Controllability of parabolic systems: the moment method Evolution - - PowerPoint PPT Presentation

Controllability of parabolic systems: the moment method

Evolution Equations: long time behavior and control Farid Ammar Khodja

Laboratoire de Mathématiques de Besançon

Mathematics in Savoie: 15-18 june 2015

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Introduction

The main goal of this course is to give a review of results relating controllability issues for parabolic systems obtained via the moment method (as used by Fattorini and Russell in the seventies). Other powerfull techniques have been developped during these last 20 years:

Carleman inequalities: Fursikov-Imanuvilov. The Lebeau-Robbiano method. The return method: J-M Coron. The transmutation method: Miller.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

The control system

Consider the following system: 8 < : (∂t D∆ A) y = Bu1ω, QT := (0, T) Ω, y = Cv1Γ0, ΣT := (0, T) ∂Ω, y (0, ) = y0, Ω, where: Ω RN is a smooth bounded domain, ω Ω is an open set, Γ0 ∂Ω is a relatively open subset; D = diag (d1, ..., dn) , A = (aij)1i,jN 2 L∞ (QT ; L (Rn)) , B = (bij) , C = (cij) 2 L∞ (QT ; L (Rm, Rn)) : control matrices.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Concepts of controllability

Approximate controllability at time T > 0 : for all ε > 0, for all

• y0, y1 2 X X, there exists (u, v) 2 L2 (QT ) L2 (ΣT ) such that
• y (T) y1
• X ε.

Null controllability at time T > 0 : for all y0 2 X , there exists (u, v) 2 L2 (QT ) L2 (ΣT ) such that y (T) = 0 in Ω. Here X is a space where the system is well-posed: for example X = L2 (Ω; Rn) when C = 0 or X = H1 (Ω; Rn) when C 6= 0 leads to a solution y 2 C ([0, T] , X) if the initial datum y0 2 X.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Dual concepts: adjoint system and observability

Consider the dual system: 8 < : (∂t + D∆ + A) ϕ = 0, QT := (0, T) Ω, ϕ = 0, ΣT := (0, T) ∂Ω, ϕ (T, ) = ϕ0, Ω, If ϕ0 2 X = L2 (Ω) , then there is a unique solution ϕ 2 C

• [0, T] ; L2

Ω; RN \ L2 0, T; H1

• Ω, RN

. If ϕ0 2 X = H1

0 (Ω) , then there is a unique solution

ϕ 2 C

• [0, T] ; H1
• Ω, RN

\ L2 0, T; H2 \ H1

• Ω, RN

.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

If C = 0 (distributed control):

Theorem

The system is approximately controllable in X = L2 Ω; RN if, and only if, the unique continuation property holds true for any solution ϕ 2 C [0, T] ; L2 Ω; RN

• f the adjoint problem:
• 1ωBϕ = 0, in QT =

) ϕ0 = 0 .

Theorem

The system is null controllable in X = L2 Ω; RN if, and only if there exists CT > 0 such that kϕ (0)k2

X CT

Z T Z

ω jBϕj2

for any ϕ0 2 X.

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If B = 0 (boundary control):

Theorem

The system is approximately controllable in X = H1 Ω; RN if, and

• nly if, the unique continuation property holds true for any solution

ϕ 2 C [0, T] ; H1

0 (Ω)

• f the adjoint problem:
• 1Γ0C ϕ = 0, in ΣT =

) ϕ0 = 0 .

Theorem

The system is null controllable in X = H1 Ω; RN if, and only if, there exists CT > 0 such that kϕ (0)k2

H 1

0 (Ω;RN ) CT

Z T Z

Γ0

• C ∂ϕ

∂ν

• 2

for any ϕ0 2 H1

• Ω; RN

.

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Results for the scalar case

Theorem

The problem 8 < : (∂t ∆ a) y = u1ω, QT := (0, T) Ω, y = 0, ΣT := (0, T) ∂Ω, y (0, ) = y0, Ω, is null and approximately controllable in X = L2 (Ω) for any open set ω Ω, provided that a 2 L∞ (QT ) . As a consequence, the problem 8 < : (∂t ∆ a) y = 0, QT := (0, T) Ω, y = v1Γ0, ΣT := (0, T) ∂Ω, y (0, ) = y0, Ω, is null and approximately controllable in X = H1 (Ω) for any relatively

• pen set Γ0 ∂Ω.

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Remarks on the scalar case

Approximate and null controllability are equivalent concepts in the scalar case.

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Remarks on the scalar case

Approximate and null controllability are equivalent concepts in the scalar case. Boundary and distributed controllability are also equivalent.

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Remarks on the scalar case

Approximate and null controllability are equivalent concepts in the scalar case. Boundary and distributed controllability are also equivalent. Unlike the hyperbolic case, the minimal time of control is Tmin = 0.

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The techniques used for the proof

Fursikov and Imanuvilov (95’) have proved a global Carleman inequality: Let η(t, x) : = s β0(x) t(T t), (t, x) 2 QT = Ω (0, T), ρ(t, x) : = s t(T t), (t, x) 2 QT . and I(τ, ϕ) =

Z

QT

ρτ1e2η jϕtj2 + j∆ϕj2 + ρ2 jrϕj2 + ρ4 jϕj2 .

