2 a trivial corollary of Theorem 1? Taking v(t) = D 1 Cy(t), we get { dy = ( Ay + Bu ) dt, t 0, y(0) = y 0 R n. Is this system exactly controllable when (A,B) satisfies the Kalman rank condition? NO. The reason is that we expect the final state could be any element in L 2 F T (Ω; R n ). (4)

22 How to prove Theorem 2? Following the idea for deterministic system, we should study the observability estimate for the adjoint system of (2).

23 How to prove Theorem 2? Following the idea for deterministic system, we should study the observability estimate for the adjoint system of (2). The adjoint system of (2) reads: { dz = ( A z + C Z ) dt + ZdW, t [0,T], z(t) = z T L 2 F T (Ω; R n ). (5)

24 How to prove Theorem 2? Following the idea for deterministic system, we should study the observability estimate for the adjoint system of (2). The adjoint system of (2) reads: { dz = ( A z + C Z ) dt + ZdW, t [0,T], z(t) = z T L 2 F T (Ω; R n ). (5) Equation (5) is a backward stochastic ODE. Such kind of equations were first introduced by J. Bismut in 1970s and studied extensively in the last twenty years by E. Pardoux, S. Peng, N. El Karoui, R. Buckdahn, X. Zhou, J. Yong, etc.

25 We only need to prove the following inequality: T z T 2 L 2 F (Ω;R n ) CE ( B z(t) 2 + Z 2 )dt. (6) T 0

26 We only need to prove the following inequality: T z T 2 L 2 F (Ω;R n ) CE ( B z(t) 2 + Z 2 )dt. (6) T Please note that we now deal with infinite dimensional space. One cannot simply mimic the proof for observability estimate for ODE to obtain (6). For example, in this case, unique continuation does not imply the observability estimate. 0

27 2. Exact controllability of Stochastic transport equations Let T > 0 and G R d (d N) a strictly convex bounded domain with the C 1 boundary Γ. Let U R d with U R d = 1. Denote by Γ S = {x Γ : U ν(x) 0}, Γ +S = Γ \ Γ S.

28 2. Exact controllability of Stochastic transport equations Let T > 0 and G R d (d N) a strictly convex bounded domain with the C 1 boundary Γ. Let U R d with U R d = 1. Denote by Γ S = {x Γ : U ν(x) 0}, Γ +S = Γ \ Γ S. Consider a linear stochastic differential equation: dy+u ydt = a 1 ydt+ [ a 2 y+v ] dw (t) in (0,T) G, y = u on (0,T) Γ S, y(0) = y 0 in G. (7) Here y 0 L 2 (G), a 1,a 2 L F (0,T;L (G)).

29 A solution to the system (7) is a stochastic process y L 2 F (Ω;C([0,T];L2 (G))) such that for every τ [0,T] and every φ C 1 (G), φ = 0 on Γ +S, it holds that y(τ, x)φ(x)dx y 0 (x)φ(x)dx = G τ 0 G τ + 0 G τ 0 G y(s,x)u φ(x,u)dxds+ G τ a 1 (s,x)y(s,x)φ(x)dxds 0 Γ S u(s,x)φ(x)u νdγ S ds [ a2 (s,x)y(s,x)+v(s,x) ] φ(x)dxdw(s),p-a.s. (8)

30 For each y 0 L 2 (G), the system (7) admits a unique solution y. Further, there is a constant C > 0 such that for every y 0, it holds that y L 2 F (Ω;C([0,T];L 2 (G))) e Cr 1 ( E y 0 L 2 (G) + u L 2 F (0,T;L 2 w (Γ S)) + v L 2 F (0,T;L2 (G))). (9) Here r 1 = a 1 2 L F (0,T;L (G)) + a 2 2 L F (0,T;L (G)) + 1.

31 System (7) is said to be exactly controllable at time T if for every initial state y 0 L 2 (G) and every y 1 L 2 F T (Ω;L 2 (G)), one can find a pair of controls (u,v) L 2 F (0,T;L2 w (Γ S)) L 2 F (0,T;L2 (G)) such that the solution y of the system (7) satisfies that y(t) = y 1, P-a.s.

