Contrôlabilité de quelques systèmes gouvernés par des equations paraboliques.
Michel Duprez
LMB, université de Franche-Comté
26 novembre 2015
Soutenance de thèse
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
Setting
Let T > 0, N ∈ N ∗ , Ω be a bounded domain in R N with a C 2 -class boundary ∂Ω and ω be a nonempty open subset of Ω.
We consider here the following system of n linear parabolic equations with m controls
∂ t y = ∆y + G · ∇ y + Ay + B 1 ω u in Q T ,
y = 0 on Σ T ,
y (0) = y 0 in Ω,
(S)
where Q T := Ω × (0, T ), Σ T := ∂Ω × (0, T ), A := (a ij ) ij , G := (g ij ) ij and B := (b ik ) ik with a ij , b ik ∈ L ∞ (Q T ) and g ij ∈ L ∞ (Q T ; R N ) for all 1 6 i, j 6 n and 1 6 k 6 m.
We denote by
P : R p × R n−p → R p ,
(y 1 , y 2 ) 7→ y 1 .
Definitions
We will say that system (S) is
null controllable on (0, T ) if for all y 0 ∈ L 2 (Ω) n , there exists a u ∈ L 2 (Q T ) m such that
y ( · , T ; y 0 , u) = 0 in Ω.
approximately controllable on (0, T ) if for all ε > 0 and y 0 , y T ∈ L 2 (Ω) n there exists a control u ∈ L 2 (Q T ) m such that
k y ( · , T ; y 0 , u) − y T ( · ) k L
2(Ω)
n6 ε.
partially null controllable on (0, T ) if for all y 0 ∈ L 2 (Ω) n , there exists a u ∈ L 2 (Q T ) m such that
Py ( · , T ; y 0 , u) = 0 in Ω.
partially approximately controllable on (0, T ) if for all ε > 0 and y 0 , y T ∈ L 2 (Ω) n there exists a control u ∈ L 2 (Q T ) m such that
k Py ( · , T ; y 0 , u) − Py T ( · ) k L
2(Ω)
p6 ε.
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
System of differential equations
Assume that A ∈ L ( R n ) and B ∈ L ( R m , R n ) and consider the system ∂ t y = Ay + Bu in (0, T ),
y (0) = y 0 ∈ R n . (SDE) P : R p × R n−p → R p , (y 1 , y 2 ) 7→ y 1 . T HEOREM (see [1])
System (SDE) is controllable on (0, T ) if and only if rank(B | AB | ... | A n−1 B) = n.
T HEOREM
System (SDE) is partially controllable on (0, T ) if and only if rank(PB | PAB | ... | PA n−1 B) = p.
[1] R. E. Kalman, P. L. Falb and M. A. Arbib, Topics in mathematical system theory.
McGraw-Hill Book Co., New-York, 1969.
Constant matrices
Assume that A ∈ L ( R n ) and B ∈ L ( R m , R n ) and consider the system
∂ t y = ∆y + Ay + B 1 ω u in Q T ,
y = 0 on Σ T ,
y (0) = y 0 in Ω.
(S)
T HEOREM (F. Ammar Khodja, A. Benabdallah, C. Dupaix and M.
González-Burgos, 2009)
System (S) is null controllable on (0, T ) if and only if rank(B | AB | ... | A n−1 B) = n.
T HEOREM (F. Ammar Khodja, F. Chouly, M.D., 2015)
System (S) is partially null controllable on (0, T ) if and only if
rank(PB | PAB | ... | PA n−1 B) = p.
Time dependent matrices
Let us suppose that A ∈ C n−1 ([0, T ]; L ( R n )) and B ∈ C n ([0, T ]; L ( R m ; R n )) and define
B 0 (t ) := B(t),
B i (t ) := A(t)B i−1 (t) − ∂ t B i−1 (t), 1 6 i 6 n − 1.
We can define for all t ∈ [0, T ], [A | B](t ) := (B 0 (t) | B 1 (t ) | ... | B n−1 (t )).
T HEOREM (F. Ammar Khodja, A. Benabdallah, C. Dupaix and M.
González-Burgos, 2009)
If there exist t 0 ∈ [0, T ] such that
rank[A | B](t 0 ) = n, then System (S) is null controllable on (0,T).
