Microlocal analysis of the HUM operator for a system of wave

Transcription

Microlocal analysis of the HUM operator for a system of wave
Microlocal analysis of the HUM operator
for a system of wave equations
Jérôme Le Rousseau
Université d’Orléans – Fédération Denis-Poisson
A joint work with
Belhassen Dehman (Faculté des Sciences de Tunis)
and
Matthieu Léautaud (Université Paris 7)
Laboratoire Jacques-Louis Lions
Université Pierre-et-Marie-Curie, October 2012
1/ 36
J. Le Rousseau
coupled wave equations
Setting
Let Ω be a Riemannian compact manifold Ω
(without boundary)
∆ is the Laplace-Beltrami operator on Ω
We consider the following system
 2
in (0, T ) × Ω,
 ∂t u1 − ∆u1 + bu2 = 0
∂t2 u2 − ∆u2 = χg
in (0, T ) × Ω,

(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (u01 , u11 , u02 , u12 ) in Ω.
χ ∈ C ∞ (Ω) and ω = {χ 6= 0}.
We wish to bring the solutions at rest at times t ≥ T :
(u1 , ∂t u1 , u2 , ∂t u2 )|t=T = (0, 0, 0, 0).
The system has a cascade structure.
The control g only acts on one of the equations.
The second equation can only be controlled through the coupling
term bu2
2/ 36
J. Le Rousseau
coupled wave equations
Setting
Let Ω be a Riemannian compact manifold Ω
(without boundary)
∆ is the Laplace-Beltrami operator on Ω
Equivalently we consider the following system
 2
in (0, T ) × Ω,
 ∂t u1 − ∆u1 + bu2 = 0
∂t2 u2 − ∆u2 = χg
in (0, T ) × Ω,

(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (0, 0, 0, 0) in Ω.
χ ∈ C ∞ (Ω) and ω = {χ 6= 0}.
We wish to bring the solutions at time t ≥ T to a prescribed state:
(u1 , ∂t u1 , u2 , ∂t u2 )|t=T = (u01 , u11 , u02 , u12 ).
3/ 36
J. Le Rousseau
coupled wave equations
Setting
We consider the following system
 2
in (0, T ) × Ω,
 ∂t u1 − ∆u1 + bu2 = 0
∂t2 u2 − ∆u2 = χg
in (0, T ) × Ω,

(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (0, 0, 0, 0) in Ω.
We wish to bring the solutions at time t ≥ T to :
(u1 , ∂t u1 , u2 , ∂t u2 )|t=T = (u01 , u11 , u02 , u12 ).
Regularity Issue
g ∈ L2 ((0, T ) × Ω),
=⇒ (u2 , ∂t u2 ) ∈ C 0 [0, T ], H01 × L2
=⇒ (u1 , ∂t u1 ) ∈ C 0 [0, T ], (H 2 ∩ H01 ) × H01
we therefore need
(u1 , ∂t u1 , u2 , ∂t u2 )|t=T ∈ (H 2 ∩ H01 ) × H01 × H01 × L2
This is the natural space for the analysis of the control problem.
First results known for weak constant coupling (symmetric system) in
[Alabau-Boussouira: 03]
4/ 36
J. Le Rousseau
coupled wave equations
Setting
Alternative formulation

1
 ∂t2 u1 − ∆u1 + b(1 − ∆) 2 u2 = 0
∂ 2 u − ∆u2 = χg
 t 2
(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (0, 0, 0, 0)
in (0, T ) × Ω,
in (0, T ) × Ω,
in Ω.
Regularity
g ∈ L2 ((0, T ) × Ω),
=⇒ (u1 , ∂t u1 ) ∈ C 0 [0, T ], H01 × L2 =⇒ (u2 , ∂t u2 ) ∈ C 0 [0, T ], H01 × L2
We therefore need
(u1 , ∂t u1 , u2 , ∂t u2 )|t=T ∈ H01 × L2 × H01 × L2
5/ 36
J. Le Rousseau
coupled wave equations
Setting

