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Partial null controllability Open problem Conclusions and perspectives

Partial controllability of parabolic systems

Michel Duprez

Laboratoire de Mathématiques de Besançon

Sept. 30th 2014

Contrôle, Problème inverse et Applications.

M. Duprez Partial controllability of parabolic systems Sept. 30th 2014 1 / 29

Motivations

Let be T > 0 and ω ⊂ Ω ⊂ RN .If we consider, for example, the following parabolic system

Find y := (y1, y2)∗ : QT → R2 such that∂ty1 = ∆y1 + y2 + 1ωu in QT := Ω× (0,T ),∂ty2 = ∆y2 in QT ,y = 0 on ΣT := ∂Ω× (0,T ),y(0) = y0 in Ω,

can we find, for all initial condition y0, a control u such that

y1(T ; y0,u) = 0 ?


Setting

We consider here the following system of n linear parabolicequations with m controls ∂ty = ∆y + Ay + B1ωu in QT := Ω× (0,T ),y = 0 on ΣT := ∂Ω× (0,T ),y(0) = y0 in Ω, (1)where A := (aij )ij and B := (bik )ik with aij , bik ∈ L∞(QT ) for all1 6 i , j 6 n and 1 6 k 6 m.

We denote byP : Rp × Rn−p → Rp,

(y1, y2)∗ 7→ y1.


Setting

We consider here the following system of n linear parabolicequations with m controls ∂ty = ∆y + Ay + B1ωu in QT := Ω× (0,T ),y = 0 on ΣT := ∂Ω× (0,T ),y(0) = y0 in Ω, (1)where A := (aij )ij and B := (bik )ik with aij , bik ∈ L∞(QT ) for all1 6 i , j 6 n and 1 6 k 6 m.

We denote byP : Rp × Rn−p → Rp,

(y1, y2)∗ 7→ y1.


Definitions

We will say that system (1) ispartially approximately controllable on (0,T ) if for all � > 0and y0, yT ∈ L2(Ω)n there exists a control u ∈ L2(QT )m such that

‖Py(·,T ; y0,u)− PyT (·)‖L2(Ω)p 6 �.

partially null controllable on (0,T ) if for all y0 ∈ L2(Ω)n, thereexists a u ∈ L2(QT )m such that

Py(·,T ; y0,u) = 0 in Ω.


Definitions

We will say that system (1) ispartially approximately controllable on (0,T ) if for all � > 0and y0, yT ∈ L2(Ω)n there exists a control u ∈ L2(QT )m such that

‖Py(·,T ; y0,u)− PyT (·)‖L2(Ω)p 6 �.

partially null controllable on (0,T ) if for all y0 ∈ L2(Ω)n, thereexists a u ∈ L2(QT )m such that

Py(·,T ; y0,u) = 0 in Ω.


Plan

1 Partial null controllabilityAutonomous case for differential equationsConstant matricesTime dependent matrices

2 Open problem

3 Conclusions and perspectives


Plan

1 Partial null controllabilityAutonomous case for differential equationsConstant matricesTime dependent matrices

2 Open problem



Autonomous case for differential equations

System of differential equations

Assume that A ∈ L(Rn) and B ∈ L(Rm,Rn) and consider the system{∂ty = Ay + Bu in (0,T ),y(0) = y0 ∈ R.

(2) P : Rp × Rn−p → Rp,(y1, y2)∗ 7→ y1.

THEOREM (Kalman, 1969)

System (2) is controllable on (0,T ) if and only if

rank(B|AB|...|An−1B) = n.

THEOREM

System (2) is partially controllable on (0,T ) if and only if

rank(PB|PAB|...|PAn−1B) = p.





(2) P : Rp × Rn−p → Rp,(y1, y2)∗ 7→ y1.




THEOREM







(2) P : Rp × Rn−p → Rp,(y1, y2)∗ 7→ y1.




THEOREM





Sketch of proof for the null controllability for n=3, m=1

⇐ • Let us first consider the variable change : z := y − ηy , where{∂ty = Ay in [0,T ],y(0) = y0 ∈ R

and η ∈ C∞[0,T ] with the following form

6

-0 T3

2T3 T

1 η

Thus, if h := −∂tηy , z is solution to{∂tz = Az + Bu + h in [0,T ],z(0) = 0.




•We recall that{∂tz = Az + Bu + h in [0,T ],z(0) = 0. (2)

If we search a solution of the form z := Kw with K := (B|AB|A2B)invertible, w satisfies

K∂tw = AKw + Bu + h.

Using the Cayley Hamilton Theorem, A3 := α0I + α1A + α2A2.

Thus{

AK = (AB|A2B|A3B) = KC,Ke1 = B,

with C =

0 0 α01 0 α10 1 α2

.