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The techniques used for the proof

Fursikov and Imanuvilov (95’) have proved a global Carleman inequality: Let η(t, x) : = s β0(x) t(T t), (t, x) 2 QT = Ω (0, T), ρ(t, x) : = s t(T t), (t, x) 2 QT . and I(τ, ϕ) =

Z

QT

ρτ1e2η jϕtj2 + j∆ϕj2 + ρ2 jrϕj2 + ρ4 jϕj2 . Carleman inequality: there exist a positive function β0 2 C 2(Ω), s0 > 0 and C > 0 such that 8s s0 and 8τ 2 R : I(τ, ϕ) C Z

QT

ρτe2η jϕt c∆ϕj2 +

Z T Z

ω ρτ+3e2η jϕj2

• ,

for any function ϕ satisfying ϕ = 0 on ΣT and for which the right hand-side is de…ned.

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For 8 < : (∂t + ∆ + a) ϕ = 0, QT := (0, T) Ω, ϕ = 0, ΣT := (0, T) ∂Ω, ϕ (T, ) = ϕ0, Ω, this inequality gives in particular

Z

QT

e2ηρ4 jϕj2

• C

Z

QT

ρe2η jaϕj2 +

Z T Z

ω ρ4e2η jϕj2

• +

Z

QT

e2ηρ4 jϕj2

• C

Z T Z

ω ρ4e2η jϕj2 , 8s s0 (kak∞) .

Using that there exists α 2 R such that t 7! R

Ω eαt ϕ2 is increasing, we get

the observability inequality kϕ (0)k2

L2(Ω) C

Z T Z

ω jϕj2

and thus the null controllability result.

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A …rst application to a parabolic system

8 > > < > > : (∂t ∆) y1 = a11y1 + a12y2 (∂t d∆) y2 = a21y1 + a22y2 + u1ω, QT := (0, T) Ω, y = (y1, y2) = 0, ΣT := (0, T) ∂Ω, y (0, ) = y0, Ω,

Theorem

If there exists ω0 ω such that a12 σ > 0 on (0, T) ω0 then the system is null (and approximately) controllable for any d > 0.

Proof.

Carleman inequality for each equation of the adjoint system and then use the assumption a12 σ > 0 on (0, T) ω0. This result generalizes to n n cascade systems.

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A natural question arises at this level: what happens if supp (a12) \ ω = ? ?

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

A natural question arises at this level: what happens if supp (a12) \ ω = ? ? What happens for the boundary control system: 8 > > > > < > > > > : (∂t ∆) y1 = a11y1 + a12y2 (∂t d∆) y2 = a21y1 + a22y2, QT := (0, T) Ω, y = y1 y2

• =

1

• v,

ΣT := (0, T) ∂Ω, y (0, ) = y0, Ω, ?

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

A natural question arises at this level: what happens if supp (a12) \ ω = ? ? What happens for the boundary control system: 8 > > > > < > > > > : (∂t ∆) y1 = a11y1 + a12y2 (∂t d∆) y2 = a21y1 + a22y2, QT := (0, T) Ω, y = y1 y2

• =

1

• v,

ΣT := (0, T) ∂Ω, y (0, ) = y0, Ω, ? There exist only partial answers to these two questions... even in the

• ne-dimensional case.

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Scalar parabolic equations in the one-dimensional case

Let us study the null controllability property of the problem: 8 < : y 0 yxx = f (x) u (t) , QT = (0, T) (0, π) yjx=0,π = 0, (0, T) yjt=0 = y0 (0, π) Here the constraint is that the control has separate variables: f 2 L2 (0, π) and u 2 L2 (0, T) . If ϕk (x) = q

2 π sin (kx) , then fϕkgk1 is an orthonormal basis of

L2 (0, π) . We look for a solution in the form y (t, x) = ∑

k1

yk (t) ϕk (x) Set f (x) = ∑

k1

fk sin (kx) , y0 = ∑

k1

y0

k sin (kx) ,

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Then y is a solution if, and only if, y 0

k = k2yk + fku (t) ,

(0, T) ykjt=0 = y0

k ,

, 8k 1. i.e. yk (t) = ek 2ty0

k + fk

Z t

0 ek 2(ts)u (s) ds, 8k 1.

There exists a control function u 2 L2 (0, T) such that the solution satis…es y (T, x) = 0 for any x 2 (0, π) if, and only if, there exists a control function u 2 L2 (0, T) such that: fk

Z T

ek 2(T s)u (s) ds = ek 2T y0

k , k 1.

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After a change of variable in the integral, we arrive to (v (t) = u (T t)): ( Find v 2 L2 (0, T) : fk R T

0 ek 2tv (t) dt = ek 2T y0 k , k 1.

This is a moment problem in L2 (0, T) . A necessary condition for the existence of a solution for any y0 2 L2 (0, π) is: fk 6= 0, k 1. If n ek 2to

k1 admits a biorthogonal family fqkgk1 in L2 (0, T) ,

• i. e.

Z T

ek 2tq` (t) dt = δk`, k, ` 1, then a formal solution is v (t) = ∑

k1

ek 2T fk y0

k qk.

The question is then: v 2 L2 (0, T)?

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Generalization: real exponential families

Let fλkg R such that 0 < λ1 < λ2 < < λk < , lim

k!∞ λk = ∞.

Let fmkgk1 2 `2 and consider the moment problem: ( Find v 2 L2 (0, T) : R T

0 eλktv (t) dt = mk, k 1.

To solve this problem, we need to answer the following two questions:

1

Does the family

• eλkt

k1 admit a biorthogonal family fqkgk1in

L2 (0, T)?

