Null controllability of non diagonalisable parabolic systems Michel Duprez

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Null controllability of non diagonalisable parabolic systems Michel Duprez

Institut de Mathématiques de Marseille

Groupe de travail contrôle et problèmes inverses

Marseille, I2M

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Setting

Let beT >0,ω⊂Ω⊂R^N and consider the system ( _∂

ty=D∆y+A(t, x)y+B1ωu inQT:= Ω×(0, T),

y= 0 on ΣT:=∂Ω×(0, T),

y(·,0) =y⁰ in Ω,

whereD∈ L(Rⁿ)non diagonalisable,A∈L^∞(QT;L(Rⁿ) and B∈ L(R^m,Rⁿ).

We are interested in

theapproximatively controllableat timeT, i.e.

∀y⁰, y^T∈L²(Ω), ε >0∃u∈L²(QT) s.t.ky(T)−y^Tk_L2(Ω)6ε, thenull controllableat timeT, i.e.

∀y⁰∈L²(Ω)∃u∈L²(QT) s.t. y(T) = 0.

Michel Duprez Sur la contrôlabilité de l’équation de la chaleur 1

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Three problems

Problem 1 : ( _∂

ty=D∆y+A(t, x)y+1ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(S1) Conjecture : (S1) always null controllable.

Problem 2 : ( _∂

ty=D∆y+Ay+B1ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(S2)

Conjecture : (S2) null controllable (resp. approx. contr.) if and only if rank[λiD−A|B] =n.

Problem 3 : ( _∂

ty1=D∆y1+A(x)y+B1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(S3) Open problem

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Problem 1 : Carleman

ty=D∆y+A(t, x)y+1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

(S1)

whereD∈ L(Rⁿ) andA∈L^∞(Q;L(Rⁿ)).

Theorem (see [1])

Let us suppose that the dimensions of the Jordan blocks of the canonical form ofDare≤4.

Then System (S1) is null controllable.

[1] Fernández-Cara, González-Burgos, de Teresa, Controllability of linear and semilinear non-diagonalisable parabolic systems

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Duality

Consider the system ( _∂

ty=D∆y+Ay+1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(S1)

Proposition

System (S1) isnull controllableat timeT iff there existsCobs>0 s.t. for allϕ⁰∈L²(Ω) the solutionϕ∈W(0, T;H₀¹(Ω), H⁻¹(Ω)) to

( _−∂

tϕ=D^∗∆ϕ+A^∗ϕ inQT,

ϕ= 0 on ΣT,

ϕ(·, T) =ϕ⁰ in Ω

(D)

satisfies theinequality of observability kϕ(·,0)k²_L2(Ω)6Cobs

Z T 0

Z

ω

ϕ².

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Carleman inequality

Consider the system ( _−∂

tϕ=d∆ϕ inQT, ϕ= 0 on ΣT, ϕ(·, T) =ϕ⁰ in Ω

(D) Let us denote byρp:=ξ^pe^−2sα, where

α(t, x) := exp(4λkη⁰k∞)−exp[2λ(kη⁰k∞+η⁰(x))]

t(T−t) , ξ(t, x) :=exp[λ(2kη⁰k∞+η⁰(x))]

t(T−t) . Here,η⁰∈ C²(Ω) is a function satisfying

|∇η⁰|>κ >0 in Ω\ω2, η⁰>0 in Ω and η⁰= 0 on∂Ω.

Proposition

The solution to system (D) satisfies Z

Q_T

{ρp−1∆ϕ²+ρp+1|∇ϕ|²+ρp+3ϕ²}6C Z

q_T

ρp+3ϕ²+λ⁻¹ Z

Q_T

ρpf²

for alls>s0 andλ>λ0.

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Proof

Let us prove the observability inequality in the simple case : ( _−∂

tϕ=D^∗∆ϕ+A^∗ϕ inQT,

ϕ= 0 on ΣT,

ϕ(·, T) =ϕ⁰ in Ω, where

D^∗:=







1 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1





 ^and^A

∗:=







0 0 0 1

0 0 0 0





^.