We recall that h := −∂tηy .Thus w is solution to

∂tw1 = α0w3 + h1 + u,∂tw2 = w1+ α1w3 + h2,∂tw3 = w2+ α2w3 + h3,w(0) = 0.

We choose w3 = 0,w2 = −h3,w1 = ∂tw2 − α1w2 − h2,

u = ∂tw1 − α0w1 − h1.

Conclusion :

wi (0) = wi (T ) = 0 for all i ∈ {1,2,3}. �


Constant matrices

Constant matrices

Assume that A ∈ L(Rn) and B ∈ L(Rm,Rn) and consider the system ∂ty = ∆y + Ay + B1ωu in QT ,y = 0 on ΣT ,y(0) = y0 in Ω. (2)THEOREM (Ammar-Khodja et al, 2009)

System (2) is null controllable on (0,T ) if and only


THEOREM

System (2) is partially null controllable on (0,T ) if and only



Constant matrices

Constant matrices




THEOREM




Constant matrices

Constant matrices




THEOREM




Constant matrices

Proof in the case n=3,m=1, p=1, P=(1,0,0)

Let us consider the system∂ty = ∆y +

a11 a12 a13a21 a22 a23a31 a32 a33

y + b1b2

b3

1ωu in QT ,y = 0 on ΣT ,

y(0) = y0 in Ω.

Goal : Find u ∈ L2(QT ) such that

y1(T ; y0,u) = 0.

Strategy : Find a variable change y = M(t)w with w solution of acascade system and use[1] González-Burgos M, de Teresa L. : Controllability results for cascade systems of m

coupled parabolic PDEs by one control force


Constant matrices



a11 a12 a13a21 a22 a23a31 a32 a33

y + b1b2

b3


y(0) = y0 in Ω.


y1(T ; y0,u) = 0.




Constant matrices



a11 a12 a13a21 a22 a23a31 a32 a33

y + b1b2

b3


y(0) = y0 in Ω.


y1(T ; y0,u) = 0.




Constant matrices


Let beK := [A|B] = (B|AB|A2B),s := rank(K ),X := span〈B,AB,A2B〉.

(i) s=1:

Since{

B 6= 0rank(B|AB|A2B) = 1) , then X = span〈B〉.

In particularAB := α1B and A2B := α2B.


Constant matrices


Let beK := [A|B] = (B|AB|A2B),s := rank(K ),X := span〈B,AB,A2B〉.

(i) s=1:

Since{

B 6= 0rank(B|AB|A2B) = 1) , then X = span〈B〉.

In particularAB := α1B and A2B := α2B.


Constant matrices


⇐ Let us suppose that rank(PK ) = 1.We define

M(t) := (B|M2(t)|M3(t)),

where {∂tM2 = AM2,M2(T ) = e2

and{∂tM3 = AM3,M3(T ) = e3.

We have b1 6= 0, indeed

rank(PB|PAB|PA2B) = rank(PB|α1PB|α2PB)= rank(b1|α1b1|α2b1) = 1.

Then M(T ) is invertible and there exists T ∗ such that M(t) isinvertible in [T ∗,T ].


Constant matrices


Matrix M satisfies{−∂tM(t) + AM(t) = (AB|0|0) = M(t)C,M(t)e1 = B,

with C given by

C :=

α1 0 00 0 00 0 0

.If T∗ = 0 : Since M(t) is invertible on [0,T ], y is the solution tosystem (2) if and only if w := (w1,w2,w3)∗ = M(t)−1y is the solutionto the cascade system ∂tw = ∆w + Cw + e11ωu in QT ,w = 0 on ΣT ,w(0) = w0 in Ω.


Constant matrices


We recall that y = M(t)w in QT and

M(T ) =

b1 0 0b2 1 0b3 0 1

.Thus y1(T ) = b1w1(T ) = 0 in Ω.And it is possible to find u ∈ L2(QT ) such that the solution to cascadesystem satisfies

w1(T ) = 0 in Ω.

If now T∗ 6= 0 :

The system is considered without control until T ∗ and we control onlyon [T ∗,T ].


Constant matrices


We recall that y = M(t)w in QT and

M(T ) =

b1 0 0b2 1 0b3 0 1

.Thus y1(T ) = b1w1(T ) = 0 in Ω.And it is possible to find u ∈ L2(QT ) such that the solution to cascadesystem satisfies

w1(T ) = 0 in Ω.

If now T∗ 6= 0 :

The system is considered without control until T ∗ and we control onlyon [T ∗,T ].


Constant matrices


⇒ If now rank(PB|PAB|PA2B) 6= 1. Let us remark that

rankP[A|B] 6 rank[A|B] = s = 1.