32 System (7) is said to be exactly controllable at time T if for every initial state y 0 L 2 (G) and every y 1 L 2 F T (Ω;L 2 (G)), one can find a pair of controls (u,v) L 2 F (0,T;L2 w (Γ S)) L 2 F (0,T;L2 (G)) such that the solution y of the system (7) satisfies that y(t) = y 1, P-a.s. Put R = max x 1 x 2 R d. x 1,x 2 G

33 System (7) is said to be exactly controllable at time T if for every initial state y 0 L 2 (G) and every y 1 L 2 F T (Ω;L 2 (G)), one can find a pair of controls (u,v) L 2 F (0,T;L2 w (Γ S)) L 2 F (0,T;L2 (G)) such that the solution y of the system (7) satisfies that y(t) = y 1, P-a.s. Put R = max x 1 x 2 R d. x 1,x 2 G Theorem 3: System (7) is exactly controllable at time T, provided that T > R.

34 We put two controls on the system. Moreover, the control v acts on the whole domain. Compared with the deterministic transport solution, it seems that our choice of controls is too restrictive. One may consider the following four weaker cases for designing the control.

35 We put two controls on the system. Moreover, the control v acts on the whole domain. Compared with the deterministic transport solution, it seems that our choice of controls is too restrictive. One may consider the following four weaker cases for designing the control. 1. Only one control is acted on the system, that is, u = 0 or v = 0 in (7).

36 We put two controls on the system. Moreover, the control v acts on the whole domain. Compared with the deterministic transport solution, it seems that our choice of controls is too restrictive. One may consider the following four weaker cases for designing the control. 1. Only one control is acted on the system, that is, u = 0 or v = 0 in (7). 2. Neither u nor v is zero. But v = 0 in (0,T) G 0, where G 0 is a nonempty open subset of G.

37 We put two controls on the system. Moreover, the control v acts on the whole domain. Compared with the deterministic transport solution, it seems that our choice of controls is too restrictive. One may consider the following four weaker cases for designing the control. 1. Only one control is acted on the system, that is, u = 0 or v = 0 in (7). 2. Neither u nor v is zero. But v = 0 in (0,T) G 0, where G 0 is a nonempty open subset of G. 3. Two controls are acted on the system. But both of them are in the drift term.

38 Theorem 4: If u 0 or v 0 in the system (7), then the system (7) is not exactly controllable at any time T.

39 Theorem 4: If u 0 or v 0 in the system (7), then the system (7) is not exactly controllable at any time T. Theorem 5: Let G 0 be a nonempty open subset of G. If v 0 (0,T) G 0, then the system (7) is not exactly controllable at any time T.

40 For the third case, we consider the following controlled equation: dy+u ydt= [ a 1 y+l ] dt+a 2 ydw (t) in (0,T) G, y = u on (0,T) Γ S, y(0) = y 0 in G. (10) Here l L 2 F (0,T;L2 (G)) is a control.

41 For the third case, we consider the following controlled equation: dy+u ydt= [ a 1 y+l ] dt+a 2 ydw (t) in (0,T) G, y = u on (0,T) Γ S, y(0) = y 0 in G. (10) Here l L 2 F (0,T;L2 (G)) is a control. Theorem 6 The system (10) is not exactly controllable for any T > 0.

42 Let us introduce the adjoint system: dz+u zdt= ( b 1 z+b 2 Z ) dt+(b 3 z+z)dw (t) in (0,T) G, z = 0 on (0,T) Γ +S, z(t) = z T in G. (11)

43 Let us introduce the adjoint system: dz+u zdt= ( b 1 z+b 2 Z ) dt+(b 3 z+z)dw (t) in (0,T) G, z = 0 on (0,T) Γ +S, z(t) = z T in G. (11) Here z T L 2 F T (Ω;L 2 (G)),b 1,b 2,b 3 L F (0,T;L (G)).

44 A solution to (11) is a pair of stochastic processes (z,z) C F ([0,T];L 2 (Ω;L 2 (G))) L 2 F(0,T;L 2 (G)) such that for every ψ C 1 (G) with ψ = 0 on Γ S and arbitrary τ [0,T], it holds that z T (x)ψ(x)dx z(τ, x)ψ(x)dx G T = + τ T G τ G T τ G G z(s, x)u ψ(x)dxds [ b1 (s,x)z(s,x)+b 2 (s,x)z(s,x) ] ψ(x)dxds [ b3 (s,x)z(s,x)+z(s,x) ] ψ(x)dxdw(s), P-a.e.