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Proof

Using the Carleman inequality : Z

Q_T

{ρ−1∆ϕ²₄+ρ3ϕ²₄}6C Z

q_T

ρ3ϕ²₄+λ⁻¹ Z

Q_T

ρ0∆ϕ²₃ Z

Q_T

{ρ0∆ϕ²3+ρ4ϕ²3}6C Z

q_T

ρ4ϕ²3+λ⁻¹ Z

Q_T

ρ1∆ϕ²2

Z

Q_T

{ρ1∆ϕ²2+ρ5ϕ²2}6C Z

q_T

ρ5ϕ²2+λ⁻¹ Z

Q_T

ρ2∆ϕ²1

Z

Q_T

{ρ2∆ϕ²₁+ρ6ϕ²₁}6C Z

q_T

ρ6ϕ²₁+λ⁻¹ Z

Q_T

ρ3ϕ²₄

We deduce that

4

X

k=1

Z

Q_T

ρ7−kϕ²_k6C

4

X

k=1

Z

q_T

ρ7−kϕ²_k

We conclude using the classical energy inequality Z

QΩ

ϕ(·,0)²6C

4

X

k=1

Z

Ω

Z 3T /4 T /4

ϕ²

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Problem 1 : Uniqueness-compactness

ty=D∆y+A(x)y+1ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

(S1)

whereD∈ L(Rⁿ) andA∈L^∞(Ω;L(Rⁿ)).

Theorem (Duprez-Olive, in preparation) Let us suppose that

• A∈L^∞(Ω;L(Rⁿ)).

• Ω satisfies (GCC).

Then System (S1) is null controllable at timeT.

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Sketch of proof

1 There existsT >0 such that the following system is null contr. at timeT :

( _∂

tty=D∆y+1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰, ∂ty(·,0) =w⁰ in Ω.

2 We can add a compact perturbation : There existsT >0 such that the following system is null contr. at timeT

( _∂

tty=D∆y+Ay+1ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰, ∂ty(·,0) =w⁰ in Ω.

3 We conclude with the transmutation method.

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Compact perturbation

LetH andU be two Hilbert spaces. LetA0:D(A0)⊂H −→H be the generator of aC0-semigroup onH and letB ∈ L(U, H). LetA1∈ L(H) and let us form the unbounded operatorA:=A0+A1withD(A) =D(A0) (we recall thatAis then the generator of aC0-semigroup onH).

Theorem (Duprez-Olive, in preparation) We assume that :

• There existsT^∗>0 such that (A0,B) is exactly controllable at time T^∗.

• A1 is compact.

• The Fattorini criterion holds for (A,B), namely, ker(λ− A^∗)∩kerB^∗={0}, ∀λ∈C.

Then, the pair (A,B) is exactly controllable at timeT for everyT > T^∗.

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Peetre Lemma

Lemma

LetH1, H2, H3 be three Banach spaces. LetL∈ L(H1, H2) and K∈ L(H1, H3) be two linear bounded operators. We assume that

(i) There existsα >0 such that

αkzkH₁≤ kLzkH₂+kKzkH₃, ∀z∈H1

(ii) K is compact.

Then

(i) kerLis finite dimensional.

(ii) If, moreover, kerL={0}, there existsβ >0 such that βkzkH₁≤ kLzkH₂, ∀z∈H1.

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Proof of the general theorem

Since (A0,B) is exactly controllable kzk²_H ≤C

Z T 0

kB^∗SA₀(t)^∗zk²_U dt

≤2C Z T

0

kB^∗SA(t)^∗zk²_U dt+ Z T

0

kB^∗SA₀(t)^∗z− B^∗SA(t)^∗zk²_U dt

.

Therefore, we want to apply the Peetre Lemma to the operators LT : H −→ L²(0, T;U)

z 7−→ B^∗SA(·)^∗z,

KT : H −→ L²(0, T;U)

z 7−→ B^∗(SA₀(·)^∗z−SA(·)^∗z). Plan :

• Step 1 :KT compact

• Step 2 :kerLT ⊂D(A^∗)

• Step 3 :kerLT isA^∗-stable

• Step 4 :kerLT={0}

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Proof of the general theorem

Step 1 :Letzn→zweakly inH. We have SA(t)^∗zn=SA₀(t)^∗zn+

Z t 0

SA₀(t−s)^∗A^∗₁SA(s)^∗znds, ∀t∈[0, T].

Lett∈[0, T]. Using the continuity of (SA(t))t≥0, we have SA(s)^∗zn−−−−−→

n→+∞ SA(s)^∗z weakly inH, ∀s∈[0, t].