T HEOREM (F. Ammar Khodja, F. Chouly, M. D., 2015) If
rank P[A | B](T ) = p,
then system (S) is partially null controllable on (0,T).
Space dependent matrices:
Example of non partial null controllability
Consider the control problem
∂ t y 1 = ∆y 1 + αy 2 + 1 ω u in Q T ,
∂ t y 2 = ∆y 2 in Q T .
y = 0 on Σ T ,
y (0) = y 0 , y 1 (T ) = 0 in Ω.
T HEOREM (F. Ammar Khodja, F. Chouly, M.D., 2015) Assume that Ω := (0, 2π) and ω ⊂ (π, 2π). Let us consider α ∈ L ∞ (0, 2π) defined by
α(x ) :=
∞
X
j=1
1
j 2 cos(15jx) for all x ∈ (0, 2π).
Then system (S 2×2 ) is not partially null controllable.
Space dependent matrices:
Example of partial null controllability
Consider the control problem
∂ t y 1 = ∆y 1 + αy 2 + 1 ω u in Q T ,
∂ t y 2 = ∆y 2 in Q T ,
y = 0 on Σ T ,
y (0) = y 0 , y 1 (T ) = 0 in Ω.
(S 2×2 )
T HEOREM (F. Ammar Khodja, F. Chouly, M.D., 2015)
Let be Ω := (0, π) and α = P α p cos(px) ∈ L ∞ (Ω) satisfying
| α p | 6 C 1 e −C
2p for all p ∈ N , where C 1 > 0 and C 2 > π are two constants.
Then system (S 2×2 ) is partially null controllable.
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
Numerical illustration
Let us consider the fonctionnal HUM J ε (u) = 1
2 Z
ω×(0,T )
u 2 dxdt + 1 2ε
Z
Ω
y 1 (T ; y 0 , u) 2 dx.
T HEOREM (Following the ideas of [1])
1
System (S 2x2 ) is partially approx. controllable from y 0 iff y 1 (T ; y 0 , u ε ) −→
ε→0 0.
2
System (S 2x2 ) is partially null controllable from y 0 iff M y 2
0,T := 2 sup
ε>0
inf
L
2(Q
T) J ε
< ∞ . In this case k y 1 (T ; y 0 , u ε ) k Ω 6 M y
0,T √
ε.
[1] F. Boyer, On the penalised HUM approach and its applications to the numerical
approximation of null-controls for parabolic problems, ESAIM Proc. 41 (2013).
10 −3 10 −2 10 −5
10 −4 10 −3 10 −2 10 −1 10 0 10 1
h
inf L
2δt
(0,T;U
h) F ε h,δt (slope =-0.103) k u ε h,δt k L
2δt(0,T ;U
h) (slope =-6.34e-2) k y ε h,δt (T ) k E
h(slope =2.74)
Figure : Case α = 1.
10 −3 10 −2 10 −4
10 −3 10 −2 10 −1 10 0 10 1
h
inf L
2δt
(0,T ;U
h) F ε h,δt (slope =-3.80) k u h,δt ε k L
2δt(0,T ;U
h) (slope =-1.70) k y ε h,δt (T ) k E
h(slope =8.34e-2)
Figure : Case α =
∞
P
j=1 1
j2
cos(15jx).
t 0.0
0.5 1.0
1.5 2.0
x 0
1 2
3 4 5 6 7
y1
−4
−3
−2
−1 0 1 2
Figure : Evolution of y
1in system (S
2x2) in the case α = 1.
t 0.0
0.5 1.0
1.5 2.0
x 0
1 2
3
4 5 6 7 y1
−0.20
−0.15
−0.10
−0.05 0.00 0.05 0.10 0.15 0.20
Figure : Evolution of y
1in system (S
2x2) in the case α =
∞
P
j=1 1
j2
cos(15jx).
