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26novembre2015Soutenancedethèse MichelDuprez Contrôlabilitédequelquessystèmesgouvernéspardesequationsparaboliques.

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(1)

Contrôlabilité de quelques systèmes gouvernés par des equations paraboliques.

Michel Duprez

LMB, université de Franche-Comté

26 novembre 2015

Soutenance de thèse

(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

(3)

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 .

(4)

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)

n

6 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)

p

6 ε.

(5)

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

(6)

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

(7)

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.

(8)

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.

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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).

(10)

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.

(11)

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

2

p for all p ∈ N , where C 1 > 0 and C 2 > π are two constants.

Then system (S 2×2 ) is partially null controllable.

(12)

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

(13)

Numerical illustration

Let us consider the fonctionnal HUM J ε (u) = 1

2 Z

ω×(0,T )

u 2 dxdt + 1 2ε

Z

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 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).

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10 −3 10 −2 10 −5

10 −4 10 −3 10 −2 10 −1 10 0 10 1

h

inf L

2

δt

(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.

(15)

10 −3 10 −2 10 −4

10 −3 10 −2 10 −1 10 0 10 1

h

inf L

2

δt

(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).

(16)

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

1

in system (S

2x2

) in the case α = 1.

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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

1

in system (S

2x2

) in the case α =

P

j=1 1

j2

cos(15jx).

(18)

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

(19)

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

(20)

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 .

(21)

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.

(22)

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 .

(23)

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

2

for 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 .

(24)

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

0

6 = 0 or a mi

0

6 = 0.

(25)

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

 .

(26)

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).

(27)

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

(28)

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.

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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 2x w 1 + a 3xx w 1 + b 1 w 2 − b 2x w 2 + b 3xx 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 2x + a 3xx , b 1 − b 2x + b 3xx ) .

(30)

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 2x + a 3xx b 1 + b 2x + b 3xx

.

(31)

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 2x + a 3xx b 1 + b 2x + b 3xx a 1x + a 2xx + a 3xxx b 1x + b 2xx + b 3xxx

 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

.

(32)

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 13x L 1 ϕ + e c 14x 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 .

(33)

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

(34)

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.

(35)

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.

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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 .

(37)

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 213

ψ 1 + (∂ t + div(d 1 ∇ ) − g 11 · ∇ + a 11 )ψ 3

.

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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 .

(39)

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 := ∂ ij 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.

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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

T

ρ −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

T

ρ 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.

(41)

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

(42)

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.

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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 .

(44)

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 .

(45)

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

(46)

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.

(47)

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

 .

(48)

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?

(49)

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

1

is the density of tumor cells, y

2

is the density of normal cells, y

3

is 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,j

are known constants.

[1] S. Chakrabarty & F. Hanson, Distributed parameters deterministic model for

treatment of brain tumours using Galerkin finite element method, 2009.

(50)

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 ).

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Comments and open problems

4

Numerical scheme using the algebraic resolvability:

Is it possible to deduce a numerical approximation of the solution to the system with one control force

 

 

 

 

t y 1 = div(d 1 ∇ y 1 ) + a 11 y 1 + a 12 y 2 + 1 ω u in Q T ,

t y 2 = div(d 2 ∇ y 2 ) + a 21 y 1 + a 22 y 2 in Q T ,

y = 0 on Σ T ,

y ( · , 0) = y 0 in Ω,

using the study of the system with two control forces

 

 

 

 

t z 1 = div(d 1 ∇ z 1 ) + a 11 z 1 + a 12 z 2 + 1 ω v 1 in Q T ,

t z 2 = div(d 2 ∇ z 2 ) + a 21 z 1 + a 22 z 2 + 1 ω v 2 in Q T ,

z = 0 on Σ T ,

z( · , 0) = z 0 in Ω

and the fictitious control method?

(52)

Merci de votre attention !

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