2

If a biorthogonal fqkgk1 family exists, is it possible to estimate kqkkL2(0,T ) as k ! ∞?

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Answer following Laurent Schwartz (1942)

As a …rst step, consider

• eλkt

k1 in L2 (0, ∞) . Then

Theorem

The family

• eλkt

k1 is

1

complete in L2 (0, ∞) if ∑k1 1/λk = ∞ and in this case, it is not minimal;

2

minimal in L2 (0, ∞) if ∑k1 1/λk < ∞ and in this case, it is not complete. The proof is based on classical properties of the Laplace transform and zeros of holomorphic functions.

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The Laplace transform

Let f 2 L2 (0, ∞) and its Laplace transform F given by: F (λ) =

Z ∞

eλtf (t) dt, Re λ > 0.

1

F 2 H (C+) , the space of holomorphic functions on C+ = fλ 2 C : Re (λ) > 0g .

2

For any ε > 0, F 2 H (Cε) the space of bounded holomorphic functions on Cε = fλ 2 C : Re (λ) > εg and moreover lim

jλj!∞,λ2Cε

F (λ) = 0.

3

The space H2 (C+) = n F 2 H (C+) : R

R jF (σ + iτ)j2 dτ < ∞, 8σ > 0

• is a

Hilbert space with norm kFkH2(C+) = R

R jF (iτ)j2 dτ

1/2 and the Laplace transform is an isometry from L2 (0, ∞) in H2 (C+) : kFkH2(C+) = kf kL2(0,∞) , f 2 L2 (0, ∞) .

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Zeros of holomorphic functions

Theorem

If f 2 H∞ (C+) is a nontrivial function and if Λ = (zk)k1 is the sequence of its zeroes in C+, then

∑

k1

R (zk) 1 + jzkj2 < ∞. (1) If (1) is satis…ed for a sequence Λ = (zk)k1 C+, then the in…nite product W (λ) = ∏

k1

1 λ/zk 1 + λ/zk , λ 2 C+ (2) converges absolutely in C+ and de…nes a function W 2 H∞ (C+) whose set of zeroes is the sequence Λ.

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Proof

Assume that ∑n1 1/λn = ∞ and let ϕ 2 L2 (0, ∞) such that: 8n 1,

Z ∞

eλnt ϕ (t) dt = 0. Let J : C+ ! C be the Laplace transform of ϕ : J (λ) =

Z ∞

eλt ϕ (t) dt. J is holomorphic on C+ and uniformly bounded on Cε = fλ 2 C : Re (λ) > εg for all ε > 0. Moreover, by the assumption on ϕ, J (λn) = 0 for all n 1. With the hypothesis on the sequence (λn) , ∑k1

λk 1+λ2

k < ∞. But this

condition is equivalent to ∑n1 1/λn < ∞. This contradicts the starting assumption and thus implies that ϕ = 0. So

• eλnt

n1 is complete in L2 (0, ∞) . However, it is not minimal

since for any k 1, the sequence (λn)n1

n6=k

still has the same properties.

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Assume now that ∑n1 1/λn < ∞. Set W (λ) = ∏

k1

1 λ/λk 1 + λ/λk , λ 2 C+ If a function J is de…ned by J (λ) = W (λ) (1 + λ)2 , λ 2 C+ then J 2 H2 (C+) . From the properties of the Laplace transform, there exists a nontrivial function ϕ 2 L2 (0, ∞) such that J (λ) =

Z ∞

eλt ϕ (t) dt, λ 2 C+,

Z ∞

jϕ (t)j2 dt =

Z +∞

∞ jJ (iτ)j2 dτ

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The λn’s are the zeros of J and thus it follows that the family

• eλnt

n1 is not complete since ϕ is non trivial and belongs to

heλnt, n 1i

?.

To prove that

• eλnt

n1 is minimal, a biorthogonal family will be

explicitly built for which an estimate of the asymptotic behavior of the norm of its element will be given.

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Set for k 1, Jk (λ) = J (λ) J0 (λk) (λ λk), λ 2 C+. where J is the previously de…ned function. It can easily be proved that Jk 2 H2 (C+) . Again, from the Laplace transform properties, there exists qk 2 L2 (0, ∞) such that: Jk (λ) =

Z ∞

eλtqk (t) dt, λ 2 C+,

Z ∞

jqk (t)j2 dt =

Z +∞

∞ jJk (iτ)j2 dτ

Since, by de…nition, Jk (λn) = δkn, the biorthogonality of the families

• eλnt

n1 and (qn)n1 follows.

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To estimate the norm of qk, we have from the previous considerations:

Z ∞

jqk (t)j2 dt =

Z +∞

∞

• J (iτ)

J0 (λk) (iτ λk)

• 2

dτ Since, jW (iτ)j = 1 for all τ 2 R, it appears that:

Z ∞

jqk (t)j2 dt = 2 jλkJ0 (λk)j2

Z ∞

dτ (1 + τ2)2

• τ

λk

• 2

+ 1

• By the Lebesgue dominated convergence,

Z ∞

dτ (1 + τ2)2

• τ

λk

• 2

+ 1 !

k!∞

Z ∞

dτ (1 + τ2)2 = π 4 . This leads immediately to:

Z ∞

jqk (t)j2 dt π 2 jλkJ0 (λk)j2

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The restriction operator

The second step in Schwartz’ work is the following: consider the closed subspace of L2 (0, T) de…ned by: A (Λ; T) = Span feλkt, k 1g

L2(0,T ), 0 < T ∞.

