Control and Observation for Stochastic Partial Differential Equations

Transcription

Outline 1. A toy model 2. Exact controllability of Stochastic transport equations 3. Some recent controlla
Control and Observation for Stochastic Partial
Differential Equations
Qi Lü
LJLL, UPMC
October 5, 2013
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• Do the controllability problems for stochastic ordinary
differential equations(SDEs for short) and stochastic partial
differential equations deserve to be studied?
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• Yes. I can find many evidences to support my conclusion.
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• 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.
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• Many excellent mathematicians have left their footprints in
the stochastic optimal control theory.
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• A very brief list includes J.-L. Lions, A. Bensoussan,
W. H. Fleming, J.-M. Bismut, P.-L. Lions, etc.
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• 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.
Outline
1. A toy model
2. Exact controllability of Stochastic transport equations
3. Some recent controllability results for SPDEs and open problems
1. A toy model
• Let us recall some basic result for deterministic ODE.
1. A toy model
• Consider the following controlled system:

 dy = Ay + Bu,
dt

y (0) = y0 .
t ∈ (0, T ),
(1)
where A ∈ Rn×n , B ∈ Rn×m , T > 0. System (1) is said to be
exactly controllable on (0, T ) if for any y0 , y1 ∈ Rn , there
exists a u ∈ L2 (0, T ; Rm ) such that y (T ) = y1 .
1. A toy model
• Consider the following controlled system:

 dy = Ay + Bu,
dt

y (0) = y0 .
t ∈ (0, T ),
(1)
where A ∈ Rn×n , B ∈ Rn×m , T > 0. System (1) is said to be
exactly controllable on (0, T ) if for any y0 , y1 ∈ Rn , there
exists a u ∈ L2 (0, T ; Rm ) such that y (T ) = y1 .
µSystem (1) is exactly controllable on (0, T )⇔
• Theorem 1
rank(B, AB, A2 B, · · · , An−1 B) = n.
• Consider a linear stochastic differential equation:
(
dy = Ay + Bu dt + CydW (t),
y (0) = y0 ∈ Rn ,
t ≥ 0,
(2)
where A, C ∈ Rn×n and B ∈ Rn×m , {W (t)}t≥0 is a standard
one dimensional Brownian motion.
(
dy = Ay + Bu dt + CydW (t),
t ≥ 0,
y (0) = y0 ∈ Rn ,
(2)
where A, C ∈ Rn×n and B ∈ Rn×m , {W (t)}t≥0 is a standard
one dimensional Brownian motion.
• System (2) is said to be exactly controllable if for any y0 ∈ Rn
and yT ∈ L2FT (Ω; Rn ), there exists a control
u(·) ∈ L2F (Ω; L2 (0, T ; Rm )) such that the corresponding
solution y (·) of (2) satisfies y (T ) = yT .
• u(·) ∈ L2F (Ω; L2 (0, T ; Rm )) 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.
• L2F (Ω; Rn ) is the natural state space at time T when there is
T
noise.
• L2F (Ω; Rn ) is the natural state space at time T when there is
T
noise.
• In virtue of a result by S. Peng (1994), system (2) is NOT
exactly controllable for any B and any m!
• Thus, we consider the following alternative system:
(
dy = Ay + Bu dt + (Cy + Dv )dW (t),
t ≥ 0,
y (0) = y0 ∈ Rn ,
where A, C , D ∈ Rn×n , B ∈ Rn×m , u ∈ L2F (0, T ; Rm ) and
v ∈ L2F (0, T ; Rn ).
(3)
• Thus, we consider the following alternative system:
(
dy = Ay + Bu dt + (Cy + Dv )dW (t),
t ≥ 0,
y (0) = y0 ∈ Rn ,
where A, C , D ∈ Rn×n , B ∈ Rn×m , u ∈ L2F (0, T ; Rm ) and
v ∈ L2F (0, T ; Rn ).
µSystem (4) is exactly controllable on (0, T )⇔
• Theorem 2
rank(B, AB, A2B, · · · , An−1 B) = n and rank(D) = n.
(3)
• 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) = y0 ∈ Rn .
Is this system exactly controllable when (A, B) satisfies the
Kalman rank condition? NO.
(4)
• Taking v (t) = D −1 Cy (t), we get
(
dy = Ay + Bu dt,
t ≥ 0,
y (0) = y0 ∈ Rn .
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 L2FT (Ω; Rn ).
(4)
• 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 ,
z(T ) = zT ∈ L2FT (Ω; Rn ).
t ∈ [0, T ],
(5)
• The adjoint system of (2) reads:
(
dz = − A⊤ z + C ⊤ Z dt + ZdW ,
t ∈ [0, T ],
z(T ) = zT ∈ L2FT (Ω; Rn ).
(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.
• We only need to prove the following inequality:
|zT |2L2
FT
(Ω;Rn )
≤ CE
Z
0
T
(|B ⊤ z(t)|2 + Z 2 )dt.
(6)
• We only need to prove the following inequality:
|zT |2L2
FT
(Ω;Rn )
≤ CE
Z
T
(|B ⊤ z(t)|2 + Z 2 )dt.
(6)
0
• 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.
• Let T > 0 and G ⊂ Rd (d ∈ N) a strictly convex bounded
domain with the C 1 boundary Γ. Let U ∈ Rd with |U|Rd = 1.
Denote by
Γ−S = {x ∈ Γ : U · ν(x) ≤ 0},
Γ+S = Γ \ Γ−S .
• Let T > 0 and G ⊂ Rd (d ∈ N) a strictly convex bounded
domain with the C 1 boundary Γ. Let U ∈ Rd with |U|Rd = 1.
Denote by
Γ−S = {x ∈ Γ : U · ν(x) ≤ 0},
Γ+S = Γ \ Γ−S .