45 For any z T L 2 (Ω, F T,P;L 2 (G)), the equation (11) admits a unique solution (z,z) satisfiing that z CF ([0,T];L 2 (Ω;L 2 (G)))+ Z L 2 F (0,T;L 2 (G)) ecr 2 z T L 2 FT (Ω;L 2 (G)), where 3 r 2 = b i 4 L F (0,T;L (G)) + 1. i=1 (12)

46 The observability estimate for (11) reads z T L 2 FT (Ω;L 2 (G)) C(b 1,b 2,b 3 ) ( z L 2 F (0,T;L 2 w (Γ S)) + Z ) (13) L 2 F (0,T;L2 (G)).

47 The observability estimate for (11) reads z T L 2 FT (Ω;L 2 (G)) C(b 1,b 2,b 3 ) ( z L 2 F (0,T;L 2 w (Γ S)) + Z ) (13) L 2 F (0,T;L2 (G)). Proposition 1: Let (z,z) C F ([0,T];L 2 (Ω;L 2 (G))) L 2 F (0,T;L2 (G)) solve the equation (11) with the terminal state z T. Then z 2 L 2 F (0,T;L2 w (Γ S)) ecr 2 E z T 2 L 2 (G).

48 The observability estimate for (11) reads z T L 2 FT (Ω;L 2 (G)) C(b 1,b 2,b 3 ) ( z L 2 F (0,T;L 2 w (Γ S)) + Z ) (13) L 2 F (0,T;L2 (G)). Proposition 1: Let (z,z) C F ([0,T];L 2 (Ω;L 2 (G))) L 2 F (0,T;L2 (G)) solve the equation (11) with the terminal state z T. Then z 2 L 2 F (0,T;L2 w (Γ S)) ecr 2 E z T 2 L 2 (G). Theorem 7: If T > R, then the inequality (13) holds.

49 We prove the inequality (13) by a global Carleman estimate.

50 We prove the inequality (13) by a global Carleman estimate. Let 0 < c < 1 such that ct > R. Put [ ( l = λ x 2 c t T ) 2 ] and θ = e l. (14) 2

51 We prove the inequality (13) by a global Carleman estimate. Let 0 < c < 1 such that ct > R. Put [ ( l = λ x 2 c t T ) 2 ] and θ = e l. (14) 2 Proposition 2: Assume that v is an H 1 (R n )-valued continuous semi-martingale. Put p = θv. We have the following equality θ(l t + U l)p [ dv + U vdt ] = 1 2 d[ (l t + U l)p 2] 1 2 U [ (l t + U l)p 2] + 1 [ ] ltt + U (U l) + 2U l t p 2 2 (15) (l t + U l)(dp) 2 + (l t + U l) 2 p 2.

52 Proof of the observability estimate: We apply Proposition 2 to the equation (11) with v = z, integrating (15) on (0,T) G, and taking mathematical expectation, then we get that T E 0 λe G G T θ 2 (l t + U l)z(dz + U zdt)dxdt (ct 2U x)θ 2 (T)z 2 (T)dx +2cλE U ν [ c(t 2t) 2U x ] θ 2 z 2 dγ S dt 0 Γ S T +2(1 c)λe θ 2 z 2 dxdt +E T 0 G 0 G θ 2 (l t +U l) [ (b 3 z + Z) 2 + 2z 2] dxdt. (16)

53 By virtue of that z solves the equation (11), we see 2λE (ct U x)θ 2 (T)z 2 (T)dx G S d 1 T +2(1 c)λe θ 2 z 2 dxdt T +E +E 3E 0 T 2cλE G 0 G T 0 G T 0 0 G θ 2 (l t + U l)(b 3 z + Z) 2 dxdt θ 2 (l t + U l) 2 z 2 dxdt θ 2 (b 2 1z 2 +b 2 2Z 2 )dxdt Γ S U ν [ c(t 2t) 2U x ] θ 2 z 2 dγ S dt. (17)

54 Taking 3 ( 2 λ 1 = b i 2 L 2(1 c) F (0,T;L (G)) + b 3 4 L F (0,T;L (G)) ), + 1 i=1 for any λ λ 1, it holds that T 3E θ 2 (b1 2 + b3 4 + b3)z 2 2 dxdt 0 G T 2(1 c)λe θ 2 z 2 dxdt. 0 G (18)

55 Since ct > R, we find that T E θ 2 (T,x)z 2 (T,x)dx E θ 2( b2 2 +2λx 2cλt) Z 2 dxdt G 0 G T 2cλE U ν [ c(t 2t) 2U x ] θ 2 z 2 dγ S dt. 0 Γ S (19)

56 Since ct > R, we find that T E θ 2 (T,x)z 2 (T,x)dx E θ 2( b2 2 +2λx 2cλt) Z 2 dxdt G 0 G T 2cλE U ν [ c(t 2t) 2U x ] θ 2 z 2 dγ S dt. 0 Γ S (19) By the definition of θ, we have e 1 2 cλt2 θ e 1 2 λr2. Thus, e 1 2 cλt2 E zt 2 dx Ce 1 2 λr2 (E G T 0 G T ) Z 2 dxdt + E 0 U νz 2 dγ S dt Γ S,

57 Lack of exact controllability We only consider a very special case of the system (7), that is, G = (0,1), a 1 = 0 and a 2 = 1. The argument for the general case is very similar.