Theorem

Assume that ∑k1 1/λk < ∞. The restriction operator RT : A (Λ, ∞) ! A (Λ, T) ϕ 7! ϕj(0,T ) is an isomorphism. In particular, there exists CT > 0 such that: kf kL2(0,∞) CT kf kL2(0,T ) , 8f 2 A (Λ, ∞) .

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If ek(t) = eλkt for t 0 and k 1, remark that RT ek = ekj(0,T ). Thus, δkj = hek, qjiL2(0,∞) =

• R1

T RT ek, qj

• L2(0,∞)

= D ek,

• R1

T

qj E

L2(0,T ) .

Therefore, the family fθkgk1 = n R1

T

qk

• k1 is biorthogonal to
• eλkt

k1 and we have the estimate C1 jλkJ 0(λk )j kθkk2 L2(0,T ) C2 jλkJ 0(λk )j, k 1.

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Anything amounts to estimate: 1 jλkJ0 (λk)j = 2 (1 + λk)2

∏n1

n6=k

• 1λk /λn

1+λk /λn

• Fattorini and Russell proved that:

Theorem

If λk = k2, then lim

k!∞

ln

1 jJ 0(λk )j

λk = 0. In other words, for all ε > 0, there exists Cε > 0 such that kθkk2

L2(0,T )

C jλkJ0 (λk)j Cεeελk , 8k 1.

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A …rst application

For the problem 8 < : y 0 yxx = f (x) u (t) , QT = (0, T) (0, π) yjx=0,π = 0, (0, T) yjt=0 = y0 (0, π) the null controllability problem had been reduced to ( Find v 2 L2 (0, T) : fk R T

0 ek 2tv (t) dt = ek 2T y0 k , k 1.

If fk 6= 0 for all k, a formal solution is v = ∑

k1

ek 2T fk y0

k θk

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The function f can be chosen suh that fk

• k!∞

C kp ) 8ε > 0, 1 jfkj = o

• eεk 2

But, for any ε > 0 : ek 2T jfkj

• y0

k

• kθkkL2 (0, T) Cεek 2T e2εk 2 = Cεek 2(T 2ε)

and the series ∑k ek 2(T 2ε) < ∞ if ε < T/2. Conclusion: v = ∑k1

ek2T fk

y0

k θk 2 L2 (0, T) and the scalar heat

equation is null controllable. The function f could be chosen such that Supp(f ) b (0, π) .

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A second application

Consider the 2 2 control system: 8 > > > < > > > : (∂t ∂xx) y1 + q(x)y2 = 0 in QT , (∂t ∂xx) y2 = v1ω y(0, ) = 0, y(π, ) = 0

• n (0, T),

y(, 0) = y0 in (0, π), q 2 L∞(0, π) is a given function, y0 is the initial datum and v 2 L2(QT ) is the control function. ω is an open subset of (0, π) . The system possesses a unique solution which satis…es y 2 L2(0, T; H1

0 (0, π; R2)) \ C 0([0, T]; L2(0, π; R2))

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Assume that ω = (a, b) (0, π) and q satis…es Supp q \ ω = ∅, i.e., Supp q [0, a] [ [b, π]. For any k 1, we associate with the function q 2 L∞(0, π) the sequences fIk(q)gk1 and fIi,k(q)gk1, i = 1, 2, given by 8 > < > : I1,k(q) :=

Z a

0 q(x)jϕk(x)j2 dx,

I2,k(q) :=

Z π

b q(x)jϕk(x)j2 dx,

Ik(q) := I1,k(q) + I2,k(q) =

Z π

0 q(x)jϕk(x)j2 dx.

where ϕk(x) = r 2 π sin(kx), 8x 2 (0, π), k 1.

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We will outline the proof of the following result:

Theorem

With the previous notations and assumption on q :

1

The system is approximately controllable at time T > 0 if and only if jIk(q)j + jI1,k(q)j 6= 0 8k 1.

2

De…ne T0(q) := lim sup minf log jI1,k(q)j, log jIk(q)jg k2 . Then:

1

If T > T0(q), the system is null controllable at time T.

2

If T < T0(q), the system is not null controllable at time T.

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Reduction to a moment problem

Set A0 = 1

• and consider the vectorial operator

L := d2 dx2 + q(x)A0 : D(L) L2(0, π; R2) ! L2(0, π; R2) with domain D(L) = H2(0, π; R2) \ H1

0 (0, π; R2) and also its adjoint L.

The spectra of L and L are given by σ(L) = σ(L) = fk2 : k 1g. Given k 1, let ψk be the unique solution of the non-homogeneous Sturm-Liouville problem: 8 > > < > > : ψxx k2ψ = [Ik(q) q(x)] ϕk in (0, π), ψ(0) = 0, ψ(π) = 0,

Z π

0 ψ(x)ϕk(x) dx = 0,

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The family B =

• Φ1,k =

ϕk

• ,

Φ2,k = ψk ϕk

• k1

satis…es (L k2Id)Φ1,k = 0 and (L k2Id)Φ2,k = Ik(q)Φ1,k The family B =

• Φ

1,k :=

ϕk ψk

• ,

Φ

2,k :=

ϕk

• k1

is biorthogonal to B and

• L k2Id
• Φ

1,k = Ik(q)Φ 2,k

and

• L k2Id
• Φ

2,k = 0).