SinceA^∗1 is compact, we obtain A^∗₁SA(s)^∗zn−−−−−→

n→+∞ A^∗₁SA(s)^∗z strongly inH, ∀s∈[0, t].

The strong continuity of (SA₀(t))t≥0finally gives SA₀(t−s)^∗A^∗1SA(s)^∗zn−−−−−→

n→+∞ SA₀(t−s)^∗A^∗1SA(s)^∗z strongly inH, ∀s∈[0, t].

Applying Lebesgue’s dominated convergence theorem, we deduce that Z ·

0

SA₀(·−s)^∗A^∗₁SA(s)^∗znds −→

n→+∞

Z · 0

SA₀(·−s)^∗A^∗₁SA(s)^∗znds str. inL²(0, T;U).

This shows thatKT is compact.

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Proof of the general theorem

Step 2 :LetT such thatT > T^∗. Letε∈(0, T −T^∗] so thatT−ε≥T^∗. Letz∈kerLT,tn→0 and

un:=(SA(tn)^∗z−z) tn

.

Letε >0. We havetn< εfor everyn≥N. Then un∈kerLT−ε, ∀n≥N.

Letµ∈ρ(A^∗) and the norm on kerLT−ε : kzk₋₁:=

(µ− A^∗)⁻¹z H. Since (µ− A^∗)⁻¹ andSA(t)^∗commute

(µ− A^∗)⁻¹un=SA(tn)^∗−Id tn

(µ− A^∗)⁻¹z−−−−−→

n→+∞ A^∗(µ− A^∗)⁻¹zinH.

Therefore, (un)n≥N is a Cauchy sequence in kerLT−ε for the normk·k₋₁. SinceT−ε≥T^∗and kerLT−ε is finite dimensional, (un)n≥N is then a Cauchy sequence for the usual normk·k_H hence converges for this norm.

Thusz∈D(A^∗).

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Proof of the general theorem

Step 3 :Letz∈kerLT :

B^∗SA(t)^∗z= 0, ∀t∈[0, T].

Sincez∈D(A^∗), we can differentiate the last equality B^∗SA(t)^∗A^∗z= 0, ∀t∈[0, T], that isA^∗z∈kerLT.

Step 4 :Consequently, the restriction ofA^∗ to kerLT is a bounded linear operator from the finite dimensional space kerLT into itself.

Suppose that kerLT6={0}.

Since kerLT⊂kerB^∗,∃λ∈Candφ6= 0 such that A^∗φ=λφ, B^∗φ= 0, which is in contradiction with the Fattorini criterion.

Thus

kerLT={0} for everyT > T^∗. Using item (ii) of Peetre Lemma,

∃C >0, kzk²_H≤C Z T

0

kB^∗SA(t)^∗zk²_U dt, ∀z∈H.

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Problem 2 : Carleman

ty=D∆y+Ay+B1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

(S2)

whereD∈ L(Rⁿ),A∈ L(Rⁿ) andB∈ L(R^m,Rⁿ).

Theorem (see [1])

Let us suppose that the dimensions of the Jordan blocks of the canonical form ofDare≤4.

Then System (S2) is null controllable if and only if rank[λiD−A|B] =n.

Proof : same idea as before...

[1] Fernández-Cara, González-Burgos, de Teresa, Controllability of linear and semilinear non-diagonalisable parabolic systems

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Problem 2 : Moment method

ty=D∆y+Ay+B1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

(S2)

whereD∈ L(Rⁿ),A∈ L(Rⁿ) andB∈ L(R^m,Rⁿ).

Theorem (Duprez-Gonzalez-Burgos-Souza, in preparation) Let us suppose that

D:=







d 1 0 0 0

0 d 1 0 0

0 0 d 1 0

0 0 0 d 1

0 0 0 0 d





 , A:=







0 0 0 0 0

1 0 0 0 0







andB:=e5.

Then System (S2) is null controllable at timeT.

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Problem 3 : Algebraic resolvability

Let beT >0,ω= (a, b)⊂(0, π) and consider the system











∂ty1=∂xxy+1ωu in (0, T)×(0, π),

∂ty2=∂xxy2+∂xxy1+ay2 in (0, T)×(0, π), y(·,0) =y(·, π) = 0 on (0, T),

y(0,·) =y⁰ in (0, π).

(S3)

Theorem

There exists a coefficienta∈ C^∞([0, π]) such that :

• There exists an open interval (a, b)⊂⊂(0, π) such that, for allT >0, System (S3) is null controllable (then approximatively controllable) at timeT.