58 Lack of exact controllability We only consider a very special case of the system (7), that is, G = (0,1), a 1 = 0 and a 2 = 1. The argument for the general case is very similar. Set { [ 1, if t (1 2 2i )T,(1 2 2i 1 )T ), i = 0,1,, η(t) = 1, otherwise

59 Lack of exact controllability We only consider a very special case of the system (7), that is, G = (0,1), a 1 = 0 and a 2 = 1. The argument for the general case is very similar. Set { [ 1, if t (1 2 2i )T,(1 2 2i 1 )T ), i = 0,1,, η(t) = 1, otherwise Lemma 1: Let ξ = T 0 η(t)dw (t). It is impossible to find ( 1, 2) L 2 F (0,T; R) L2 F (0,T; R) and x R with such that lim E 2(t) (T) 2 = 0, t T T T ξ = x + 1(t)dt + 2(t)dW (t). (20) 0 0

60 Put V = {v : v L 2 F(0,T;L 2 (0,1)), v = 0 in (0,T) G 0 }. Let ξ be given by Lemma 1. Choose a ψ C0 (G 0) such that ψ L 2 (G) = 1 and set y T = ξψ. We will show that y T cannot be attained for any y 0 R, u L 2 F (0,T; R) and v V.

61 Put V = {v : v L 2 F(0,T;L 2 (0,1)), v = 0 in (0,T) G 0 }. Let ξ be given by Lemma 1. Choose a ψ C0 (G 0) such that ψ L 2 (G) = 1 and set y T = ξψ. We will show that y T cannot be attained for any y 0 R, u L 2 F (0,T; R) and v V. If there exist a u L 2 F (0,T; R) and a v V such that y(t) = y T, then by Itô s formula, we obtain ξ = y T ψdx = = G G G y 0 ψdx y 0 ψdx+ T 0 G T 0 ( ψy x dxdt+ G T 0 ) ψ x ydx dt+ G T 0 ψ(y + v)dxdw(t) ( G ) ψydx dw (t). (21)

62 It is clear that both L 2 F G ψ x ydx and G ψydx belong to (0,T; R). Further, lim E ψy(t)dx ψy(t)dx 2 t T G G = lim ψ [ y(t) y(t) ] dx 2 = 0. t T E G These, together with (21), contradict Lemma 1.

63 3. Some recent controllability results for SPDEs and open problems Let us consider the following forward stochastic parabolic equation dy (a ij y i ) j dt = [ α, y + βy + χ G0 f ]dt i,j +(qy + g)dw (t) in Q, y = 0 on Σ, y(0) = y 0 in G, (22) with suitable coefficients α,β,p and q. In (22), the initial state y 0 L 2 F 0 (Ω;L 2 (G)). The control variable consists of the pair (f,g) L 2 F (0,T;L2 (G 0 )) L 2 F (0,T;L2 (G)).

64 The null controllability of (22) is established by S. Tang and X. Zhang(2009, SICON). From which, they proved that system (22) is null controllable with two controls, i.e., one act on the drift term and the other act on the diffusion term. The control in the diffusion term have to be acted on the whole domain where the solution of system (22) defined!!!

65 The null controllability of (22) is established by S. Tang and X. Zhang(2009, SICON). From which, they proved that system (22) is null controllable with two controls, i.e., one act on the drift term and the other act on the diffusion term. The control in the diffusion term have to be acted on the whole domain where the solution of system (22) defined!!! Is it possible to use one control to drive the solution to zero? Positive answer is given by Lü(2011,JFA) for a special case of system (22).

66 Consider the following controlled stochastic heat equation with one control: n dy (a ij y xi ) xj dt = a(t)ydw + χ E χ G0 fdt in Q, i,j=1 (23) y = 0 on Σ, y(0) = y 0 in G, where a(t) L F (0,T), E is a measurable subset in (0,T) with a positive Lebesgue measure (i.e., m(e) > 0), χ E is the characteristic function of E, y 0 L 2 (Ω, F 0,P;L 2 (G)), the control f belongs to L F (0,T;L2 (Ω;L 2 (G))).

