Partial controllability of parabolic systems
Michel Duprez
Laboratoire de Mathématiques de Besançon
9 mars 2015
Besançon
Colloquium on dispersive PDEs & related problems
Example
Let be T > 0 and ω ⊂ Ω ⊂ R N .
If we consider, for example, the following parabolic system Find y := (y 1 , y 2 ) ∗ : Q T → R 2 such that
∂ t y 1 = ∆y 1 + y 2 + 1 ω u in Q T := Ω × (0, T ),
∂ t y 2 = ∆y 2 in Q T ,
y = 0 on Σ T := ∂Ω × (0, T ), y (0) = y 0 in Ω.
Can we find, for all initial condition y 0 , a control u such that
y 1 (T ; y 0 , u) = 0 ?
Example
Let be T > 0 and ω ⊂ Ω ⊂ R N .
If we consider, for example, the following parabolic system Find y := (y 1 , y 2 ) ∗ : Q T → R 2 such that
∂ t y 1 = ∆y 1 + y 2 + 1 ω u in Q T := Ω × (0, T ),
∂ t y 2 = ∆y 2 in Q T ,
y = 0 on Σ T := ∂Ω × (0, T ), y (0) = y 0 in Ω.
Can we find, for all initial condition y 0 , a control u such that
y 1 (T ; y 0 , u) = 0 ?
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Motivations
∂ 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
1
y 1 is the density of tumor cells,
2
y 2 is the density of normal cells,
3
y 3 is the drug concentration,
4
u is the rate at which the drug is being injected (control),
5
d i , a i , k i , α i ,j , κ i ,j are known constants.
[1] Chakrabarty, Hanson: Distributed parameters deterministic model
for treatment of brain tumors using Galerkin finite element method,
Setting
We consider here the following system of n linear parabolic equations with m controls
∂ t y = ∆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 and
B := (b ik ) ik with a ij , b ik ∈ L ∞ (Q T ) 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 .
Setting
We consider here the following system of n linear parabolic equations with m controls
∂ t y = ∆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 and
B := (b ik ) ik with a ij , b ik ∈ L ∞ (Q T ) 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
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 > 0 and y 0 , y T ∈ L 2 (Ω) n there exists a control u ∈ L 2 (Q T ) m such that
kPy (·, T ; y 0 , u) − Py T (·)k L
2(Ω)
p6 .
Definitions
We will say that system (S) is
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 > 0 and y 0 , y T ∈ L 2 (Ω) n there exists a control u ∈ L 2 (Q T ) m such that
kPy (·, T ; y 0 , u) − Py T (·)k L
2(Ω)
p6 .
Plan
1 Autonomous case for differential equations
2 Constant matrices
3 Time dependent matrices
4 A 2 x 2 system with a space dependent matrix
5 Conclusions and comments
Plan
1 Autonomous case for differential equations
2 Constant matrices
3 Time dependent matrices
4 A 2 x 2 system with a space dependent matrix
5 Conclusions and comments
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 (Kalman, 1969)
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.
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 (Kalman, 1969)
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.
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 (Kalman, 1969)
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.
Sketch of proof for the null controllability for n=3, m=1
⇐ • Let us first consider the variable change z := y − ηy , where ∂ t y = Ay in [0, T ],
y (0) = y 0 ∈ R and η ∈ C ∞ [0, T ] with the following form
6
0 T 3 2T 3 T -
1 η
Thus, if h := −∂ t ηy , z is solution to
∂ t z = Az + Bu + h in [0, T ],
z (0) = 0.
Sketch of proof for the null controllability for n=3, m=1
• We recall that
∂ t z = Az + Bu + h in [0, T ], z (0) = 0.
If we search a solution of the form z := Kw with K := (B|AB|A 2 B) invertible, w satisfies
K ∂ t w = AKw + Bu + h.
Using the Cayley Hamilton Theorem, A 3 := α 0 I + α 1 A + α 2 A 2 .
Thus
AK = (AB|A 2 B|A 3 B) = KC,
Ke 1 = B, with C =
0 0 α 0 1 0 α 1 0 1 α 2
.
Sketch of proof for the null controllability for n=3, m=1
We recall that h := −∂ t ηy . Thus w is solution to
∂ t w 1 = α 0 w 3 + h 1 + u,
∂ t w 2 = w 1 + α 1 w 3 + h 2 ,
∂ t w 3 = w 2 + α 2 w 3 + h 3 , w(0) = 0.