In particular, if Ik 6= 0 then k2 is a simple eigenvalue and Φ1,k and Φ2,k (resp., Φ

2,k and Φ 1,k) are, respectively, an eigenfunction and a

generalized eigenfunction of the operator L (resp., L) associated to k2, while if Ik = 0 then Φ1,k and Φ2,k are both eigenfunctions of L (resp., L) associated to k2.

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B and B are Riesz bases in L2(0, π; R2) and for any y0 2 L2(0, π; R2) y0 = ∑

k1

• y0, Φ

1,k

• Φ1,k +
• y0, Φ

2,k

• Φ2,k
• = ∑

k1

• y0

1,kΦ1,k + y0 2,kΦ2,k

• .

If we look for the solution of the system in the form: y (t) = ∑

k1

fy1,k (t) Φ1,k + y2,k (t) Φ2,kg we readily get the system 8 < : y 0

1,k + k2y1,k + Ik (q) y2,k =

• Bv1ω, Φ

1,k

• y 0

2,k + k2y2,k =

• Bv1ω, Φ

2,k

• (y1,k, y2,k)jt=0 =
• y0

1,k, y0 2,k

• with B =

1

• .

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Solving this system, we get by setting vi,k (t) =

• Bv1ω, Φ

i,k

• for i = 1, 2 :

y1,k (T) = ek 2T y0

1,k TIky0 2,k

• +

Z T

ek 2(T t) [v1,k (t) (T t) Ikv2,k (t)] dt, y2,k (T) = ek 2T y0

2,k +

Z T

ek 2(T t)v2,k (t) dt. Then, y (T) = 0 if and only if ( R T

0 ek 2(T t) [v1,k (t) (T t) Ikv2,k (t)] dt = ek 2T

y0

1,k + TIky0 2,k

• ,

R T

0 ek 2(T t)v2,k (t) dt = ek 2T y0 2,k.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

If we look for v(x, t) = f1(x)v1(T t) + f2(x)v2(T t), (x, t) 2 QT , where v1, v2 2 L2(0, T) are new controls, only depending on t, and f1, f2 2 L2(0, π) are appropriate functions satisfying the condition Supp f1, Supp f2 ω = (a, b), then we get the system: 8 > > > > > > > > < > > > > > > > > : f1,k

Z T

v1(t)ek 2t dt + f2,k

Z T

v2(t)ek 2t dt = ek 2T y0, Φ

2,k

• e

f1,k

Z T

v1(t)ek 2t dt + e f2,k

Z T

v2(t)ek 2t dt Ik(q)f1,k

Z T

v1(t) tek 2t dt Ik(q)f2,k

Z T

v2(t) tek 2t dt = ek 2T y0, Φ

1,k

TIk(q)

• y0, Φ

2,k

• ,

where, for k 1, e f1,k, e f2,k are given by e fi,k :=

Z π

0 fi(x)ψk(x) dx, fi,k :=

Z π

0 fi(x)ϕk(x) dx, i = 1, 2.

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Remark that, if we …x k 1, it is a linear system of two equations and four unknown quantities:

Z T

vi(t)ek 2t dt,

Z T

vi(t) tek 2t dt, i = 1, 2.

Theorem

The moment problem has the form 8 > > < > > :

Z T

vi(t)ek 2t dt = ek 2T M(k)

1,i (y0),

Z T

vi(t) tek 2t dt = ek 2T M(k)

2,i (y0),

where the quantities M(k)

i,j (y0) 2 R, with k 1 and 1 i, j 2, satisfy

the following property: for any ε > 0 there exists a positive constant Cε (only depending on ε) such that

• M(k)

i,j (y0)

• Cεek 2(T0(q)+2ε)ky0kL2(0,π;R2),

8k 1, 1 i, j 2.

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The conclusion giving the positive null controllability result is based on the following two observations: The family n ek2t, tek 2to

k1 is minimal in L2 (0, T) .

There exists a biorthogonal family fq1,k, q2,kgk1 in L2 (0, T) such that kqi,kk2

L2(0,T ) Cεeεk 2

The formal solution of the moment problem is given by vi (t) = ∑

k1

n ek 2T M(k)

1,i (y0)q1,k + ek 2T M(k) 2,i (y0)

• , i = 1, 2.

With the previous estimates: vi 2 L2 (0, T) for T > T0 (q) and leads to the second point of the theorem.

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The negative null controllability result

Let T < T0 (q) and consider the adjoint problem: 8 > < > : θt θxx + q(x)A

0θ = 0

in QT , θ(0, ) = 0, θ(π, ) = 0

• n (0, T),

θ(, T) = θ0 in (0, π), The idea, is to …nd a sequence of initial data

• θ0

k

• k1 for which

R R

ω(0,T ) jBθk(x, t)j2 dx dt

kθk(, 0)k2

L2(0,π;R2)

!

k!θ 0

where θk is the solution of the adjoint problem associated with θ0

k.