• There exists an open interval (a, b)⊂⊂(0, π) such that, for allT >0, System (S3) is not approximatively controllable (then not null controllable) at timeT.

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Proof

Fictitious control method :

Let^ωb^⊂⊂ω. Consider (b^y,bu) a solution to











∂t^yb¹⁼^∂^xx^yb¹⁺¹^ωb^u¹ ⁱⁿ^Q^T^,

∂t^yb²⁼^∂^xx^yb¹⁺^∂^xxb^y²⁺âb^y²⁺¹^ωbû² ⁱⁿ^Q^T^, by= 0 on ΣT, by(·) =y⁰, y(·, T) = 0 in Ω, Supp(bû)^⊂⊂b^ω^×^{(0, T}^).

If we find a solution to







∂tz1=∂xxz1+az+v+b^u¹ ⁱⁿ^Q^T^,

∂tz2=∂xxz1+∂xxz2+az2+b^u² ⁱⁿ^Q^T^, Supp(z, v)⊂⊂ω×(0, T),

(1)

then(b^y⁻^z,^−v)will be a solution of the initial problem.

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Proof

Algebraic problem : Rewrite (1) as

L(z, v) =b^u Let us search a differential operatorMsuch that

L ◦ M=Id (2) Thus (z, v) :=M(bu) will be a solution to (1).

Remark :b^uhave to be regular enough.

We will solve the formal adjoint to (2) :

M^∗◦ L^∗ψ=ψ (3) where

L^∗(ψ) = L^∗₁ψ L^∗2ψ L^∗3ψ

!

=

−∂tψ1−∂xxψ1−∂xxψ2

−∂tψ2−∂xxψ2−aψ2

−ψ1

! .

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Proof

Resolution of the algebraic problem :

We recall the expression toL^∗: L^∗(ψ) =

L^∗1ψ L^∗2ψ L^∗3ψ

!

=

−∂tψ1−∂xxψ1−∂xxψ2

−∂tψ2−∂xxψ2−aψ2

−ψ1

! .

We have

L^∗4ψ:=−L1ψ+ (∂t+∂xx)◦ L3ψ=∂xxψ2. The commutator ofL^∗4 andL^∗2 is :

L^∗₅ψ:= [L^∗₄:L^∗₂]ψ= 2∂x(a)∂xψ2+∂xx(a)ψ2. Moreover

L^∗₆ψ:=∂x◦ L^∗₅ψ−2∂x(a)L^∗₄ψ= 3∂x(a)∂xψ2+∂xxx(a)ψ2

So

L^∗7ψ:= 3∂x(a)L^∗5ψ−2∂x(a)L6^∗ψ= [3∂x(a)∂xx(a)−2∂x(a)∂xxx(a)]ψ2.

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Proof

Thus

M^∗1◦ L^∗1+M^∗2◦ L^∗2=Id, if

3∂x(a)∂xx(a)−2∂x(a)∂xxx(a)6= 0.

That is

a6∈ h1, x, e^3x/2i.

We have proved that system











∂ty1=∂xxy+1ωu in (0, T)×(0, π),

∂ty2=∂xxy2+∂xxy1+ay2 in (0, T)×(0, π), y(·,0) =y(·, π) = 0 on (0, T),

y(0,·) =y⁰ in (0, π).

(S3)

is null controllable if

a6∈ h1, x, e^3x/2i.

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Fattorini Criterion

Let Ω := (0, π),ω:= (5π/12,7π/12) and the system











∂ty1=∂xxy1+1ωu inQT,

∂ty2=∂xxy2+∂xxy1+ay2 inQT,

y= 0 on ΣT,

y(0,·) =y⁰ in Ω,

(S3)

whereu∈L²(QT) is the control anda∈C^∞(Ω) will be specified later.

Theorem

System (S3) isapproximately controllableon the time interval (0, T), if and only if for everys∈Cand everyϕ∈D(∆), we have

−∂xxϕ−∂xxψ=sϕ in Ω

−∂xxψ−aψ=sψ in Ω

ϕ= 0 inω







⇒(ϕ, ψ) = (0,0).