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
Setting
Let T > 0, Ω be a bounded domain in R N (N ∈ N ∗ ) of class C 2 and ω be an nonempty open subset of Ω. Consider the system
∂ t y 1 = div(d 1 ∇ y 1 ) + g 11 · ∇ y 1 + g 12 · ∇ y 2 + a 11 y 1 + a 12 y 2 + 1 ω u in Q T ,
∂ t y 2 = div(d 2 ∇ y 2 ) + g 21 · ∇ y 1 + g 22 · ∇ y 2 + a 21 y 1 + a 22 y 2 in Q T ,
y = 0 on Σ T ,
y( · , 0) = y 0 in Ω,
(S full ) where y 0 ∈ L 2 (Ω; R 2 ), u ∈ L 2 (Q T ), g ij ∈ L ∞ (Q T ; R N ), a ij ∈ L ∞ (Q T ) for all i, j ∈ { 1, 2 } , Q T := Ω × (0, T ), Σ T := (0, T ) × ∂Ω and
( d l ij ∈ W ∞ 1 (Q T ), d l ij = d l ji in Q T ,
N
X
i,j =1
d l ij ξ i ξ j > d 0 | ξ | 2 in Q T , ∀ ξ ∈ R N .
Cascade condition
T HEOREM (see [1] and [2]) Let us suppose that
g 21 ≡ 0 in q T := ω × (0, T ) and
(a 21 > a 0 in q T or a 21 < − a 0 in q T ), for a constant a 0 > 0 and q T := ω × (0, T ).
Then System (S full ) is null controllable at time T .
[1] F. Ammar Khodja, A. Benabdallah and C. Dupaix, Null-controllability of some reaction-diffusion systems with one control force, J. Math. Anal. Appl., 2006.
[2] M. González-Burgos and L. de Teresa, Controllability results for cascade systems of
m coupled parabolic PDEs by one control force. Port. Math., 2010.
Condition on the dimension
T HEOREM (S. Guerrero, 2007)
Let N = 1, d 1 , d 2 , a 11 , g 11 , a 22 , g 22 be constant and suppose that g 21 · ∇ + a 21 = P 1 (θ · ) in Ω × (0, T ),
where θ ∈ C 2 (Ω) with | θ | > C in ω 0 ⊂ ω and P 1 := m 0 · ∇ + m 1 , for m 0 , m 1 ∈ R. Moreover, assume that
m 0 6 = 0.
Then System (S full ) is null controllable at time T .
Boundary condition
T HEOREM (A. Benabdallah, M. Cristofol, P. Gaitan, L. De Teresa, 2014)
Assume that a ij ∈ C 4 (Q T ), g ij ∈ C 3 (Q T ) N , d i ∈ C 3 (Q T ) N
2for all i, j ∈ { 1, 2 } and
∃ γ 6 = ∅ an open subset of ∂ω ∩ ∂Ω,
∃ x 0 ∈ γ s.t. g 21 (t, x 0 ) · ν (x 0 ) 6 = 0 for all t ∈ [0, T ], where ν is the exterior normal unit vector to ∂Ω.
Then System (S full ) is null controllable at time T .
Necessary and sufficient condition
T HEOREM 1 (M. D., P. Lissy, 2015 )
Let us assume that d i , g ij and a ij are constant in space and time for all i, j ∈ { 1, 2 } .
Then System (S full ) is null controllable at time T if and only if g 21 6 = 0 or a 21 6 = 0.
1
The term g 21 can be different from zero in the control domain
2
Theorem 1 is true for all N ∈ N ∗
3
We have no geometric restriction
4
Concerning the controllability of a system with m equations controlled by m − 1 forces, the last condition becomes:
∃ i 0 ∈ { 1, ..., m − 1 } s.t. g mi
06 = 0 or a mi
06 = 0.
Additional result
T HEOREM 2 (M. D., P. Lissy, 2015 ) Let us assume that Ω := (0, L) with L > 0.
Then System (S full ) is null controllable at time T if for an open subset (a, b) × O ⊆ q T one of the following conditions is verified:
(i) d i , g ij , a ij ∈ C 1 ((a, b), C 2 ( O )) for i, j = 1, 2 and g 21 = 0 and a 21 6 = 0 in (a, b) × O ,
1
a
21∈ L ∞ ( O ) in (a, b).
(ii) d i , g ij , a ij ∈ C 3 ((a, b), C 7 ( O )) for i = 1, 2 and
| det(H(t , x )) | > C for every (t , x ) ∈ (a, b) × O , where
H:=
−a21+∂x g21 g21 0 0 0 0
−∂x a21+∂xx g21 −a21+2∂x g21 0 g21 0 0
−∂t a21+∂tx g21 ∂t g21 −a21+∂x g21 0 g21 0
−∂xx a21+∂xxx g21 −2∂x a21+3∂xx g21 0 −a21+3∂x g21 0 g21
−a22+∂x g22 g22−∂x d2 −1 −d2 0 0
−∂x a22+∂xx g22 −a22+2∂x g22−∂xx d2 0 g22−2∂x d2 −1 −d2
.