In this way, the observability inequality kθ(, 0)k2

L2(0,π;R2) C

Z

ω(0,T ) jBθ(x, t)j2 dx dt

will not be satis…ed.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

For θ0

k = akΦ 1,k + bkΦ 2,k, with (ak, bk) 2 R2, the solution of the adjoint

problem is given by: θk (t, x) = akek 2(T t) Φ

1,k (T t)Ik (q) Φ 2,k

+ bkek 2(T t)Φ

2,k

Computing, we get kθk(, 0)k2

L2(0,π;R2)

= e2k 2T n jakj2 + h jakj2j kψkk2

L2(0,π) + (bk TakIk(q))2io

• e2k 2T jakj2

and

Z T Z

ω jBθ(x, t)j2 =

Z T Z

ω e2k 2t jakψk(x) + (bk takIk(q))ϕk(x)j2 dx

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

It can be proved that for any x 2 ω ψk(x) = τk ϕk (x) Ik(q)gk (x) r π 2 1 k I1,k(q) cos(kx) Thus:

Z T Z

ω jBθ(x, t)j2

=

Z T Z

ω e2k 2t j(akτk + bk) ϕk (x)

akIk(q) (gk (x) + tϕk (x)) r π 2 ak k I1,k(q) cos(kx)

• 2

dxdt Choosing ak = 1 and bk = τk gives the inequality:

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Z T Z

ω jBθ(x, t)j2 C

• jI1,k(q)j2 + jIk(q)j2

and to summarize: ek 2T

• C
• jI1,k(q)j2 + jIk(q)j2
• Ce2k 2[ 1

k2 min( logjI1,k (q)j, logjIk (q)j)]

The contradiction follows with a suitable choice of a subsequence (kn) ...

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Application 3

Consider the system 8 > > > > > > > > > < > > > > > > > > > : y 0

1 = ∂2y1

∂x2 QT = (0, T) (0, π) y 0

2 = d ∂2y2

∂x2 QT yi (t, 0) = biv (t) , yi (t, π) = 0 i = 1, 2 yi (0, ) = y0

i

i = 1, 2 where 0 < d < 1, bi 2 R (i = 1, 2) and v 2 L2 (0, T) . For y0 =

• y0

1 , y0 2

2 H1 0, π; R2 , there exists a unique solution y 2 C [0, T] ; H1 0, π; R2 \ L2 (QT )

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Indeed, the solution is given by: yi = ∑

k1

yi,k ϕk where ϕk (x) = q

2 π sin (kx) and

y1,k (t) = ek 2ty0

1,k +

r 2 πkb1

Z t

0 ek 2(ts)v (s) ds

y2,k (t) = edk2ty0

2,k +

r 2 πkb2

Z t

0 edk 2(ts)v (s) ds

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Thus the null controllability issue reduces to: …nd v 2 L2 (0, T) such that 8 > > < > > : q

2 πkb1

R T

0 ek 2tv (T t) ds = ek 2T y0 1,k

q

2 πkb2

R T

0 edk 2tv (T t) ds = edk 2T y0 2,k

, k 1 A …rst necessary condition for solvability: bi 6= 0 for i = 1, 2. A second necessary condition: dk2 6= `2 for all k, ` 1

• ,

p d / 2 Q

• .

With these two necessary conditions, we now have to solve: 8 > > > < > > > : R T

0 ek 2tv (T t) ds = ek 2T y 0

1,k

p 2

π kb1

R T

0 edk2tv (T t) ds = edk2T y 0

2,k

p 2

π kb2

, k 1

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So, this time, we are dealing with the family n ek 2t, edk 2to . Set λ2k = dk2, λ2k+1 = k2, k 1. Then clearly ∑n1 1/λn < ∞. It follows that, following Schwartz,

• eλnt

is minimal in L2 (0, T) and a biothogonal family fθng can be found such that C1 jλnJ0 (λn)j kθnk2

L2(0,T )

C2 jλnJ0 (λn)j, n 1. with 1 jλnJ0 (λn)j = 2 (1 + λn)2

∏`1

n6=`

• 1λn/λ`

1+λn/λ`

• (LMB)

Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Again, a formal solution is: v (T t) = ∑

n1

eλnT mnθn and as in the Fattorini-Russell example (heat equation), we need an estimate of 1/ jλnJ0 (λn)j or, in other words, to compute lim sup ln 1 jλnJ0 (λn)j =? Remember that if λn = n2, we had lim sup ln 1 jn2J0 (n2)j = 0. Indeed, it has been proved that for an increasing sequence jλn λmj α jn λmj = ) lim sup ln 1 jλnJ0 (λn)j = 0 But for the sequence

• λ2k = dk2, λ2k+1 = k2

this condition cannot be satis…ed for any p d 2 Q.

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Indeed:

Theorem

For any c 2 [0, ∞] , there exists p d / 2 Q such that for

• λ2k = dk2, λ2k+1 = k2

lim sup ln 1 jλnJ0 (λn)j = c Actually, the number lim sup ln 1 jλnJ0 (λn)j associated with the sequence fλng has... a name: it is the index of condensation of the sequence fλng .

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Before saying more about this index, let us conclude with a controllability result related to the considered system:

Theorem

Assume that bi 6= 0 (i = 1, 2) and p d / 2 Q. Then, if c is the condensation index of

• λ2k = dk2, λ2k+1 = k2

:

1

If T > c, then the system is null controllable.

2

If T < c, then the system is not null controllable.

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The index of condensation

De…nition

Let Λ = (λk)k1 be an increasing sequence of real numbers. A condensation grouping of Λ is any sequence of sets G = (Gk)k1 satisfyng the following properties:

1

Λ \ Gk 6= ? for all k 1 and Λ = [k1 (Λ \ Gk) .

2

If Λ \ Gk = fλnk , λnk +1, ..., λnk +pk g , k 1 then lim

k!∞

pk λnk = lim

k!∞

λnk +pk λnk = 1 The second item of this de…nition characterizes what is meant by condensation.