We choose
w 3 = 0, w 2 = −h 3 ,
w 1 = ∂ t w 2 − α 1 w 2 − h 2 , u = ∂ t w 1 − α 0 w 1 − h 1 . Conclusion :
w i (0) = w i (T ) = 0 for all i ∈ {1, 2, 3}.
Plan
1 Autonomous case for differential equations
2 Constant matrices
3 Time dependent matrices
4 A 2 x 2 system with a space dependent matrix
5 Conclusions and comments
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 (Ammar-Khodja et al 2009)
System (S) is null controllable on (0, T ) if and only rank(B|AB|...|A n−1 B) = n.
T HEOREM (Ammar-Khodja, Chouly, D 2015)
System (S) is partially null controllable on (0, T ) if and only
rank(PB|PAB|...|PA n−1 B) = p.
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 (Ammar-Khodja et al 2009)
System (S) is null controllable on (0, T ) if and only rank(B|AB|...|A n−1 B) = n.
T HEOREM (Ammar-Khodja, Chouly, D 2015)
System (S) is partially null controllable on (0, T ) if and only
rank(PB|PAB|...|PA n−1 B) = p.
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 (Ammar-Khodja et al 2009)
System (S) is null controllable on (0, T ) if and only rank(B|AB|...|A n−1 B) = n.
T HEOREM (Ammar-Khodja, Chouly, D 2015)
System (S) is partially null controllable on (0, T ) if and only
rank(PB|PAB|...|PA n−1 B) = p.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Let us consider the system
∂ t y = ∆y +
a 11 a 12 a 13
a 21 a 22 a 23 a 31 a 32 a 33
y +
b 1
b 2 b 3
1 ω u in Q T ,
y = 0 on Σ T ,
y (0) = y 0 in Ω.
Goal : Find u ∈ L 2 (Q T ) such that
y 1 (T ; y 0 , u) = 0.
Strategy : Find a variable change y = M(t)w with w solution of a cascade system and use
[1] González-Burgos M, de Teresa L. : Controllability results for cascade systems of m
coupled parabolic PDEs by one control force
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Let us consider the system
∂ t y = ∆y +
a 11 a 12 a 13
a 21 a 22 a 23 a 31 a 32 a 33
y +
b 1
b 2 b 3
1 ω u in Q T ,
y = 0 on Σ T ,
y (0) = y 0 in Ω.
Goal : Find u ∈ L 2 (Q T ) such that
y 1 (T ; y 0 , u) = 0.
Strategy : Find a variable change y = M(t)w with w solution of a cascade system and use
[1] González-Burgos M, de Teresa L. : Controllability results for cascade systems of m
coupled parabolic PDEs by one control force
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Let us consider the system
∂ t y = ∆y +
a 11 a 12 a 13
a 21 a 22 a 23 a 31 a 32 a 33
y +
b 1
b 2 b 3
1 ω u in Q T ,
y = 0 on Σ T ,
y (0) = y 0 in Ω.
Goal : Find u ∈ L 2 (Q T ) such that
y 1 (T ; y 0 , u) = 0.
Strategy : Find a variable change y = M(t)w with w solution of a cascade system and use
[1] González-Burgos M, de Teresa L. : Controllability results for cascade systems of m
coupled parabolic PDEs by one control force
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Let be
K := [A|B] = (B|AB|A 2 B), s := rank(K ),
X := spanhB, AB, A 2 Bi.
(i) s=1:
Since
B 6= 0
rank(B|AB|A 2 B) = 1) , then X = spanhBi.
In particular
AB := α 1 B and A 2 B := α 2 B.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Let be
K := [A|B] = (B|AB|A 2 B), s := rank(K ),
X := spanhB, AB, A 2 Bi.
(i) s=1:
Since
B 6= 0
rank(B|AB|A 2 B) = 1) , then X = spanhBi.
In particular
AB := α 1 B and A 2 B := α 2 B.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
⇐ Let us suppose that rank(PK ) = 1.
We define
M (t) := (B|M 2 (t )|M 3 (t)), where
∂ t M 2 = AM 2 , M 2 (T ) = e 2 and
∂ t M 3 = AM 3 , M 3 (T ) = e 3 . We have b 1 6= 0, indeed
rank(PB|PAB|PA 2 B) = rank(PB|α 1 PB|α 2 PB)
= rank(b 1 |α 1 b 1 |α 2 b 1 ) = 1.
Then M(T ) is invertible and there exists T ∗ such that M (t) is
invertible in [T ∗ , T ].
