Sur la contrôlabilité de l’équation de la chaleur avec potentiel Michel Duprez

Sur la contrôlabilité de l’équation de la chaleur avec potentiel

Michel Duprez

Institut de Mathématiques de Marseille

Séminaire Analyse Appliquée (AA)

Marseille, I2M

Problem

How to act on the temperature from a given region ?

Controllability of the heat equation

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

ty= ∆y+1^ωu inQT := Ω×(0, T), y= 0 on ΣT :=∂Ω×(0, T), y(·,0) =y⁰ in Ω.

Such a system is said :

approximatively controllableat timeT if

∀y⁰, y^T∈L²(Ω), ε >0∃u∈L²(QT) s.t.ky(T)−y^TkL²(Ω)6ε, exactly controllableat timeT if

∀y⁰, y^T ∈L²(Ω)∃u∈L²(QT) s.t.y(T) =y^T, null controllableat timeT if

∀y⁰∈L²(Ω)∃u∈L²(QT) s.t. y(T) = 0.

Remark :

•The system is not exactly contr.

•Null contr.⇒approx. contr.

Controllability of the heat equation

Consider the system ( _∂

ty= ∆y+1ωu inQT,

y= 0 on ΣT,

y(·,0) =y0 in Ω.

Approximate controllability å 58’ Mizohata Null controllabilityN = 1

å 74’ Fattorini and Russell Null controllabilityN >1

å 95’ Lebeau and Robbiano å 95’ Fursikov and Imanuvilov

Controllability of a parabolic equation

Consider the parabolic equation ( _∂

ty= div(d∇y) +g· ∇y+ay+1ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

(G)

wherea∈L^∞(QT),g∈L^∞(0, T;W∞¹(Ω))^N and for a constantd0>0











d^ij∈W∞¹(QT), d^ij=d^ji_l inQT,

N

P

i,j=1

d^ijξiξj>d0|ξ|² inQT, ∀ξ∈R^N.

å System (G) is null controllable at any timeT.

Problem

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

ty= ∆y+∂x1(1ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

Ifω⊂⊂Ω,∂x1(1ωu) can be seen as a control with average equal to zero

Preliminary study for the Fokker-Planck control problem ( _∂

tρ=ε²∆ρ−div(ρ(v+1ωu)), ρ(·,0) =ρ⁰,

Supp(u)⊂ω

å Change of variables : div→∂x₁

Null controllability when∂Ω∩∂ω6=∅

å Benabdallah, Cristofol, Gaitan, De Teresa, 2014 Null controllability in general

å Duprez, Lissy, 2016

Problem

General system







∂ty= div(d∇y) +g· ∇y+ay+∂x1(u) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂ω.

(G)

Toy model







∂ty= ∆y+ay+∂x₁(u) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂ω.

(S)

We will see thataplay a critical role for the controllability of these system ! ! !

Duality

Consider the system ( _∂

ty= ∆y+ay+1^ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(S)

Proposition

(i) System (S) 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= ∆ϕ+aϕ 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.

(ii) System (S) isapproximately controllableat timeT iff for all ϕ⁰∈L²(Ω) the solutionϕsatisfies

ϕ= 0 inω×(0, T) ⇒ϕ= 0 in Ω×(0, T).

Duality

Consider the system ( _∂

ty= ∆y+ay+∂x₁(1^ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(S)

Proposition

(i) System (S) 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= ∆ϕ+aϕ 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

∂x₁ϕ².

(ii) System (S) isapproximately controllableat timeT iff for all ϕ⁰∈L²(Ω) the solutionϕsatisfies

∂x1ϕ= 0 inω×(0, T) ⇒ϕ= 0 in Ω×(0, T).

Proof

Let

St: L²(Ω) → L²(Ω)

y⁰ 7→ y(t;y⁰,0) and Lt: L²(QT) → L²(Ω) u 7→ y(t; 0, u).

(i) System (S) is null controllable at timeT iff

_∀y0∈L²(Ω), ∃u∈L²(QT) s.t.

STy⁰=−LTu.

⇔ ImST⊆ImLT.

⇔

_{∃C >0 : ∀ϕ}0 ∈L²(Ω), kS_T^∗ϕ⁰k²_L2(Ω)6CkL^∗_Tϕ⁰k²_L2(Q_T). (ii) System (S) is approximatively controllable at timeT iff

_∀y0, y^T ∈L²(Ω), ε >0, ∃u∈L²(QT) s.t.

kLTu+STy⁰−y^TkL²(Ω)6ε.

⇔ LT(L²(QT)) =L²(Ω).

⇔ ker (L^∗_T) ={0}

⇔ ∀ϕ⁰∈L²(Ω) : L^∗_Tϕ⁰ ⇒ϕ≡0.