dy +U ·∇ydt = a1 ydt + a2 y +v dW (t)



y =u



y (0) = y0
Here y0 ∈
L2 (G ),
a1 , a2 ∈
in (0, T )×G ,
on (0, T )×Γ−S ,
in G .
∞
L∞
F (0, T ; L (G )).
(7)
• A solution to the system (7) is a stochastic process
y ∈ L2F (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
Z
Z
y (τ, x)φ(x)dx −
y0 (x)φ(x)dx
G
G
Z Z
Z τZ
τ
−
y (s,x)U ·∇φ(x,U)dxds+
u(s,x)φ(x)U ·νdΓ−S ds
=
0 G
τ Z
Z
0
+
Z
0
0
Γ−S
a1 (s, x)y (s, x)φ(x)dxds
G
τZ
G
a2 (s,x)y (s,x)+v (s,x) φ(x)dxdW (s), P-a.s.
(8)
• For each y0 ∈ L2 (G ), the system (7) admits a unique solution
y . Further, there is a constant C > 0 such that for every y0 ,
it holds that
|y |L2
2
F (Ω;C ([0,T ];L (G )))
≤ e Cr1 E|y0 |L2 (G ) + |u|L2
2
F (0,T ;Lw (Γ−S ))
+ |v |L2
2
F (0,T ;L (G ))
Here r1 = |a1 |2L∞ (0,T ;L∞ (G )) + |a2 |2L∞ (0,T ;L∞ (G )) + 1.
F
F
.
(9)
• System (7) is said to be exactly controllable at time T if for
every initial state y0 ∈ L2 (G ) and every y1 ∈ L2FT (Ω; L2 (G )),
one can find a pair of controls
(u, v ) ∈ L2F (0, T ; L2w (Γ−S )) × L2F (0, T ; L2 (G ))
such that the solution y of the system (7) satisfies that
y (T ) = y1 , P-a.s.
(u, v ) ∈ L2F (0, T ; L2w (Γ−S )) × L2F (0, T ; L2 (G ))
y (T ) = y1 , P-a.s.
• Put R = max |x1 − x2 |Rd .
x1 ,x2 ∈G
(u, v ) ∈ L2F (0, T ; L2w (Γ−S )) × L2F (0, T ; L2 (G ))
y (T ) = y1 , P-a.s.
• Put R = max |x1 − x2 |Rd .
x1 ,x2 ∈G
• Theorem 3: System (7) is exactly controllable at time T ,
provided that T > R.
• 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).
v = 0 in (7).
• 2. Neither u nor v is zero. But v = 0 in (0, T ) × G0 , where
G0 is a nonempty open subset of G .
v = 0 in (7).
• 2. Neither u nor v is zero. But v = 0 in (0, T ) × G0 , where
G0 is a nonempty open subset of G .
• 3. Two controls are acted on the system. But both of them
are in the drift term.
• 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 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 G0 be a nonempty open subset of G . If
v ≡ 0 (0, T ) × G0 , then the system (7) is not exactly
controllable at any time T .
• For the third case, we consider the following controlled
equation:

dy +U ·∇ydt= a1 y +ℓ dt +a2 ydW (t)



y =u



y (0) = y0
Here ℓ ∈
L2F (0, T ; L2 (G ))
is a control.
in (0, T ) × G ,
on (0, T ) × Γ−S ,
in G .
(10)
• For the third case, we consider the following controlled
equation:

dy +U ·∇ydt= a1 y +ℓ dt +a2 ydW (t)



y =u



y (0) = y0
Here ℓ ∈
L2F (0, T ; L2 (G ))
in (0, T ) × G ,
on (0, T ) × Γ−S ,
in G .
(10)
is a control.
• Theorem 6 The system (10) is not exactly controllable for
any T > 0.
• Let us introduce the adjoint system:

dz +U ·∇zdt= b1 z +b2 Z dt+(b3 z +Z )dW (t) in (0, T )×G ,



z =0
on (0,T )×Γ+S ,



z(T ) = zT
in G .
(11)
• Let us introduce the adjoint system:

dz +U ·∇zdt= b1 z +b2 Z dt+(b3 z +Z )dW (t) in (0, T )×G ,



z =0
on (0,T )×Γ+S ,



z(T ) = zT
in G .
(11)
∞
• Here zT ∈ L2F (Ω; L2 (G )),b1 , b2 , b3 ∈ L∞
F (0, T ; L (G )).
T
• A solution to (11) is a pair of stochastic processes
(z, Z ) ∈ CF ([0, T ]; L2 (Ω; L2 (G ))) × L2F (0, T ; L2 (G ))
such that for every ψ ∈ C 1 (G ) with ψ = 0 on Γ−S and
arbitrary τ ∈ [0, T ], it holds that
Z
Z
zT (x)ψ(x)dx −
z(τ, x)ψ(x)dx
G
−
G
Z
τ
=
T
Z
Z TZ
Z TZ
τ
G
τ
+
z(s, x)U · ∇ψ(x)dxds
G
G
b1 (s,x)z(s,x)+b2 (s,x)Z (s,x) ψ(x)dxds
b3 (s,x)z(s,x)+Z (s,x) ψ(x)dxdW (s), P-a.e.
• For any zT ∈ L2 (Ω, FT , P; L2 (G )), the equation (11) admits a
unique solution (z, Z ) satisfiing that
|z|CF ([0,T ];L2 (Ω;L2 (G ))) +|Z |L2 (0,T ;L2 (G )) ≤ e Cr2 |zT |L2
F
where
△
r2 =
3
X
i =1
FT
|bi |4L∞ (0,T ;L∞ (G )) + 1.
F
(Ω;L2 (G )) ,
(12)
• The observability estimate for (11) reads
|zT |L2
FT
(Ω;L2 (G ))
≤ C(b1 ,b2 ,b3 ) |z|L2 (0,T ;L2w (Γ−S )) +|Z |L2 (0,T ;L2 (G )) .
F
F
(13)
|zT |L2
FT
(Ω;L2 (G ))
F
(13)
F
• Proposition 1: Let (z, Z ) ∈ CF ([0, T ]; L2 (Ω; L2 (G )))
×L2F (0, T ; L2 (G )) solve the equation (11) with the terminal
state zT . Then
|z|2L2
2
F (0,T ;Lw (Γ−S ))
≤ e Cr2 E|zT |2L2 (G ) .
|zT |L2
FT
(Ω;L2 (G ))
F
(13)
F
• Proposition 1: Let (z, Z ) ∈ CF ([0, T ]; L2 (Ω; L2 (G )))
×L2F (0, T ; L2 (G )) solve the equation (11) with the terminal
state zT . Then
|z|2L2
2
F (0,T ;Lw (Γ−S ))
≤ e Cr2 E|zT |2L2 (G ) .
• Theorem 7: If T > R, then the inequality (13) holds.
• We prove the inequality (13) by a global Carleman estimate.
• Let 0 < c < 1 such that cT > R. Put
h
T 2 i
l = λ |x|2 − c t −
and θ = e l .
2
(14)
• Let 0 < c < 1 such that cT > R. Put
h
T 2 i
l = λ |x|2 − c t −
and θ = e l .
2
(14)
• Proposition 2: Assume that v is an H 1 (Rn )-valued
continuous semi-martingale. Put p = θv . We have the
following equality
−θ(lt + U · ∇l )p dv + U · ∇vdt
1
1 = − d (lt + U · ∇l )p 2 − U · ∇ (lt + U · ∇l )p 2
2
2
1
+ ltt + U · ∇(U · ∇l ) + 2U · ∇lt p 2
2
1
+ (lt + U · ∇l )(dp)2 + (lt + U · ∇l )2 p 2 .
2
(15)
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
−E
Z
T
0
≥ λE
Z
θ 2 (lt + U · ∇l )z(dz + U · ∇zdt)dxdt
G
Z