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Proof

The idea will be to construct the functionψas a perturbation of x7→sin(3x). Considerψa function ofC^∞(Ω)∩D(∆) satisfying







ψ(x) = sin(2x) +C1θ1+C2θ2+C3θ3+C4θ4

for allx∈[0,4π/12]∪[8π/12, π], ψ(x) =−2xfor allx∈ω,

whereθ1, θ2, θ3, θ4 are three nontrivial functions ofC^∞(Ω) which will be chosen later on satisfying











Supp(θ1)⊂⊂(π/12,2π/12), Supp(θ2)⊂⊂(2π/12,3π/12), Supp(θ3)⊂⊂(8π/12,9π/2), Supp(θ4)⊂⊂(9π/12,10π/12), θ1, θ2, θ3, θ4>0 in Ω,

ε >0 small enough andC1, C2, C3, C4 are four positive constants to determined

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Proof

The functionψwill have one of the two following forms

0 0.5 1 1.5 2 2.5 3

−1 0

1 sin(2x)

π−2x

Figure– Example of functionψ on [0, π]

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Proof

For aα∈Rto determined, the functionϕ∈ C^∞(Ω) defined for allx∈Ω by ϕ(x) := αsin(2x)−¹₂

Z x 0

sin(3(x−y))∂xxψ(y)dy is solution to

−∂xxϕ−∂xxψ= 2ϕ.

We searchC1,C2 andαsuch thatϕ= 0 inω. Sinceψ=π−2xinω, ϕ(x) =

α+ 1−cos(^5π₆)−sin(^5π₆ )^π₆ + 2 Z 5π/12

0

cos(2y)ψ(y)dy

sin(2x) +

sin(^5π₆ ) + cos(^5π₆ )^π₆ + 2 Z 5π/12

0

sin(2y)ψ(y)dy

cos(2x), for allx∈ω.

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Proof

Let us distinguish two cases :

1 If sin(5π

6 )+cos(5π 6 )π

6+2 Z 4π/12

0

cos(2y) sin(2y)dy+2 Z 5π/12

4π/12

sin(2y)ψ(y)dy (1) is negative, since sin(2x), cos(2x)>0 in (2π/12,3π/12), forC2= 0, one can choseC1 such that

sin(5π

6 ) + cos(5π 6 )π

6+ 2 Z 5π/12

0

sin(3y)ψ(y)dy= 0. (2)

2 If the quantity (1) is positive, since sin(2x)>0 and cos(2x)<0 in (3π/12,4π/12), forC1= 0, one can choseC2>0 such that (2) holds.

Thus, forαgiven by α:=−1 + cos(5π

6 ) + sin(5π 6 )π

6 −2 Z 5π/12

0

cos(2y)ψ(y)dy, we obtainϕ= 0 inω.

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Proof

By definition ofϕ, we haveϕ(0) = 0.

Now we searchC2, C3such thatϕ(π) = 0.

We remark that

ϕ(π) = 2 Z π

0

sin(2y)ψ(y)dy.

Let us distinguish two cases :

1 If

1 3

Z 8π/12 0

sin(2y)ψ(y)dy+1 3

Z π 8π/12

sin(2y) sin(2y)dy (3) is positive, then, using the fact that sin(2x), cos(2x)<0 for all x∈(8π/12,9π/12), one can chooseC4:= 0 and find some some C3>0 such thatϕ(π) = 0.

2 If now the quantity (3) is negative, since sin(2x)>0 and cos(2x)<0 for allx∈(9π/12,10π/12), one can chooseC3:= 0 and find some someC4>0 such thatϕ(π) = 0.

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Proof

We define the functiona∈ C^∞(Ω) as follows a:=−∆ψ−4ψ

ψ ∈ C^∞(Ω).

Thus the three functionsϕ,ψ,a∈ C^∞(Ω) satisfy











−∂xxϕ−∂xxψ= 4ϕ in Ω,

−∂xxψ−aψ= 4ψ in Ω,

ϕ(0) =ϕ(π) =ψ(0) =ψ(π) = 0,

ϕ= 0 inω,

(ϕ, ψ)6= 0 in Ω.

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Using Fattorini Criterion, System (S3) is not approximately controllable on the time interval (0, T).

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Open problems

Problem 1 :

å Is System (S1) always null controllable ? å Carleman inequality

Problem 2 :

å Is (S2) null controllable (resp. approx. contr.) if and only if rank[λiD−A|B] =n?

Problem 3 :

å Necessary and sufficient conditions å Minimal time of controllability

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Null controllability of non diagonalisable parabolic systems Michel Duprez