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Let G = (Gk)k1 a condensation grouping of Λ. For all k 1, the index of condensation of Gk = fλnk , ..., λnk +pk g is the number δ (Gk) = sup

0lpk

1 jλnk +lj ln B B B B B @ pk!

∏

0jpk j6=l

j(λnk +l λnk +j)j 1 C C C C C A The index of condensation of G = (Gk)k1 is the number : δ (G) = limk!∞ δ (Gk) .

De…nition

The index of condensation of Λ = (λk)k1 is the number δ (Λ) de…ned to be the supremum of the set fδ (G)g where G may be any condensation grouping.

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Example

Let Λ = (λn) and set Gn = fλng . G = (Gn) is a condensation grouping with pn = 0 and δ (Gn) = 0, n 1. Conclusion: δ (Λ) 0 for any Λ

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Example

Let α 1 and β > 0 and set Λ = n λ2n = nα, λ2n+1 = nα + enβ, n 1

• .

De…ne: Gn =

• λ2n 1

2; λ2n + 1 2

• , n 1.

G = (Gn) is a condensation grouping with pn = 1 and δ (Gn) = max 1 nα ln 1! enβ , 1 nα + enβ ln 1! enβ

• =

1 nα ln 1 enβ = nβα. Thus: δ (G) = lim sup

n!∞

δ (Gn) = 8 < : 0, 0 < β < α, 1, β = α, ∞, β > α. .

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Optimal condensation grouping

De…nition

Let Λ = (λn)n1 be a real increasing sequence.A sequence Λ is measurable if there exists D 2 [0, ∞[ such that limn!!∞ n λn = D.The number D when it exists is the density of Λ.

Example

Let α 1 and β > 0 and set Λ = n λ2n = nα, λ2n+1 = nα + enβ, n 1

• . Then

D = 8 < : 0, if α > 1 1 if α = 1 ∞ if α < 1

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Optimal condensation grouping

De…nition

For any subset E R, NΛ (E) = card (Λ \ E) .

Lemma

Let Λ = (λk)k1 be an increasing sequence of real numbers, with …nite density D, and let 0 < q < 1/ (2D + η) where η > 0 is some arbitrary …xed number. For all λ 2 Λ and any integer r 1, let I (λ, rq) = ]λ rq, λ + rq[ . Then there exists a greatest integer p (λ) such that NΛ (I (λ, p (λ) q)) p (λ) .

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With the previous lemma it can be proved that:

Theorem

Let Λ = (λn)n1 be an increasing sequence of positive real numbers, with density D and 0 < q < 1/ (2D + η) where η > 0 is a …xed arbitrary number. Then there exists a condensation grouping G = (Gk)k1 such that :

1

δ (G) = δ (Λ) .

2

Fix k 1. For all λ 2 Λ \ Gk and (µn)1nm Λn (Λ \ Gk) , we have

m

∏

n=1

jλ µnj qmm!

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Proof.

[Sketch of the proof] By successive applications of the previous Lemma, de…ne a sequence of intervals in the following way. Let: G1 = I (λ1, p1q) , n1 = 1, p1 = p (λn1) . Denote by λn2 the smallest element of Λ not belonging to I1 (q) and set G2 = I (λn2, p2q) , p2 = p (λn2) Let k 1 and suppose constructed the intervals (Gj)1jk . Denote by λnk+1 the smallest element of Λ not belonging to [1jkGj. We then de…ne Gk+1 = I (λnk+1, pk+1q) , pk+1 = p (λnk+1) .

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Example

Let α 1 and β > 0 and set Λ = n λ2n = nα, λ2n+1 = nα + enβ, n 1

• .

De…ne: Gn =

• λ2n 1

2; λ2n + 1 2

• , n 1.

G = (Gn) is a condensation grouping with pn = 1 and δ (Gn) = max 1 nα ln 1! enβ , 1 nα + enβ ln 1! enβ

• =

1 nα ln 1 enβ = nβα. Thus: δ (G) = lim sup

n!∞

δ (Gn) = 8 < : 0, 0 < β < α, 1, β = α, ∞, β > α. . With this previous construction, it can be checked that G is optimal and thus: δ (G) = δ (Λ) .

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Interpolating function

To the sequence Λ = (λn)n1 with density D 0, is associated the interpolating function C (λ) = ∏

n1

1 λ2 λ2

n

! , λ 2 C.

Theorem

Let Λ = (λn)n1 be an increasing real sequence of positive numbers, measurable with …nite density D 0. Then its index of condensation δ (Λ) is given by: δ (Λ) = lim sup

k!∞

ln

1 jC 0(λk )j

λk .

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

The property that links with the boundary control problem:

Theorem

If Λ = (λn)n1 is an increasing sequence of positive numbers such that ∑n1 1/λn < ∞, then δ (Λ) = lim

k!∞

ln

1 jJ 0(λk )j

λk where

• J0 (λk)
• =

∏n1

n6=k

• 1λk /λn

1+λk /λn

• 2λk (1 + λk)2 .

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Example

De…ne Λ = (λn)n1 by λk 2+l = k2 + lek β, k 1, 0 l 2k, (β > 0) and prove that: δ (Λ) = 8 < : ∞, β > 1 2, β = 1 0, 0 < β < 1 .