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Matrix M satisfies
−∂ t M (t) + AM (t) = (AB|0|0) = M (t )C, M (t)e 1 = B,
with C given by
C :=
α 1 0 0
0 0 0
0 0 0
.
If T ∗ = 0 : Since M (t ) is invertible on [0, T ], y is the solution to system (S) if and only if w := (w 1 , w 2 , w 3 ) ∗ = M (t) −1 y is the solution to the cascade system
∂ t w = ∆w + Cw + e 1 1 ω u in Q T ,
w = 0 on Σ T ,
w(0) = w 0 in Ω.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
We recall that y = M (t)w in Q T and
M (T ) =
b 1 0 0 b 2 1 0 b 3 0 1
.
Thus y 1 (T ) = b 1 w 1 (T ) = 0 in Ω.
And it is possible to find u ∈ L 2 (Q T ) such that the solution to cascade system satisfies
w 1 (T ) = 0 in Ω.
If now T ∗ 6= 0 :
The system is considered without control until T ∗ and we control only
on [T ∗ , T ].
Proof in the case n=3,m=1, p=1, P=(1,0,0)
We recall that y = M (t)w in Q T and
M (T ) =
b 1 0 0 b 2 1 0 b 3 0 1
.
Thus y 1 (T ) = b 1 w 1 (T ) = 0 in Ω.
And it is possible to find u ∈ L 2 (Q T ) such that the solution to cascade system satisfies
w 1 (T ) = 0 in Ω.
If now T ∗ 6= 0 :
The system is considered without control until T ∗ and we control only
on [T ∗ , T ].
Proof in the case n=3,m=1, p=1, P=(1,0,0)
⇒ If now rank(PB|PAB|PA 2 B) 6= 1. Let us remark that rankP[A|B] 6 rank[A|B] = s = 1.
Thus rank(PB|PAB|PA 2 B) = 0 and PB = PAB = PA 2 B = 0.
Since B 6= 0, let us suppose that b 2 6= 0.
We define
M := (B|e 1 |e 3 ) =
0 1 0
b 2 0 0 b 3 0 1
.
We have Ae 1 := c 12 B + c 22 e 1 + c 32 e 3 , Ae 3 := c 13 B + c 23 e 1 + c 33 e 3 . Hence
AM = (α 1 B|Ae 1 |Ae 3 ) = MC, Ke 1 = B,
with C given by C :=
α 1 c 12 c 13
0 c 22 c 23
.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Since K is invertible, y is the solution to (S) if and only if w := (w 1 , w 2 , w 3 ) ∗ = M −1 y is the solution to cascade system
∂ t w = ∆w + Cw + e 1 1 ω u in Q T ,
w = 0 on Σ T ,
w(0) = M −1 y 0 in Ω.
Since M is given by
M := (B|e 1 |e 3 ) =
0 1 0
b 2 0 0 b 3 0 1
.
Thus we have
y 1 (T ) = 0 in Ω ⇔ w 2 (T ) = 0 in Ω.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
Since K is invertible, y is the solution to (S) if and only if w := (w 1 , w 2 , w 3 ) ∗ = M −1 y is the solution to cascade system
∂ t w = ∆w + Cw + e 1 1 ω u in Q T ,
w = 0 on Σ T ,
w(0) = M −1 y 0 in Ω.
Since M is given by
M := (B|e 1 |e 3 ) =
0 1 0
b 2 0 0 b 3 0 1
.
Thus we have
y 1 (T ) = 0 in Ω ⇔ w 2 (T ) = 0 in Ω.
Remark on the general dual system :
−∂ t ϕ = ∆ϕ + A ∗ ϕ in Q T ,
ϕ = 0 on Σ T ,
ϕ(T , ·) = P ∗ ϕ T in Ω,
with P ∗ : R p → R p × R n−p ϕ T 7→ (ϕ T , 0 n−p ) ∗ . P ROPOSITION
1
System (S) is partially null controllable on (0, T ), if and only if there exists C obs > 0 such that for all ϕ T ∈ L 2 (Ω) p
kϕ(0)k 2 L
2(Ω)
n6 C obs
Z T 0
kB ∗ ϕk 2 L
2(ω)
mwith ϕ the solution to the dual system.
2
System (S) is partially approx. controllable on (0, T ), if and only if for all ϕ T ∈ L 2 (Ω) p the solution to the dual system satisfies
B ∗ ϕ = 0 in ω × (0, T ) ⇒ ϕ = 0 in Q T .