Remarks
Even though the last condition seems complicated, it can be simplified in some cases :
1
System (S full ) is null controllable at time T if there exists an open subset (a, b) × O ⊆ q T such that
g 21 ≡ κ ∈ R ∗ in (a, b) × O , a 21 ≡ 0 in (a, b) × O ,
∂ x a 22 6 = ∂ xx g 22 in (a, b) × O .
2
If the coefficients depend only on the time variable, the condition becomes :
∃ (a, b) ⊂ (0, T ) s.t.:
g 21 (t)∂ t a 21 (t ) 6 = a 21 (t )∂ t g 21 (t ) in (a, b).
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
The notion of the algebraic resolvability can be found in:
[1] M. Gromov, Partial differential relations, Springer-Verlag , 1986.
And was used for the first time in the control theory of pde’s in:
[2] J.-M. Coron & P. Lissy, Local null controllability of the three-dimensional
Navier-Stokes system with a distributed control having two vanishing components,
Invent. Math. 198 (2014), 833–880.
Let f ∈ C c ∞ ([0, 1]). Consider the problem ( Find (w 1 , w 2 ) ∈ C c ∞ ([0, 1]; R 2 ) s. t. :
a 1 w 1 − a 2 ∂ x w 1 + a 3 ∂ xx w 1 + b 1 w 2 − b 2 ∂ x w 2 + b 3 ∂ xx w 2 = f , (1) with a 1 , a 2 , a 3 , b 1 , b 2 , b 3 ∈ R .
Problem (1) can be rewritten as follows : L
w 1
w 2
= f , where the operator L is given by
L := (a 1 − a 2 ∂ x + a 3 ∂ xx , b 1 − b 2 ∂ x + b 3 ∂ xx ) .
If there exists a differential operator M := P M
i=0 m i ∂ x i with m 0 , ..., m M ∈ R, M ∈ N such that
L ◦ M = Id ,
then (w 1 , w 2 ) := M f is a solution to problem (1). The last equality is formally equivalent to
M ∗ ◦ L ∗ = Id , with
L ∗ (ϕ) =
a 1 + a 2 ∂ x + a 3 ∂ xx b 1 + b 2 ∂ x + b 3 ∂ xx
.
Consider the operator defined for all ϕ ∈ C c ∞ ([0, 1]) by
L ∗ 1
L ∗ 2
∂ x L ∗ 1
∂ x L ∗ 2
ϕ :=
a 1 + a 2 ∂ x + a 3 ∂ xx b 1 + b 2 ∂ x + b 3 ∂ xx a 1 ∂ x + a 2 ∂ xx + a 3 ∂ xxx b 1 ∂ x + b 2 ∂ xx + b 3 ∂ xxx
ϕ = C
ϕ
∂ x ϕ
∂ xx ϕ
∂ xxx ϕ
,
where
C :=
a 1 a 2 a 3 0 b 1 b 2 b 3 0 0 a 1 a 2 a 3
0 b 1 b 2 b 3
.
Let us suppose that C is invertible, denote by
C −1 :=
e c 11 e c 12 e c 13 e c 14
e c 21 e c 22 e c 23 e c 24
e c 31 e c 32 e c 33 e c 34
e c 41 e c 42 e c 43 e c 44
.
Then we have
C −1
L ∗ 1
L ∗ 2
∂ x L ∗ 1
∂ x L ∗ 2
ϕ =
ϕ
∂ x ϕ
∂ xx ϕ
∂ xxx ϕ
,
where the first line is given by
e c 11 L ∗ 1 ϕ + c e 12 L ∗ 2 ϕ + e c 13 ∂ x L ∗ 1 ϕ + e c 14 ∂ x L ∗ 2 ϕ = ϕ.
Thus the problem is solved if we define M ∗ for all (ψ 1 , ψ 2 ) ∈ C ∞ c ([0, 1]; R 2 ) by
M ∗ ψ 1
ψ 2
:= e c 11 ψ 1 + e c 12 ψ 2 + c e 13 ∂ x ψ 1 + e c 14 ∂ x ψ 2 .