1
 ∂t2 u1 − ∆u1 + b(1 − ∆) 2 u2 = 0
∂ 2 u − ∆u2 = χg
 t 2
(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (0, 0, 0, 0)
in (0, T ) × Ω,
in (0, T ) × Ω,
in Ω.
Control space g ∈ L2 ((0, T ) × Ω) → state space
(u1 , ∂t u1 , u2 , ∂t u2 ) ∈ H01 × L2 × H01 × L2 .
We also need conditions on
both the supports of the control function and the coupling term;
the time T .
6/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation I
Consider the control problem

2

in
∂t u − ∆u = χg
u=0
on


(u(0), ∂t u(0)) = (0, 0)
(0, T ) × Ω
(0, T ) × ×∂Ω
Wave symbol: p = −τ 2 + R(x, ξ)
(flat case: R(x, ξ) = |ξ|2 , here R(x, ξ) = |ξ|2x )
Hamiltonian vector field:
Hp = (∂τ p)∂t − (∂t p)∂τ + (∂ξ p)∂x − (∂x p)∂ξ
Integral curves for Hp
dx
dt
= ∂τ p(t, x, τ, ξ)
= ∂ξ p(t, x, τ, ξ)
ds
ds
dτ
dξ
= −∂t p(t, x, τ, ξ) = 0
= −∂x p(t, x, τ, ξ)
ds
ds
Rays or bicharacteristics: integral curves for Hp within the
characteristic set p = 0.
Singularities travel along such bicharacteristics (Hörmander)
7/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation I
Geometric control condition [Rauch-Taylor ’74,
Bardos-Lebeau-Rauch ’92]
Case of manifold without boundary.
(ω, T ) is said to satisfy GCC at time T if all bicharacteristics
starting from (x, ξ) at time t = 0 enter ω = {χ 6= 0} before t = T .
Case of manifold with boundary.
(ω, T ) is said to satisfy GCC at time T if all generalized
bicharacteristics starting from (x, ξ) at time t = 0 enter
ω = {χ 6= 0} before t = T .
Generalized bicharacteristics are described in [Melrose-Sjöstrand ’78,
’82]
Theorem (Bardos-Lebeau-Rauch ’92, Burq-Gérard’ 97)
GCC is equivalent to the exact controllability of the wave equation at
time T
8/ 36
J. Le Rousseau
coupled wave equations
Back to the system of two wave equations...
The system is

1
 ∂t2 u1 − ∆u1 + b(1 − ∆) 2 u2 = 0
∂ 2 u − ∆u2 = χg
 t 2
(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (0, 0, 0, 0)
in (0, T ) × Ω,
in (0, T ) × Ω,
in Ω.
with b ≥ 0.
ω = {χ 6= 0},
O = {b > 0}
Necessary condition to controllability
(ω, Tω ) satisfies GCC
(O, TO ) satisfies GCC
Definition
Given two sets ω and O both satisfying GCC, we set Tω→O→ω the
infimum of times T > 0 s.t.
every bicharacteristics traveling at speed one in Ω meets ω in a time
t0 < T , meets O in a time t1 ∈ (t0 , T ) and meets ω again in a time
t2 ∈ (t1 , T ).
Observe that max TO , Tω ≤ Tω→O→ω ≤ 2Tω + TO .
J. Le Rousseau
coupled wave equations
9/ 36
System of two wave equations
T
Tω→O→ω
t2
t1
t0
t=0
O
ω
Ω
Figure: Geometric condition and time Tω→O→ω .
10/ 36
J. Le Rousseau
coupled wave equations
System of two wave equations
O
ω
O
ω
Figure: Examples of open sets (Ω, ω, O) s.t. ω and O both satisfy GCC in
Ω: case (a), Ω is the flat torus (or the square), case (b), Ω is the disk.
11/ 36
J. Le Rousseau
coupled wave equations
System of two wave equations