Proof

Lety⁰, ϕ⁰∈L²(Ω) andu∈L²(QT) : hSTy⁰, ϕ⁰iL²(Ω) = hy(T;y⁰,0), ϕ(T)iL²(Ω)

= Z T

0

h∂ty(s;y⁰,0), ϕ(s)i_H−1,H₀¹ds

+ Z T

0

h∂tϕ(s), y(s;y⁰,0)i_H−1,H¹

0ds+hy⁰, ϕ(0)i_L2(Ω)

= hy⁰, ϕ(0)iL²(Ω)

and

hLTu, ϕ⁰i_L2(Ω) = hy(T; 0, u), ϕ(T)i_L2(Ω)

= Z T

0

n

h∂ty(s; 0, u), ϕ(s)i_H−1,H¹₀+h∂tϕ(s), y(s; 0, u)i_H−1,H₀¹

o ds

= h∂x₁u, ϕi_L2(0,T;H⁻¹(Ω)),L²(0,T;H₀¹(Ω))=−hu, ∂x₁ϕi_L2(Q_T). Thus

S_T^∗ : L²(Ω) → L²(Ω)

ϕ⁰ 7→ ϕ(·,0) and L^∗_T : L²(Ω) → L²(QT).

ϕ⁰ 7→ −∂x₁ϕ

Boundary condition

Consider







∂ty= ∆y+ay+∂x₁(u) inQT,

y= 0 on ΣT,

y(·,0) =y0 in Ω,

Supp(u)⊂ω

(S)

Theorem

If∂Ω∩∂ω6=∅, then System (S) isnull controllableat timeT.

Suppose that∂Ω∩∂ω∩(Ox1)^⊥6=∅. If (0, R)×e^ω⊂ω, let (y, v) solution of ( _∂

ty= ∆y+ay+1(0,R)טωv inQT,

y= 0 on ΣT,

y(·,0) =y⁰, y(·, T) = 0 in Ω.

We take

u:=

Z x1 R

v.

Or equivalently Z

Ω

ϕ(·,0)²6C Z

ω×(0,T)

ϕ²6C Z

ω×(0,T)

(∂x₁ϕ)².

Boundary condition : General parabolic equation

Consider the parabolic system ( _∂

ty= div(d∇y) +g· ∇y+ay+∂x₁(1^ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(G)

Theorem (Benabdallah, Cristofol, Gaitan, De Teresa, 2014)

Assume that∂Ω∩∂ω6=∅. Then System (G) isnull controllableon the time interval (0, T).

General parabolic operator

Consider the system







∂ty= ∆y+p(x)∂xu+q(x)u in (0, π)×(0, T), y(0,·) =y(π,·) = 0 on (0, T),

y(·,0) =y⁰ in (0, π),

Supp(u)⊂ω.

(S’)

Theorem (Duprez, 2016) Suppose that

Supp(p, q)∩ω6=∅.

Then System (S’) isnull controllableon the time interval (0, T).

Constant coefficient(s)

Consider the system







∂ty= ∆y+ay+∂x1(u) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂ω.

(S)

Theorem ( D., Lissy, 2016)

Ifais constant, then System (S) isnull controllableat timeT. Sketch of proof :

Up to the initial condition and the boundary condition,∂x1ϕsatisfies the same system asϕ:

( _−∂

tϕ= ∆ϕ+aϕ inQT,

ϕ= 0 on ΣT,

ϕ(·, T) =ϕ⁰ in Ω.

Using the Carleman estimates, for some weightsρ1, ρ2, ρ3, Z

Ω

ϕ(·,0)²6C Z

Q_T

ρ1ϕ²6C Z

Q_T

ρ2(∂x₁ϕ)²6C Z

q_T

ρ3(∂x₁ϕ)².

Constant coefficients : General parabolic equation

Consider the system ( _∂

ty= div(d∇y) +g· ∇y+ay+∂x₁(1^ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(G)

Theorem ( D., Lissy, 2016)

Suppose thatd, gandaare constant in space and in time.

Then System (G) isnull controllableon the time interval (0, T).

Algebraic resolvability

Let us recall the system











∂ty= ∆y+ay+∂x1(1ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂⊂ω.

(S)

Theorem (D., Lissy, 2016, submitted)

Leta∈ C¹(Q_T). If there exists (x0, t0)∈ω×(0, T) such that (∂x₁a)(x0, t0)6= 0,

then System (S) isnull controllable(hence approximately controllable) on the time interval (0, T).

Proof : first approach

Fictitious control method :

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











∂tb^{y= ∆}b^y+ayb^{+ ˆu inQT,}

yb= 0 on ΣT,

y(·) =b y⁰, y(·, T) = 0 in Ω, Supp(^u)b ^⊂⊂b^{ω×(0, T).}

If we find a solution to

∂tz= ∆z+az+∂x₁(v) +^ub ^inQT, Supp(z, v)⊂⊂ω×(0, T), (S1) then(b^y−z,−v)will be a solution of the initial problem.