(cT − 2U · x)θ 2 (T )z 2 (T )dx
G
+2cλE
Z
T
0
Z
+2(1 − c)λE
Γ−S
Z
0
+E
Z
0
TZ
G
U · ν c(T − 2t) − 2U · x θ 2 z 2 dΓ−S dt
TZ
θ 2 z 2 dxdt
G
θ 2 (lt +U · ∇l ) (b3 z + Z )2 + 2z 2 dxdt.
(16)
By virtue of that z solves the equation (11), we see
Z Z
(cT − U · x)θ 2 (T )z 2 (T )dx
2λE
S d−1
G
+2(1 − c)λE
+E
Z
+E
Z
T
0
0
≤ 3E
Z
TZ
Z
−2cλE
0
T
0
Z
θ 2 z 2 dxdt
G
θ 2 (lt + U · ∇l )(b3 z + Z )2 dxdt
G
(17)
θ 2 (lt + U · ∇l )2 z 2 dxdt
G
TZ
G
Z TZ
0
Z
θ 2 (b12 z 2 +b22 Z 2 )dxdt
U ·ν c(T−2t)−2U · x θ 2 z 2 dΓ−S dt.
Γ−S
Taking
λ1 =
2
X
3
|bi |2L∞ (0,T ;L∞ (G )) + |b3 |4L∞ (0,T ;L∞ (G )) + 1 ,
F
F
2(1 − c)
i =1
for any λ ≥ λ1 , it holds that
3E
Z
0
T
Z
G
θ 2 (b12 + b34 + b32 )z 2 dxdt
≤ 2(1 − c)λE
Z
T
0
Z
(18)
θ 2 z 2 dxdt.
G
• Since cT > R, we find that
E
Z
2
2
θ (T , x)z (T , x)dx ≤ E
0
G
−2cλE
Z TZ
Z TZ
0
θ 2 b22 +2λx −2cλt Z 2 dxdt
G
U ·ν c(T −2t)−2U ·x θ 2 z 2 dΓ−S dt.
Γ−S
(19)
• Since cT > R, we find that
E
Z
2
2
θ (T , x)z (T , x)dx ≤ E
Z TZ
0
G
−2cλE
Z TZ
0
2
Γ−S
e − 2 cλT E
Z
G
≤ Ce
1
λR 2
2
G
U ·ν c(T −2t)−2U ·x θ 2 z 2 dΓ−S dt.
• By the definition of θ, we have e
1
θ 2 b22 +2λx −2cλt Z 2 dxdt
Z
E
≤θ≤e
1
λR 2
2
(19)
. Thus,
zT2 dx
T
0
− 12 cλT 2
Z
2
Z dxdt + E
G
Z
T
0
Z
U · νz dΓ−S dt ,
2
Γ−S
Lack of exact controllability
• We only consider a very special case of the system (7), that
is, G = (0, 1), a1 = 0 and a2 = 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
• Set
(
1, if t ∈ (1 − 2−2i )T , (1 − 2−2i −1 )T , i = 0, 1, · · · ,
η(t) =
−1, otherwise
Z T
• Lemma 1: Let ξ =
η(t)dW (t). It is impossible to find
0
(̺1 , ̺2 ) ∈ L2F (0, T ; R) × L2F (0, T ; R) and x ∈ R with
lim E|̺2 (t) − ̺(T )|2 = 0,
t→T
such that
ξ=x+
Z
T
̺1 (t)dt +
0
Z
T
̺2 (t)dW (t).
0
(20)
• Put
△
V = {v : v ∈ L2F (0, T ; L2 (0, 1)), v = 0 in (0, T ) × G0 }.
Let ξ be given by Lemma 1. Choose a ψ ∈ C0∞ (G0 ) such that
|ψ|L2 (G ) = 1 and set yT = ξψ. We will show that yT cannot
be attained for any y0 ∈ R, u ∈ L2F (0, T ; R) and v ∈ V.
• Put
△
V = {v : v ∈ L2F (0, T ; L2 (0, 1)), v = 0 in (0, T ) × G0 }.
Let ξ be given by Lemma 1. Choose a ψ ∈ C0∞ (G0 ) such that
|ψ|L2 (G ) = 1 and set yT = ξψ. We will show that yT cannot
be attained for any y0 ∈ R, u ∈ L2F (0, T ; R) and v ∈ V.
• If there exist a u ∈ L2F (0, T ; R) and a v ∈ V such that
y (T ) = yT , then by Itô’s formula, we obtain
Z
ξ =
yT ψdx
G
=
Z
=
Z
G
G
Z TZ
Z TZ
y0 ψdx −
ψyx dxdt +
ψ(y + v )dxdW (t)
0
0
G
G
Z T Z
Z T Z
y0 ψdx +
ψx ydx dt +
ψydx dW (t).
0
G
0
G
(21)
• It is clear that both
Z
ψx ydx and
G
Z
ψydx belong to
G
L2F (0, T ; R). Further,
Z
Z
2
lim E
ψy (t)dx −
ψy (T )dx t→T
G
G
Z 2
= lim E
ψ y (t) − y (T ) dx = 0.
t→T
G
These, together with (21), contradict Lemma 1.
3. Some recent controllability results for SPDEs and
open problems
Let us consider the following forward stochastic parabolic equation