Thus rank(PB|PAB|PA2B) = 0 and PB = PAB = PA2B = 0.Since B 6= 0, let us suppose that b2 6= 0.We define

M := (B|e1|e3) =

0 1 0b2 0 0b3 0 1

.We have Ae1 := c12B + c22e1 + c32e3, Ae3 := c13B + c23e1 + c33e3.Hence {

AM = (α1B|Ae1|Ae3) = MC,Ke1 = B,

with C given by C :=

α1 c12 c130 c22 c230 c32 c33

.M. Duprez Partial controllability of parabolic systems Sept. 30th 2014 17 / 29

Constant matrices


Since K is invertible, y is the solution to (2) if and only ifw := (w1,w2,w3)∗ = M−1y is the solution to cascade system

∂tw = ∆w + Cw + e11ωu in QT ,w = 0 on ΣT ,w(0) = M−1y0 in Ω.

Since M is given by

M := (B|e1|e3) =

0 1 0b2 0 0b3 0 1

.Thus we have

y1(T ) = 0 in Ω⇔ w2(T ) = 0 in Ω.


Constant matrices


Since K is invertible, y is the solution to (2) if and only ifw := (w1,w2,w3)∗ = M−1y is the solution to cascade system

∂tw = ∆w + Cw + e11ωu in QT ,w = 0 on ΣT ,w(0) = M−1y0 in Ω.

Since M is given by

M := (B|e1|e3) =

0 1 0b2 0 0b3 0 1

.Thus we have

y1(T ) = 0 in Ω⇔ w2(T ) = 0 in Ω.


Constant matrices

Remark on the general dual system : −∂tϕ = ∆ϕ+ A∗ϕ in QT ,

ϕ = 0 on ΣT ,ϕ(T , ·) = P∗ϕT in Ω.

with P∗ : Rp → Rp × Rn−pϕT 7→ (ϕT ,0n−p)∗.

PROPOSITION

1 System (1) is partially null controllable on (0,T ), if and only ifthere exists Cobs > 0 such that for all ϕT ∈ L2(Ω)p

‖ϕ(0)‖2L2(Ω)n 6 Cobs∫ T

0‖B∗ϕ‖2L2(ω)m

with ϕ the solution to the dual system.2 System (1) is partially approx. controllable on (0,T ), if and

only if for all ϕT ∈ L2(Ω)p the solution to the dual system satisfies

B∗ϕ = 0 in ω × (0,T ) ⇒ ϕ = 0 in QT .


Constant matrices


ϕ = 0 on ΣT ,ϕ(T , ·) = P∗ϕT in Ω.


PROPOSITION


‖ϕ(0)‖2L2(Ω)n 6 Cobs∫ T

0‖B∗ϕ‖2L2(ω)m



B∗ϕ = 0 in ω × (0,T ) ⇒ ϕ = 0 in QT .


Constant matrices


ϕ = 0 on ΣT ,ϕ(T , ·) = P∗ϕT in Ω.


PROPOSITION


‖ϕ(0)‖2L2(Ω)n 6 Cobs∫ T

0‖B∗ϕ‖2L2(ω)m



B∗ϕ = 0 in ω × (0,T ) ⇒ ϕ = 0 in QT .


Constant matrices


The previous cascade system is partially null controllable if and only ifthere exists C > 0 such that for all ϕT2 ∈ L2(Ω) the solution to the dualsystem

−∂tϕ1 = ∆ϕ1 + α1ϕ1 in QT ,−∂tϕ2 = ∆ϕ2 + c12ϕ1 + c22ϕ2 + c32ϕ3 in QT ,−∂tϕ3 = ∆ϕ3 + c13ϕ1 + c23ϕ2 + c33ϕ3 in QT ,ϕ = 0 on ΣT ,ϕ(T ) = (0, ϕT2 ,0)

∗ in Ω

satisfies the observability inequality∫Ω

ϕ(0)2 6 C∫ω×(0,T )

ϕ21.

We have ϕ1(T ) = 0 in Ω⇒ ϕ1 = 0 in QTand ϕT2 6≡ 0 in Ω⇒ (ϕ2(0), ϕ3(0))∗ 6≡ 0 in Ω.Thus the observability inequality is not satisfy.


Constant matrices


The previous cascade system is partially null controllable if and only ifthere exists C > 0 such that for all ϕT2 ∈ L2(Ω) the solution to the dualsystem

−∂tϕ1 = ∆ϕ1 + α1ϕ1 in QT ,−∂tϕ2 = ∆ϕ2 + c12ϕ1 + c22ϕ2 + c32ϕ3 in QT ,−∂tϕ3 = ∆ϕ3 + c13ϕ1 + c23ϕ2 + c33ϕ3 in QT ,ϕ = 0 on ΣT ,ϕ(T ) = (0, ϕT2 ,0)

∗ in Ω

satisfies the observability inequality∫Ω

ϕ(0)2 6 C∫ω×(0,T )

ϕ21.