1
 ∂t2 u1 − ∆u1 + b(1 − ∆) 2 u2 = 0
∂ 2 u − ∆u2 = χg
 t 2
(u1 , ∂t u1 , u2 , ∂t u2 )|t=0 = (0, 0, 0, 0)
with b ≥ 0.
in (0, T ) × Ω,
in (0, T ) × Ω,
in Ω.
Theorem (Dehman, LR, Léautaud)
Let Ω be a compact manifold without boundary. If
ω = {χ 6= 0} satisfies GCC
O = {b > 0} satisfies GCC
T > Tω→O→ω
then the system is exactly controllable.
If either
ω does not satisfy GCC
or O does not satisfy GCC
or T < Tω→O→ω
the system is NOT controllable.
J. Le Rousseau
12/ 36
coupled wave equations
System of two wave equations
Existing results
[F. Alabau-Boussouira - M. Leautaud ’11]
symmetric systems,
weak coupling, long control time.
[L. Rosier - L. de Teresa ’11]
1-D, geometric but not sharp control
time.
[F. Alabau-Boussouira]
Cascade of N equations, long control time.
13/ 36
J. Le Rousseau
coupled wave equations
System of two wave equations
Adjoint system:
2
(∂t − ∆)w1 = 0
1
(∂t2 − ∆)w2 = −b(x)(1 − ∆) 2 w1
in (0, T ) × Ω
in (0, T ) × Ω.
(Adj)
Controllability is equivalent to the observability inequality:
T
e0 (w1 (0)) + e0 (w2 (0)) ≤ C ∫ ∫ |χw2 |2 dx dt,
(Obs)
0Ω
for all (w1 , w2 ) solutions of (Adj).
Here e0 (w) = kwk2L2 (Ω) + k∂t wk2H −1 (Ω) .
14/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
Adjoint equation for the wave equation:
(∂t2 − ∆)v = 0
in (0, T ) × Ω
We set V = (v, ∂t v)
Controllability is equivalent to the observability inequality:
T
e0 (V (0)) ≤ C ∫ ∫ |χv|2 dx dt,
0Ω
for all v solution of the adjoint equation.
Here
e0 (V (0)) = kv(0)k2L2 (Ω) + k∂t v(0)k2H −1 (Ω)
15/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
Controlled equation: (∂t2 − ∆)u = χg
U (t) = (u(t), ∂t u(t)) ∈ H 1 (Ω) × L2 (Ω).
Adjoint equation: (∂t2 − ∆)v = 0
in (0, T ) × Ω
in (0, T ) × Ω.
With an integration by parts we have:
=0
hχg, viL2 ((0,T )×Ω) = h∂t u(T ), v(T )iL2
z
}|
{
− h∂t u(0), v(0)iL2
− hu(T ), ∂t v(T )iH 1 ,H −1 − hu(0), ∂t v(0)iH 1 ,H −1
|
{z
}
=0
We introduce the maps
L : L2 (Ω) × H −1 (Ω) → L2 ((0, T ) × Ω)
VT = V (T ) = (v(T ), ∂t v(T )) 7→ χv
and
M : L2 ((0, T ) × Ω) → H 1 (Ω) × L2 (Ω)
g 7→ (−u(T ), ∂t u(T ))
16/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
In the HUM approach we seek the control as a solution of the
backward wave equation: we set g = L(VT ) = χv.
Then (u(T ), ∂t u(T )) = M ◦ L(VT )
We then find
kL(VT )k2L2 ((0,T )×Ω) = h∂t u(T ), v(T )iL2 − hu(T ), ∂t v(T )iH 1 ,H −1
= hU (T ), VT i∗ = hM ◦ L(VT ), VT i∗
where
hU, V i∗ = hU2 , V1 iL2 +hU1 , V2 iH 1 ,H −1 ,
U ∈ H 1 ×L2 , V ∈ L2 ×H −1
G = M ◦ L is the Gramian operator.
17/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
We have
kL(VT )k2L2 ((0,T )×Ω) = hG (VT ), VT i∗
Note that G : L2 (Ω) × H −1 (Ω) → H 1 (Ω) × L2 (Ω)
18/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
We have
kL(VT )k2L2 ((0,T )×Ω) = hG (VT ), VT i∗
Note that G : L2 (Ω) × H −1 (Ω) → H 1 (Ω) × L2 (Ω)
Observability:
kVT kL2 ×H −1 ≤ CkL(VT )kL2 ((0,T )×Ω)
We have
controllability
⇔ observability ⇔ invertibility of G