67 Theorem 8(Qi Lü, 2011, JFA): System (1) is null controllable at time T, i.e., for each initial datum y 0 L 2 F 0 (Ω;L 2 (G)), there is a control f L F (0,T;L2 (Ω;L 2 (G))) such that the solution y of system (1) satisfies y(t) = 0 in G, P-a.s. Moreover, the control f satisfies the following estimate: f 2 L F (0,T;L2 (Ω;L 2 (G))) CE y 0 2 L 2 (G). (24)

68 Theorem 8(Qi Lü, 2011, JFA): System (1) is null controllable at time T, i.e., for each initial datum y 0 L 2 F 0 (Ω;L 2 (G)), there is a control f L F (0,T;L2 (Ω;L 2 (G))) such that the solution y of system (1) satisfies y(t) = 0 in G, P-a.s. Moreover, the control f satisfies the following estimate: f 2 L F (0,T;L2 (Ω;L 2 (G))) CE y 0 2 L 2 (G). (24) For general stochastic heat equations, the answer is unknown.

69 Consider the following stochastic Schrödinger equation: idy+ ydt =(a 1 y+a 2 y)dt+(a 3 y+g)dw (t) in Q, y = 0 on Σ \ Σ 0, y = u on Σ 0, y(0) = y 0 in G. (25)

70 Consider the following stochastic Schrödinger equation: idy+ ydt =(a 1 y+a 2 y)dt+(a 3 y+g)dw (t) in Q, y = 0 on Σ \ Σ 0, y = u on Σ 0, y(0) = y 0 in G. (25) Here, the initial state y 0 H 1 (G), the control u L 2 F(0,T;L 2 (Γ 0 )), g L 2 (0,T;H 1 (G)), a k (k = 1,2,3) are suitable coefficients.

71 The solution to (25) is defined in the sense of transposition solution, which is not studied in the literature of stochastic PDEs.

72 The solution to (25) is defined in the sense of transposition solution, which is not studied in the literature of stochastic PDEs. Theorem 9(Qi Lü, 2013, JDE): System (1) is exactly controllable at any time T > 0.

73 The main difficulty of the study of the control and observation problems for SPDEs. 1. The lack of desired theory, i.e., the well-posedness results of the nonhomogeneous boundary value problems for SPDEs, the propagation of singularities results of the solution for SPDEs, etc.

74 The main difficulty of the study of the control and observation problems for SPDEs. 1. The lack of desired theory, i.e., the well-posedness results of the nonhomogeneous boundary value problems for SPDEs, the propagation of singularities results of the solution for SPDEs, etc. 2. The stochastic settings lead some useful methods invalid, for example, the lost of the compact embedding for the state spaces, i.e., although L 2 (Ω;H 1 0 (G)) L2 (Ω;L 2 (G)), the embedding is not compact, which violates the compactness -uniqueness argument. Another example is that the irregularity of the solution with respect to the time variable, for example, it is very hard to estimate the time derivative of the solutions, which is an obstacle to follow the method by Fernández Cara-Guerrero-Imanuvilov-Puel to study the null controllability of the stochastic Navier-Stokes equation.

75 The main difficulty of the study of the control and observation problems for SPDEs. 3. Generally speaking, the adjoint system is a backward SPDE, whose solution is constituted by two stochastic process. For example, let us recall the adjoint system of (4) is as follows: { dz = ( A z + C Z ) dt + ZdW, t [0,T], z(t) = z T L 2 F T (Ω; R n ). (26) When studying the observability problems, people now usually treat Z as a nonhomogeneous term. It is hard to use the fact that Z is a part of the solution.

76 This area is full of open problems. In my opinion, some of them are as follows. The observability estimate for stochastic wave equations under the Geometric Control Conditions.

77 This area is full of open problems. In my opinion, some of them are as follows. The observability estimate for stochastic wave equations under the Geometric Control Conditions. The null controllability for stochastic heat equations with only one control.

78 This area is full of open problems. In my opinion, some of them are as follows. The observability estimate for stochastic wave equations under the Geometric Control Conditions. The null controllability for stochastic heat equations with only one control. The null controllability of stochsatic wave equations, stochastic Schrödinger equations, etc. For this, one need to show that either one control is enough or, as the exact controllability problems, two controls are necessary.

79 . Thank you!