Proof : first approach

Algebraic problem : Rewrite (S1) as

L(z, v) =b^u where

L(z, v) =∂tz−∆z−az−∂x₁(v).

Let us search a differential operatorMsuch that L ◦ M=Id (S2) Thus (z, v) :=M(bu) will be a solution to (S1).

Remark :b^uhave to be regular enough.

We will solve the formal adjoint to (S2) :

M^∗◦ L^∗ψ=ψ (S3) where

L^∗(ψ) = L^∗₁ψ

L^∗2ψ

=

−∂tψ−∆ψ−aψ

−∂x₁ψ

.

Proof : first approach

Resolution of the algebraic problem :

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

L^∗₁ψ L^∗2ψ

=

−∂tψ−∆ψ−aψ

−∂x₁ψ

.

The commutator ofL^∗₁ andL^∗₂ is :

[L^∗1:L^∗2]ψ=∂x₁(a)ψ.

So we have

M^∗1◦ L^∗1+M^∗2◦ L^∗2=Id, where

M^∗₁=∂x₁(a)⁻¹∂x₁ andM^∗₂=∂x₁(a)⁻¹[∂t+ ∆ +a].

Proof : second approach

Resolvability + Inequality of observability :

Letϕthe solution to the dual system ( _−∂

tϕ= ∆ϕ+aϕ inQT,

ϕ= 0 on ΣT,

ϕ(·, T) =ϕ⁰ in Ω.

Using the algebraic resolvability :

ϕ=M^∗1◦(−∂t−∆−a)ϕ+M2^∗◦(−∂x₁)ϕ=M^∗2◦(−∂x₁)ϕ Hence

kϕ(·,0)k²_L2(Ω)6C Z

q_T

ϕ²=C Z

q_T

(M^∗₂◦∂x1ϕ)² Thus we have a solution to the control problem :

( _∂

ty= ∆y+ay+∂x₁(M2u) inQT,

y= 0 on ΣT,

y(·,0) =y⁰, y(·, T) = 0 in Ω.

Proof : first approach

Regular control with the fictitious control method :

Let^ωb^{⊂⊂ωand (y,}bb^{u),y,θ,ηsuch that}







∂t^yb^{= ∆}b^y+ayb^{+1ωˆuˆ inQT,} by= 0 on ΣT, by(·,0) =y⁰, b^{y(·, T) = 0 in Ω}







∂ty= ∆y+ay inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω

( _{Supp(θ)⊂ω,} θ= 1 inb^ω, 0≤θ≤1

and

( _η= 1 in [0, T /3], η= 0 in [2T /3, T], 0≤η≤1.

Theny:= (1−θ)b^y+ηθyis solution to







∂ty= ∆y+ay+1ωu inQT,

y= 0 on ΣT,

y(·,0) =y⁰, y(·, T) = 0 in Ω.

withu:= ∆θb^{y+ 2∇θ· ∇}b^{y+∂t(η)θy−η∆θy−2η∇θ· ∇y.}

We remark thatyis more regular thanb^y.

General parabolic equation

Consider the system ( _∂

ty= div(d∇y) +g· ∇y+ay+∂x₁(1ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

(G)

Theorem (D., Lissy, 2016, submitted)

Letd^kl_i , g^k_ij∈ C^N2+3(ωT) andaij∈ C^N2+2(ωT) for everyi, j∈ {1,2}and k, l∈ {1, ..., N}. Assume∃ωT ⊂qT s.t.

(

e^ais not an element of theC_t,x⁰ _2,...,xN(ωT)-module 1,e^g

2, ...,e^g

N, d²², ..., d^{N N}

C⁰_t,x

2,...,xN(ω_T), where







egⁱ:=gⁱ−

N

P

j=1

∂x_jd^ij,

ea:=−a+ div(g).

Then System (G) isnull controllableon the time interval (0, T).

Sketch of proof

Suppose thatN= 2 ( _∂

ty=∂x₁x₁(d1y) +∂x₂x₂(d2y) +∂x₁(^ge^{1y) +∂x2(g}e^{2y) +ay}e ^{+∂x1(1ωu) inQT,}

y= 0 on ΣT,

y(·,0) =y⁰ in Ω.

Algebraic resolution : Expression toL^∗: L^∗₁ψ

L^∗2ψ

=

∂tψ−d1∂x1x1(ψ)−d2∂x2x2(ψ) +^ge^{1∂x1(ψ) +g}e^{2∂x2(ψ) +}e^aψ

−∂x₁ψ

.