Remark on the general dual system :
−∂ t ϕ = ∆ϕ + A ∗ ϕ in Q T ,
ϕ = 0 on Σ T ,
ϕ(T , ·) = P ∗ ϕ T in Ω,
with P ∗ : R p → R p × R n−p ϕ T 7→ (ϕ T , 0 n−p ) ∗ . P ROPOSITION
1
System (S) is partially null controllable on (0, T ), if and only if there exists C obs > 0 such that for all ϕ T ∈ L 2 (Ω) p
kϕ(0)k 2 L
2(Ω)
n6 C obs
Z T 0
kB ∗ ϕk 2 L
2(ω)
mwith ϕ the solution to the dual system.
2
System (S) is partially approx. controllable on (0, T ), if and only if for all ϕ T ∈ L 2 (Ω) p the solution to the dual system satisfies
B ∗ ϕ = 0 in ω × (0, T ) ⇒ ϕ = 0 in Q T .
Remark on the general dual system :
−∂ t ϕ = ∆ϕ + A ∗ ϕ in Q T ,
ϕ = 0 on Σ T ,
ϕ(T , ·) = P ∗ ϕ T in Ω,
with P ∗ : R p → R p × R n−p ϕ T 7→ (ϕ T , 0 n−p ) ∗ . P ROPOSITION
1
System (S) is partially null controllable on (0, T ), if and only if there exists C obs > 0 such that for all ϕ T ∈ L 2 (Ω) p
kϕ(0)k 2 L
2(Ω)
n6 C obs
Z T 0
kB ∗ ϕk 2 L
2(ω)
mwith ϕ the solution to the dual system.
2
System (S) is partially approx. controllable on (0, T ), if and only if for all ϕ T ∈ L 2 (Ω) p the solution to the dual system satisfies
B ∗ ϕ = 0 in ω × (0, T ) ⇒ ϕ = 0 in Q T .
Proof in the case n=3,m=1, p=1, P=(1,0,0)
The previous cascade system is partially null controllable if and only if there exists C > 0 such that for all ϕ T 2 ∈ L 2 (Ω) the solution to the dual system
−∂ t ϕ 1 = ∆ϕ 1 + α 1 ϕ 1 in Q T ,
−∂ t ϕ 2 = ∆ϕ 2 + c 12 ϕ 1 + c 22 ϕ 2 + c 32 ϕ 3 in Q T ,
−∂ t ϕ 3 = ∆ϕ 3 + c 13 ϕ 1 + c 23 ϕ 2 + c 33 ϕ 3 in Q T ,
ϕ = 0 on Σ T ,
ϕ(T ) = (0, ϕ T 2 , 0) ∗ in Ω satisfies the observability inequality
Z
Ω
ϕ(0) 2 6 C Z
ω×(0,T )
ϕ 2 1 .
We have ϕ 1 (T ) = 0 in Ω ⇒ ϕ 1 = 0 in Q T
and ϕ T 2 6≡ 0 in Ω ⇒ (ϕ 2 (0), ϕ 3 (0)) ∗ 6≡ 0 in Ω.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
The previous cascade system is partially null controllable if and only if there exists C > 0 such that for all ϕ T 2 ∈ L 2 (Ω) the solution to the dual system
−∂ t ϕ 1 = ∆ϕ 1 + α 1 ϕ 1 in Q T ,
−∂ t ϕ 2 = ∆ϕ 2 + c 12 ϕ 1 + c 22 ϕ 2 + c 32 ϕ 3 in Q T ,
−∂ t ϕ 3 = ∆ϕ 3 + c 13 ϕ 1 + c 23 ϕ 2 + c 33 ϕ 3 in Q T ,
ϕ = 0 on Σ T ,
ϕ(T ) = (0, ϕ T 2 , 0) ∗ in Ω satisfies the observability inequality
Z
Ω
ϕ(0) 2 6 C Z
ω×(0,T )
ϕ 2 1 .
We have ϕ 1 (T ) = 0 in Ω ⇒ ϕ 1 = 0 in Q T
and ϕ T 2 6≡ 0 in Ω ⇒ (ϕ 2 (0), ϕ 3 (0)) ∗ 6≡ 0 in Ω.
Proof in the case n=3,m=1, p=1, P=(1,0,0)
(ii) s=2: We take M(t ) := (B|AB|M 3 (t)) with M 3 (T ) = e 2 or e 3 . (iii) s=3: We take M := K = (B|AB|A 2 B).
Remark : We prove also the partial approximate controllability.