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

An interpolation formula due to Jensen assures that if f is a holomorphic function on a convex domain Ω C and A = fajg0jq Ω is a set of distinct points, then there exists θ 2 [1, 1] and ξ 2 Conv (A), the convex hull of A, such that

q

∑

j=0

f (aj) P0

A(aj) = θ

q! dqf dzq (ξ), where for any …nite set F the function PF is de…ned by: PF (λ) = ∏

µ2F

(λ µ) As a consequence of this formula, we have:

Theorem

Let Λ = fλkgk1 be an increasing sequence of real numbers and G = fGkgk1 any condensation grouping associated with Λ. Then: lim

k!∞

Z ∞

• ∑

λn2Gk

pk! P0

Gk (λn)eλnt

• 2

dt = 0.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Back to the boundary control problem

The associated observability inequality is of the form:

Z T

• ∑

k1

ckeλkt

• 2

dt CT ∑

k1

eλkT c2

k , 8c 2 `2.

Let G = (Gk) an optimal condensation grouping. Fix k 1 and set ck

n =

8 < : pk! P0

Gk (λn),

if λn 2 Gk.

• therwise,

(3) Then, from a previous theorem:

Z T

• ∑

n1

ck

n eλnt

• 2

dt =

Z T

• ∑

λn2Gk

pk! P0

Gk (λn)eλnt

• 2

dt !

k!∞ 0

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

On the other hand: if T < δ (Λ)

∑

n1

• ck

n

• 2

eλnT =

∑

λn2Gn

• pk!

P0

Gk (λn)eλnT

• 2
• eλnk (δ(Λ)εT )
• 2

! ∞ This implies that the observability inequality does not hold... and concludes the proof of the theorem.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

References

• V. Bernstein, Leçons sur les Progrès Récents de la Théorie des Séries

de Dirichlet, Gauthier-Villars, Paris, 1933.

• L. Schwartz. Étude des sommes d’exponentielles réelles. Hermann

(1943). H.O. Fattorini, D. L. Russell, Arch. Rational Mech. Anal. 43 (1971), 272–292 & Quart. Appl. Math. 32 (1974/75), 45–69.

• A. V. Fursikov, O. Yu. Imanuvilov, Controllability of evolution

equations, Lecture Notes Series, 34. Seoul National University, Research Institute of Mathematics, Global Analysis Research Center, Seoul, 1996.

• G. Lebeau, L. Robbiano, Contrôle exact de l’équation de la

chaleur, Comm. Partial Di¤erential Equations 20 (1995), no. 1-2, 335–356.

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• S. Dolecki, Observability for the one-dimensional heat equation,

Studia Math. 48 (1973), 291–305.

• F. Ammar Khodja, A. Benabdallah, C. Dupaix,

Null-controllability of some reaction-di¤usion systems with one control force, J. Math. Anal. Appl. 320 (2006), no. 2, 928–943.

• M. González-Burgos, L. de Teresa, Controllability results for

cascade systems of m coupled parabolic PDEs by one controll force,

• Port. Math. 67 (2010), no. 1, 91–113.
• F. Alabau-Boussouira, M. Léautaud, Indirect controllability of

locally coupled wave-type systems and applications, J. Math. Pures

• Appl. (9) 99 (2013), no. 5, 544–576.
• F. Boyer and G. Olive, Approximate controllability conditions for

some linear 1D parabolic systems with space-dependent coe¢cients,

• Math. Control Relat. Fields 4 (2014), no 3, 263–287.

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• L. Rosier, , L. de Teresa, Exact controllability of a cascade

system of conservative equations, C. R. Math. Acad. Sci. Paris 349 (2011), no. 5-6, 291–296.

• F. Ammar Khodja, A. Benabdallah, M. González-Burgos,
• L. de Teresa, Recent results on the controllability of linear coupled

parabolic problems: a survey, Math. Control Relat. Fields 1 (2011),

• no. 3, 267–306.
• F. Ammar Khodja, A. Benabdallah, M. González-Burgos,
• L. de Teresa, Minimal time of controllability of two parabolic

equations with disjoint control and coupling domains, C. R. Math.

• Acad. Sci. Paris, Ser. I 352 (2014), no. 5, 391–396.
• F. Ammar Khodja, A. Benabdallah, M. González-Burgos,
• L. de Teresa, Minimal time for the null controllability of parabolic

systems: the e¤ect of the condensation index of complex sequences,

• J. Funct. Anal. 267 (2014), no. 7, 2077–2151.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Open problems

The null controllability issue for the problems 8 > > > < > > > : (∂t ∆) y1 + q(x)y2 = 0 in QT = (0, T) Ω, (∂t ∆) y2 = v1ω y = 0,

• n ΣT = (0, T) ∂Ω,

y(, 0) = y0 in Ω, where supp(q) \ ω = ? and 8 > > < > > : y 0

1 = ∆y1

QT = (0, T) Ω y 0

2 = d∆y2

QT y = Bv1Γ0 ΣT y (0, ) = y0 Ω. are open. In higher space dimension, the moment method does not work. It seems that Lebeau-Robbiano and Fursikov-Imanuvilov methods cannot be applied.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

The following problem has been recently studied by Michel Duprez: 8 > > > < > > > : (∂t ∆) y1 + (p.r + q) y2 = 0 in QT = (0, T) Ω, (∂t ∆) y2 = v1ω y = 0,

• n ΣT = (0, T) ∂Ω,

y(, 0) = y0 in Ω, with quite complete results in the one-dimensional case.

(LMB) Control parabolic Mathematics in Savoie: 15-18 june 2015 / 72

Thank you for your attention.

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