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
Let us recall system (S full ) and Theorem 1. Consider the system of two parabolic equations controlled by one force
∂ t y = div(D ∇ y ) + G · ∇ y + Ay + e 1 1 ω u in Q T ,
y = 0 on Σ T ,
y ( · , 0) = y 0 in Ω,
(S full )
where y 0 ∈ L 2 (Ω; R 2 ), D := (d 1 , d 2 ) ∈ L ( R 2N , R 2N ), G := (g ij ) ∈ L ( R 2N , R 2 ), A := (a ij ) ∈ L ( R 2 ).
T HEOREM 1 (M. D., P. Lissy, 2015 )
Let us assume that d i , g i,j and a i,j are constant in space and time for all i, j ∈ { 1, 2 } .
Then System (S full ) is null controllable at time T if and only if
g 2,1 6 = 0 or a 2,1 6 = 0.
Strategy: fictitious control method
Analytic problem
Find (z, v ) with Supp(v ) ⊂ ω × (ε, T − ε) s.t. :
∂ t z = div(D ∇ z ) + G · ∇ z + Az + N ( 1 ω v ) in Q T ,
z = 0 on Σ T ,
z (0, · ) = y 0 , z(T , · ) = 0 in Ω, where the operator N is well chosen.
Difficulties: • the control is in the range of the operator N ,
• v has to be regular enough.
Algebraic problem:
Find ( b z, b v ) with Supp( b z , v b ) ⊂ ω × (ε, T − ε) s.t. :
∂ t b z = div(D ∇b z) + G · ∇b z + A z b + B b v + N f in Q T , Conclusion:
The couple (y , u) := (z − z b , −b v ) will be a solution to system (S full )
and will satisfy y (T ) ≡ 0 in Ω.
Resolution of the algebraic problem
The algebraic problem can be rewritten as:
For f := 1 ω v , find ( b z , v b ) with Supp( b z, b v ) ⊂ ω × (ε, T − ε) such that L ( b z, b v ) = N f ,
where
L ( b z , v b ) := ∂ t b z − div(D ∇b z) − G · ∇b z − A b z − B b v .
This problem is solved if we can find a differential operator M such that
L ◦ M = N .
The last equality is equivalent to
M ∗ ◦ L ∗ = N ∗ , where L ∗ is given for all ϕ ∈ C c ∞ (Q T ) 2 by
L ∗ ϕ :=
L ∗ 1 ϕ L ∗ 2 ϕ L ∗ 3 ϕ
=
− ∂ t ϕ 1 − div(d 1 ∇ ϕ 1 ) + P 2
j=1 { g j1 · ∇ ϕ j − a j1 ϕ j }
− ∂ t ϕ 2 − div(d 1 ∇ ϕ 2 ) + P 2
j=1 { g j2 · ∇ ϕ j − a j2 ϕ j } ϕ 1
.
We remark that
(g 21 · ∇ − a 21 ) L ∗ 1 ϕ
L ∗ 1 ϕ + (∂ t + div(d 1 ∇ ) − g 11 · ∇ + a 11 ) L ∗ 3 ϕ
= N ∗ ϕ, where
N ∗ := g 21 · ∇ + a 21 . Thus the algebraic problem is solved by taking
M ∗
ψ 1 ψ 2 ψ 3
:=
(g 21 · ∇ − a 21 )ψ 3
ψ 1 + (∂ t + div(d 1 ∇ ) − g 11 · ∇ + a 11 )ψ 3
.
Resolution of the analytic problem: sketch of the proof
The system
∂ t z = div(D ∇ z ) + G · ∇ z + Az + N ( 1 ω v ) in Q T ,
z = 0 on Σ T ,
z (0, · ) = y 0 , z(T , · ) = 0 in Ω,
is null controllable if for all ψ 0 ∈ L 2 (Ω; R 2 ) the solution of the adjoint system
− ∂ t ψ = div(D ∗ ∇ ψ) − G ∗ · ∇ ψ + A ∗ ψ in Q T ,
ψ = 0 on Σ T ,
ψ(T , · ) = ψ 0 in Ω
satisfies the following inequality of observability Z
Ω | ψ(0, x ) | 2 dx 6 C obs Z T
0
Z
ω
ρ 0 |N ∗ ψ(t , x ) | 2 dxdt ,
where ρ 0 can be chosen to be exponentially decreasing at times t = 0
and t = T .