Plan
1 Autonomous case for differential equations
2 Constant matrices
3 Time dependent matrices
4 A 2 x 2 system with a space dependent matrix
5 Conclusions and comments
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 (Ammar-Khodja et al 2009)
1
If there exist t 0 ∈ [0, T ] such that
rank[A|B](t 0 ) = n, then System (S) is null controllable on (0,T).
2
System (S) is null controllable on every interval (T 0 , T 1 ) with T 0 < T 1 6 T if and only if there exists a dense subset E of (0, T ) such that rank[A|B](t) = n for every t ∈ E .
T HEOREM (Ammar-Khodja, Chouly, D 2015) If
rankP[A|B](T ) = p,
then system (S) is partially null controllable on (0,T).
T HEOREM (Ammar-Khodja et al 2009)
1
If there exist t 0 ∈ [0, T ] such that
rank[A|B](t 0 ) = n, then System (S) is null controllable on (0,T).
2
System (S) is null controllable on every interval (T 0 , T 1 ) with T 0 < T 1 6 T if and only if there exists a dense subset E of (0, T ) such that rank[A|B](t) = n for every t ∈ E .
T HEOREM (Ammar-Khodja, Chouly, D 2015) If
rankP[A|B](T ) = p,
then system (S) is partially null controllable on (0,T).
Plan
1 Autonomous case for differential equations
2 Constant matrices
3 Time dependent matrices
4 A 2 x 2 system with a space dependent matrix
5 Conclusions and comments
Controllability with A = A(t )
Let us 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 )
If α := α(t ), let us consider the change of variable z 1 := y 1 + µy 2
(∂ t − ∆)z 1 = (∂ t µ + α)y 2 + 1 ω u in Q T . If we choose µ solution to
∂ t µ = −α in [0, T ], µ(T ) = 0,
the new system is not coupled. And thus it is partially null controllable.
Difficulty when A = A(x)
Let us consider the same 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.
If now α := α(x ), let us consider the change of variable z 1 := y 1 + µy 2 (∂ t − ∆)z 1 = (α − ∂ t µ + ∆µ + 2∇µ · ∇)y 2 + 1 ω u in Q T
What kind of µ can be chosen in this situation ???
[1] Benabdallah A.; Cristofol M.; Gaitan P.; De Teresa, Luz : Controllability to
trajectories for some parabolic systems of three and two equations by one control force (2014).
[2] Coron J-M, Lissy P. : Local null controllability of the three-dimensional Navier-Stokes
system with a distributed control having two vanishing components (2014)
Difficulty when A = A(x)
Let us consider the same 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.
If now α := α(x ), let us consider the change of variable z 1 := y 1 + µy 2 (∂ t − ∆)z 1 = (α − ∂ t µ + ∆µ + 2∇µ · ∇)y 2 + 1 ω u in Q T
What kind of µ can be chosen in this situation ???
[1] Benabdallah A.; Cristofol M.; Gaitan P.; De Teresa, Luz : Controllability to
trajectories for some parabolic systems of three and two equations by one control force (2014).
[2] Coron J-M, Lissy P. : Local null controllability of the three-dimensional Navier-Stokes
system with a distributed control having two vanishing components (2014)
Partial approximate controllability
Let α ∈ L ∞ (Ω). The dual system associated to (S 2x2 ) is the following
−∂ t ϕ 1 = ∆ϕ 1 in Q T .
−∂ t ϕ 2 = ∆ϕ 2 + αϕ 1 in Q T ,
ϕ = 0 on Σ T ,
ϕ T 1 (T ) = ϕ T , ϕ T 2 (T ) = 0 in Ω.
If ϕ 1 ≡ 0 in ω × (0, T ), using the approximate controllability of the heat equation,
ϕ 1 ≡ 0 in Ω × (0, T ), and necessarly
ϕ 2 ≡ 0 in Ω × (0, T ).
Thus (S 2x 2 ) is partially approximately controllable for all
α ∈ L ∞ (Ω).
Example of non partial null controllability
T HEOREM (Ammar-Khodja, Chouly, 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.
More precisely:
there exists k 1 ∈ {1, 7} such that, for (y 1 0 , y 2 0 ) = (0, sin(k 1 x )) and all
u ∈ L 2 (Q T ), we have y 1 (T ) 6≡ 0 in Ω.
Example of non partial null controllability
T HEOREM (Ammar-Khodja, Chouly, 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.
More precisely:
there exists k 1 ∈ {1, 7} such that, for (y 1 0 , y 2 0 ) = (0, sin(k 1 x )) and all
u ∈ L 2 (Q T ), we have y 1 (T ) 6≡ 0 in Ω.