Step 1 : The derivatives inx1

L^∗3ψ:=L^∗1ψ−(d1∂x₁+b^g1)L

∗

2ψ=∂tψ−d2∂x₂x₂(ψ) +^ge^2∂x2(ψ) +e^aψ Step 2 : The derivative in time

L^∗4ψ:= [L^∗3 :L2^∗]ψ=−∂x₁(d2)∂x₂x₂(ψ) +∂x₁(^ge^{2)∂x2(ψ) +∂x1(}e^a)ψ

Sketch of proof

Step 3 : The other terms

•If|∂x₁(d2)| ≥δ1, then

L^∗5ψ:= [∂x₁(d2)⁻¹L^∗4 :L^∗2]ψ=∂x₁

∂x₁(^ge²⁾

∂x₁(d2)

∂x₂(ψ) +∂x₁

∂x₁(e^a)

∂x₁(d2)

ψ

•If|∂x₁

∂x1(^ge²⁾

∂_x1(d₂)

| ≥δ2, then

L^∗₆ψ:= [∂x1

∂x1(^ge²⁾

∂x1(d2) ⁻¹

L^∗₅ :L^∗₂]ψ=∂x1









∂x1

∂_x1(

e^a)

∂x1(d2)

∂x1

∂_x1( ge₂)

∂x1(d2)







 ψ

Sketch of proof

Step 4 : Conclusion

∂x₁









∂x1

∂x1(

e^a)

∂x1(d2 )

∂_x1

∂x1(^ge^{2 )}

∂x1(d2 )









= 0 ⇔

∂x1

∂x1(

e^a)

∂x1(d2 )

∂_x1

∂x1(^ge^{2 )}

∂x1(d2 )

=λ1(t, x2)

⇔ ∂x1

∂_x1(

e^a−λ1ge²⁾

∂x1(d2)

= 0

⇔ ^∂x1(e^a−λ1ge²⁾

∂_x1(d₂) =λ2(t, x2)

⇔ ∂x1(e^a−λ1ge^{2−λ2d2) = 0}

⇔ e^a=λ1ge^2+λ2d2+λ3

Summary

Let us recall the system











∂ty= ∆y+ay+∂x₁(1^ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂⊂ω.

(S)

aconstant

å Null controllable

There exists (x0, t0)∈ω×(0, T) s.t.∂x₁a6= 0 å Null controllable

Is System (S) always null controllable ?

Example of non controllability

Let us recall the system











∂ty= ∆y+ay+∂x₁(u) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂ω.

(S)

Theorem (D., Lissy, 2016, submitted)

Ifω⊂⊂Ω, then there existsa∈ C^∞(Ω) such that System (S) isnot approximately controllable(hence not null controllable) at timeT.

Fattorini criterion

Let us recall the system











∂ty= ∆y+ay+∂x₁(1ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂⊂ω.

(S)

Theorem (Fattorini criterion)

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

−∆ϕ−aϕ=sϕ in Ω

∂x₁ϕ= 0 inω

⇒ϕ= 0.

å Olive, 2014

Proof

Suppose thatN= 1, Ω := (0, π) andω1:= (2π/5,3π/5).

Consider

a:=−∆ϕ−ϕ

ϕ .

Thus







−∆ϕ−aϕ=ϕ in Ω,

∂x₁ϕ= 0 inω, ϕ6= 0.

Using the Fattorini Criterion, the system isnot approximately controllableon the time interval (0, T).

Letϕ∈ C²([0, π]) satisfying







ϕ(x) = sin(x) ∀x∈[0, π/5]∪[4π/5, π], ϕ(x) = 1 ∀x∈[2π/5,3π/5], ϕ > δ >0 in [π/5,4π/5].

0 1 2 3

0 0.5 1

sin(x) 1

Geometrical condition

Let us recall the system











∂ty= ∆y+ay+∂x₁(u) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂ω.

(S)

Theorem (D., Lissy, 2016, submitted) There existsa∈ C^∞(Ω) such that :

(i) There existsω⊂⊂Ω such that, for allT >0, System (S) isnull controllable(then approximatively controllable) at timeT. (ii) There existsω⊂⊂Ω such that, for allT >0, System (S) isnot

approximatively controllable(then not null controllable) at time T.

Conclusions and perspectives

Let us recall the system











∂ty= ∆y+ay+∂x₁(1^ωu) inQT,

y= 0 on ΣT,

y(·,0) =y⁰ in Ω,

Supp(u)⊂⊂ω.

(S)

Conclusions :

System (S) isnull controllableunder one of the following condition : å ∂Ω∩∂ω6=∅

å constant coefficients

å ∂_x1a(x₀, t₀)6= 0 for a (t₀, x₀)∈q_T

There existsa∈ C^∞(QT) s.t. System (S) isnot approximately controllable

Perspectives :

Find a general condition...

Study the Fokker-Planck equation

Thank you for your attention !