This inequality is obtained by applying the differential operator
∇∇N ∗ = ∇∇ ( − a 21 + g 21 · ∇ )
to the adjoint system. More precisely we study the solution φ ij := ∂ i ∂ j N ∗ ψ of the system
− ∂ t φ ij = D∆φ ij − G ∗ · ∇ φ ij + A ∗ φ ij in Q T ,
∂φ
∂n = ∂(∇∇N ∂n
∗ψ) on Σ T ,
φ(T , · ) = ∇∇N ∗ ψ 0 in Ω.
Remark : We need that ∇∇N ∗ commutes with the other operators of
the system, what is possible with some constant coefficients.
Following the ideas of Barbu developed in [1], the control is chosen as the solution of the problem
minimize J k (v ) := 1 2 Z
Q
Tρ −1 0 | v | 2 dxdt + k 2
Z
Ω | z (T ) | 2 dx, v ∈ L 2 (Q T , ρ −1/2 0 ) 2 .
We obtain the regularity of the control with the help of the following Carleman inequality
3
P
j =1
x
Q
Tρ j |∇ j ψ | 2 dxdt 6 C T
Z T 0
Z
ω
ρ 0 |N ∗ ψ(t , x ) | 2 dxdt ,
where ρ j are some appropriate weight functions.
[1] V. Barbu, Exact controllability of the superlinear heat equation, Appl. Math. Optim.
42 (2000), 73–89.
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
Consider the following system
∂ t z 1 = ∂ xx z 1 + 1 ω u in (0, π) × (0, T ),
∂ t z 2 = ∂ xx z 2 + p(x )∂ x z 1 + q(x )z 1 in (0, π) × (0, T ), z(0, · ) = z(π, · ) = 0 on (0, T ), z( · , 0) = z 0 in (0, π),
(S simpl. )
where z 0 ∈ L 2 ((0, π); R 2 ), u ∈ L 2 ((0, π) × (0, T )), p ∈ W ∞ 1 (0, π), q ∈ L ∞ (0, π) and ω := (a, b) ⊂ (0, π). Consider the two following quantities
I a,k (p, q) :=
Z a 0
q − 1 2 ∂ x p ϕ 2 k ,
I k (p, q) :=
Z π 0
q − 1 2 ∂ x p ϕ 2 k ,
where for all k ∈ N ∗ and x ∈ (0, π), we denote by ϕ k (x ) :=
q 2
π sin(kx ) the eigenvector of the Laplacian operator, with
Dirichlet boundary condition.
When the supports of the control and the coupling terms are disjoint in System (S simpl . ), following the ideas F. Ammar Khodja et al, we obtain a minimal time of null controllability:
T HEOREM (M. D., 2015)
Let p ∈ W ∞ 1 (0, π), q ∈ L ∞ (0, π). Suppose that | I k | + | I a,k | 6 = 0 for all k ∈ N and
(Supp(p) ∪ Supp(q)) ∩ ω = ∅ . Let T 0 (p, q) be given by
T 0 (p, q) := lim sup
k→∞
min( − log | I k (p, q) | , − log | I a,k (p, q) | )
k 2 .
One has
1
If T > T 0 (p, q), then System (S simpl. ) is null controllable at time T .
2
If T < T 0 (p, q), then System (S simpl . ) is not null controllable at
time T .
Recall the studied system
∂ t z 1 = ∂ xx z 1 + 1 ω u in (0, π) × (0, T ),
∂ t z 2 = ∂ xx z 2 + p(x)∂ x z 1 + q (x )z 1 in (0, π) × (0, T ), z (0, · ) = z (π, · ) = 0 on (0, T ), z ( · , 0) = z 0 in (0, π),
where z 0 ∈ L 2 ((0, π); R 2 ), u ∈ L 2 ((0, π) × (0, T )), p ∈ W ∞ 1 (0, π), q ∈ L ∞ (0, π) and ω := (a, b) ⊂ (0, π).
T HEOREM (M. D., 2015)
Suppose that p ∈ W ∞ 1 (0, π) ∩ W ∞ 2 (ω), q ∈ L ∞ (0, π) ∩ W ∞ 1 (ω) and (Supp(p) ∪ Supp(q)) ∩ ω 6 = ∅ .
Then the system is null controllable at time T .