X

(aij yi )j dt = [hα, ∇y i + βy + χG0 f ]dt
 dy −



i ,j



+(qy + g ) dW (t) in Q, (22)



y =0
on Σ,





y (0) = y0
in G ,
with suitable coefficients α, β, p and q. In (22), the initial state
y0 ∈ L2F0 (Ω; L2 (G )). The control variable consists of the pair
(f , g ) ∈ L2F (0, T ; L2 (G0 )) × L2F (0, T ; L2 (G )).
• 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!!!
• 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).
Consider the following controlled stochastic heat equation with one
control:

n
X



dy −
(aij yxi )xj dt = a(t)ydW + χE χG0 fdt in Q,



i ,j=1
(23)

y =0
on Σ,





y (0) = y0
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), χE is the
characteristic function of E , y0 ∈ L2 (Ω, F0 , P; L2 (G )), the control
2
2
f belongs to L∞
F (0, T ; L (Ω; L (G ))).
• Theorem 8(Qi Lü, 2011, JFA): System (1) is null controllable
at time T , i.e., for each initial datum y0 ∈ L2F0 (Ω; L2 (G )),
2
2
there is a control f ∈ L∞
F (0, T ; L (Ω; L (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 |2L∞ (0,T ;L2 (Ω;L2 (G ))) ≤ C E|y0 |2L2 (G ) .
F
(24)
• Theorem 8(Qi Lü, 2011, JFA): System (1) is null controllable
at time T , i.e., for each initial datum y0 ∈ L2F0 (Ω; L2 (G )),
2
2
there is a control f ∈ L∞
F (0, T ; L (Ω; L (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 |2L∞ (0,T ;L2 (Ω;L2 (G ))) ≤ C E|y0 |2L2 (G ) .
F
•
(24)
For general stochastic heat equations, the
answer is unknown.
• Consider the following stochastic Schrödinger equation:

idy +∆ydt = (a1 ·∇y +a2 y )dt +(a3 y +g )dW (t)





y =0


y =u



y (0) = y0
in Q,
on Σ \ Σ0 ,
on Σ0 ,
in G .
(25)
• Consider the following stochastic Schrödinger equation:

idy +∆ydt = (a1 ·∇y +a2 y )dt +(a3 y +g )dW (t)





y =0


y =u



y (0) = y0
in Q,
on Σ \ Σ0 ,
on Σ0 ,
in G .
(25)
• Here, the initial state y0 ∈ H −1 (G ), the control
u ∈ L2F (0, T ; L2 (Γ0 )),
g ∈ L2 (0, T ; H −1 (G )),
ak (k = 1, 2, 3) are suitable coefficients.
• The solution to (25) is defined in the sense of transposition
solution, which is not studied in the literature of stochastic
PDEs.
• 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.
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.
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 L2 (Ω; H01 (G )) ⊂ L2 (Ω; L2 (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.
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 ,
z(T ) = zT ∈ L2FT (Ω; Rn ).
t ∈ [0, T ],
(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.
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.
are as follows.
• The null controllability for stochastic heat equations with only
one control.
are as follows.
• 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.
.
Thank you!

Control and Observation for Stochastic Partial Differential Equations

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