Example of partial null controllability
Let (w k ) k and (µ k ) k be the normed eigenfunctions and eigenvalues of
−∆ in Ω.
T HEOREM (Ammar-Khodja, Chouly, D 2015)
Let be Ω ⊂ R. Let us suppose that α ∈ L ∞ (Ω) satisfies
Z
Ω
αw k w l
6 C 1 e −C
2|µ
k−µ
l| for all k , l ∈ N ∗ , (C) where C 1 and C 2 are two positive constants.
Then system (S 2×2 ) is partially null controllable.
Remark: If Ω = (0, π) and α = P α p cos(px), then
(C) ⇔ |α p | 6 C 1 e −C
2p
2for all p ∈ N .
Example of partial null controllability
Let (w k ) k and (µ k ) k be the normed eigenfunctions and eigenvalues of
−∆ in Ω.
T HEOREM (Ammar-Khodja, Chouly, D 2015)
Let be Ω ⊂ R. Let us suppose that α ∈ L ∞ (Ω) satisfies
Z
Ω
αw k w l
6 C 1 e −C
2|µ
k−µ
l| for all k , l ∈ N ∗ , (C) where C 1 and C 2 are two positive constants.
Then system (S 2×2 ) is partially null controllable.
Remark: If Ω = (0, π) and α = P α p cos(px), then
(C) ⇔ |α p | 6 C 1 e −C
2p
2for all p ∈ N .
Numerical illustration
Let us consider the fonctionnal HUM J ε (u) = 1
2 Z
ω×(0,T )
u 2 dxdt + 1 2
Z
Ω
y 1 (T ; y 0 , u) 2 dx.
T HEOREM
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 ky 1 (T ; y 0 , u ε )k Ω 6 M y
0,T √ ε.
[1] Boyer F.: On the penalised HUM approach and its applications to the numerical
Numerical illustration
Let us consider the fonctionnal HUM J ε (u) = 1
2 Z
ω×(0,T )
u 2 dxdt + 1 2
Z
Ω
y 1 (T ; y 0 , u) 2 dx.
T HEOREM
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 ky 1 (T ; y 0 , u ε )k Ω 6 M y
0,T √ ε.
[1] Boyer F.: On the penalised HUM approach and its applications to the numerical
Numerical illustration
Let us consider the fonctionnal HUM J ε (u) = 1
2 Z
ω×(0,T )
u 2 dxdt + 1 2
Z
Ω
y 1 (T ; y 0 , u) 2 dx.
T HEOREM
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 ky 1 (T ; y 0 , u ε )k Ω 6 M y
0,T √ ε.
[1] Boyer F.: On the penalised HUM approach and its applications to the numerical
10-3 10-2 10-1 h =ǫ1/4
10-6 10-5 10-4 10-3 10-2 10-1 100 101 102
||y1,ǫh,δt(T)||Eh (slope =2.23)
||uǫh,δt||Lδt2(0,T;Uh) (slope =−0.08) infuh,δt∈Lδt2(0,T;Uh)Jǫh,δt(uh,δt) (slope =−0.13)
Example of partial null controllability : α = 1.
10-3 10-2 10-1 h =ǫ1/4
10-5 10-4 10-3 10-2 10-1 100 101
||y1,ǫh,δt(T)||Eh (slope =0.31)
||uǫh,δt||L2
δt(0,T;Uh) (slope =−0.71) infuh,δt∈Lδt2(0,T;Uh)Jǫh,δt(uh,δt) (slope =−3.26)
Example of non partial null controllability : α = P
p>1 1
p
2cos(15px).
Plan
1 Autonomous case for differential equations
2 Constant matrices
3 Time dependent matrices
4 A 2 x 2 system with a space dependent matrix
5 Conclusions and comments
Conclusions :
1
Necessary and sufficient conditions in the constant case.
2
Sufficient conditions in the time dependent case.
3
Same conditions concerning the (partial) approximate controllability.
4
The regularity in space of the coupling matrix seems important.
Perspectives :
1
More general conditions in the space dependent case.
2
Partial controllability of semilinear parabolic systems.
Conclusions :
1
Necessary and sufficient conditions in the constant case.
2
Sufficient conditions in the time dependent case.
3
Same conditions concerning the (partial) approximate controllability.
4
The regularity in space of the coupling matrix seems important.
Perspectives :
1
More general conditions in the space dependent case.
2