1 Partial controllability of parabolic systems Results
Numerical simulations
2 Indirect controllability of parabolic systems coupled with a first order term
Full system
Local inversion of a differential operator Proof in the constant case
Simplified system
3 Conclusion
Works in progress with P. Lissy
1
A general condition for the null controllability by m − 1 forces of m parabolic equations coupled by zero and first order terms:
∂ t y 1 = div(d 1 ∇ y 1 ) + g 11 · ∇ y 1 + g 12 · ∇ y 2 + a 11 y 1 + a 12 y 2 + 1 ω u in Q T ,
∂ t y 2 = div(d 2 ∇ y 2 ) + g 21 · ∇ y 1 + g 22 · ∇ y 2 + a 21 y 1 + a 22 y 2 in Q T ,
y = 0 on Σ T ,
y ( · , 0) = y 0 in Ω.
(S full ) System (S full ) is null controllable at time T if
g 21 6 = 0 in q T and a 22 belong to an appropriate space.
Using a non linear variable change it can be proved that the
second condition is artificial.
Works in progress with M. Gonzalez-Burgos and D.
Souza
2
The null controllability for a parabolic systems with non-diagonalizable diffusion matrix:
Consider the parabolic systems
y t − Dy xx + Ay = B 1 ω u in (0, π) × (0, T ), y ( · , 0) = 0 on (0, T ),
y (0, · ) = y 0 in (0, π),
where y 0 ∈ L 2 (0, π; R n ) is the initial data, u ∈ L 2 (Q T ) is the control and D, A ∈ L ( R n ) and B ∈ L ( R , R n ) are given, for d > 0, by
D =
d 1 0 · · · 0 0 . .. ... ... ...
.. . . .. ... ... 0 .. . . .. ... 1 0 · · · · 0 d
, A =
0 0 · · · 0 .. . .. . .. . 0 .. . .. . 1 0 · · · 0
and B =
0 .. . 0 1
.
Comments and open problems
1
Partial null controllability of a parabolic system:
Consider the system
∂ t y 1 = ∆y 1 + αy 2 in Q T ,
∂ t y 2 = ∆y 2 + 1 ω u in Q T .
y = 0 on Σ T ,
y (0) = y 0 in Ω.
When Supp(α) ∩ ω = ∅ , there exists a minimal time T 0 for the null controllability.
If T < T 0 , is it possible, for all y 0 ∈ L 2 (Ω; R 2 ), to find a control
u ∈ L 2 (Q T ) such that y 1 (T ) = 0 in Ω?
Comments and open problems
2
Partial controllability of semilinear systems:
Let Ω ⊂ R 3 , T > 0, Q T := Ω × (0, T ) and Σ T := ∂Ω × (0, T ).
Consider the system
∂ t y 1 = d 1 ∆y 1 + a 1 (1 − y 1 /k 1 )y 1 − (α 1,2 y 2 + κ 1,3 y 3 )y 1 in Q T ,
∂ t y 2 = d 2 ∆y 2 + a 2 (1 − y 2 /k 2 )y 2 − (α 2,1 y 1 + κ 2,3 y 3 )y 2 in Q T ,
∂ t y 3 = d 3 ∆y 3 − a 3 y 3 + u in Q T ,
∂ n y i := ∇ y i · ~ n = 0 ∀ 1 6 i 6 3 on Σ T ,
y (x , 0) = y 0 in Ω,
where
y
1is the density of tumor cells, y
2is the density of normal cells, y
3is the drug concentration,
u is the rate at which the drug is being injected (control), d
i, a
i, k
i, α
i,j, κ
i,jare known constants.
[1] S. Chakrabarty & F. Hanson, Distributed parameters deterministic model for
treatment of brain tumours using Galerkin finite element method, 2009.
Comments and open problems
3
Null controllability of a general parabolic system:
Consider the system of n equations controlled by m forces
∂ t y = ∆y + G · ∇ y + Ay + B 1 ω u in Q T ,
y = 0 on Σ T ,
y ( · , 0) = y 0 in Ω,
where y 0 ∈ L 2 (Ω; R n ), G ∈ L ∞ (Q T ; L ( R nN , R n )),
A ∈ L ∞ (Q T ; L ( R n )), B ∈ L ∞ (Q T ; L ( R m , R n )) and u ∈ L 2 (Q T ; R m ).
Comments and open problems
4