We have ϕ1(T ) = 0 in Ω⇒ ϕ1 = 0 in QTand ϕT2 6≡ 0 in Ω⇒ (ϕ2(0), ϕ3(0))∗ 6≡ 0 in Ω.Thus the observability inequality is not satisfy.


Constant matrices


(ii) s=2: We take M(t) := (B|AB|M3(t)) with M3(T ) = e2 or e3.(iii) s=3: We take M := K = (B|AB|A2B).

�Remark : We prove also the partial approximate controllability.


Time dependent matrices


Let us suppose that A ∈ Cn−1([0,T ];L(Rn)) andB ∈ Cn([0,T ];L(Rm;Rn)) and define{

B0(t) := B(t),Bi (t) := A(t)Bi−1(t)− ∂tBi−1(t), 1 6 i 6 n − 1.

We can define for all t ∈ [0,T ]

[A|B](t) := (B0(t)|B1(t)|...|Bl−1(t)).



THEOREM (Ammar-Khodja et al 09)1 If there exist t0 ∈ [0,T ] such that

rank[A|B](t0) = n,

then system (1) is null controllable on (0,T).2 System (1) is null controllable on every interval (T0,T1) with

T0 < T1 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.

THEOREMIf

rankP[A|B](T ) = p,

then system (1) is partially null controllable on (0,T).



THEOREM (Ammar-Khodja et al 09)1 If there exist t0 ∈ [0,T ] such that

rank[A|B](t0) = n,

then system (1) is null controllable on (0,T).2 System (1) is null controllable on every interval (T0,T1) with

T0 < T1 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.

THEOREMIf

rankP[A|B](T ) = p,

then system (1) is partially null controllable on (0,T).


Plan

1 Partial null controllability

2 Open problem



An example with A = A(t)

Let us consider the control problem∂ty1 = ∆y1 + ay2 + 1ωu in QT ,∂ty2 = ∆y2 in QT .y = 0 on ΣT ,y(0) = y0, y1(T ) = 0 in Ω.

If a := a(t), let us consider the change of variable z1 := y1 + µy2

(∂t −∆)z1 = (∂tµ+ a)y2 + 1ωu in QT .

If we choose µ solution to{∂tµ = −a in [0,T ],µ(T ) = 0,

the system is not coupled. And thus it is partially null controllable.


Difficulty when A = A(x)

Let us consider the same control problem∂ty1 = ∆y1 + ay2 + 1ωu in QT .∂ty2 = ∆y2 in QT ,y = 0 on ΣT ,y(0) = y0, y1(T ) = 0 in Ω.

If now a := a(x), let us consider the change of variable z1 := y1 + µy2

(∂t −∆)z1 = (a− ∂tµ+ ∆µ+ 2∇µ · ∇)y2 + 1ωu in QT

What kind of µ can be chosen in this situation ???[1] Benabdallah A.; Cristofol M.; Gaitan P.; De Teresa, Luz : Controllability totrajectories 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-Stokessystem with a distributed control having two vanishing components (2014)


Difficulty when A = A(x)

Let us consider the same control problem∂ty1 = ∆y1 + ay2 + 1ωu in QT .∂ty2 = ∆y2 in QT ,y = 0 on ΣT ,y(0) = y0, y1(T ) = 0 in Ω.

If now a := a(x), let us consider the change of variable z1 := y1 + µy2

(∂t −∆)z1 = (a− ∂tµ+ ∆µ+ 2∇µ · ∇)y2 + 1ωu in QT

What kind of µ can be chosen in this situation ???[1] Benabdallah A.; Cristofol M.; Gaitan P.; De Teresa, Luz : Controllability totrajectories 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-Stokessystem with a distributed control having two vanishing components (2014)


Plan

1 Partial null controllability

2 Open problem



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.

Perspectives :1 Example of non partial controllability in the space dependent

case.2 Numerical simulations and analysis.3 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.

Perspectives :1 Example of non partial controllability in the space dependent

case.2 Numerical simulations and analysis.3 Partial controllability of semilinear parabolic systems.


Thank you for your attention !


Partial null controllabilityAutonomous case for differential equationsConstant matrices Time dependent matrices

Open problemConclusions and perspectives

Partial controllability of parabolic systemsmath.univ-bpclermont.fr/~munch/duprez14.pdfPartial null...

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Transcript of Partial controllability of parabolic systemsmath.univ-bpclermont.fr/~munch/duprez14.pdfPartial null...