18/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
We have
kL(VT )k2L2 ((0,T )×Ω) = hG (VT ), VT i∗
Note that G : L2 (Ω) × H −1 (Ω) → H 1 (Ω) × L2 (Ω)
Observability:
kVT kL2 ×H −1 ≤ CkL(VT )kL2 ((0,T )×Ω)
We have
controllability
⇔ observability ⇔ invertibility of G
In such case we can solve G (VT ) = UT with UT = (−u0 , u1 ) ∈ H 1 × L2
The HUM operator is precisely G −1
18/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
We have
kL(VT )k2L2 ((0,T )×Ω) = hG (VT ), VT i∗
Note that G : L2 (Ω) × H −1 (Ω) → H 1 (Ω) × L2 (Ω)
Observability:
kVT kL2 ×H −1 ≤ CkL(VT )kL2 ((0,T )×Ω)
We have
⇔ observability ⇔ invertibility of G
controllability
In such case we can solve G (VT ) = UT with UT = (−u0 , u1 ) ∈ H 1 × L2
The HUM operator is precisely G −1
The controlled equation is then
(∂t2 − ∆)u = χL(VT )
in (0, T ) × Ω
(u(0), ∂t u(0)) = (0, 0)
and we obtain (u(T ), ∂t u(T )) = (u0 , u1 ).
18/ 36
J. Le Rousseau
coupled wave equations
Review of the controllability of the wave equation II
We have
kL(VT )k2L2 ((0,T )×Ω) = hG (VT ), VT i∗
Note that G : L2 (Ω) × H −1 (Ω) → H 1 (Ω) × L2 (Ω)
Observability:
kVT kL2 ×H −1 ≤ CkL(VT )kL2 ((0,T )×Ω)
We have
⇔ observability ⇔ invertibility of G
controllability
In such case we can solve G (VT ) = UT with UT = (−u0 , u1 ) ∈ H 1 × L2
The HUM operator is precisely G −1
The controlled equation is then
(∂t2 − ∆)u = χL(VT )
in (0, T ) × Ω
(u(0), ∂t u(0)) = (0, 0)
and we obtain (u(T ), ∂t u(T )) = (u0 , u1 ).
Question: can we prove directly the invertibility of the Gramian
operator G ?
18/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
In part we follow the approach of Dehman-Lebeau
Here, to expose their method, we consider a simplified model to avoid
technicalities
We consider the half-wave first-order equation
√
∂t u − iλu = χg ∈ L2 (Ω),
λ = −∆,
u(0) = 0
λ is pseudo-differential of order 1 with principal symbol |ξ|
We have u ∈ C 0 ([0, T ]; L2 (Ω)) ∩ C 1 ([0, T ]; H −1 (Ω))
Note that (∂t − iλ)(∂t + iλ) = ∂t2 − ∆. We have factorized the wave
equation
The adjoint equation is
∂t v − iλv = 0 ∈ L2 (Ω),
v(T ) = vT
19/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
∂t u − iλu = χg ∈ L2 (Ω),
u(0) = 0
The adjoint equation is
∂t v − iλv = 0 ∈ L2 (Ω),
v(T ) = vT
Proceeding as above we find
hχg, viL2 ((0,T )×Ω) = hu(T ), v(T )iL2 .
We introduce the maps
L : L2 (Ω) → L2 ((0, T ) × Ω)
vT = v(T ) 7→ χv
and
M : L2 ((0, T ) × Ω) → L2 (Ω)
g 7→ u(T )
and set G = M ◦ L.
20/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
∂t u − iλu = χg ∈ L2 (Ω),
u(0) = 0
The adjoint equation is
∂t v − iλv = 0 ∈ L2 (Ω),
v(T ) = vT
Setting g = L(vT ) we have
kL(vT )k2L2 ((0,T )×Ω) = hG vT , vT iL2 .
controllability
⇔ observability ⇔ invertibility of G
21/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
∂t u − iλu = χg ∈ L2 (Ω),
u(0) = 0
Changing t → T − t in the adjoint equation
∂t v + iλv = 0 ∈ L2 (Ω),
v(0) = v0
We have
kL(v0 )k2L2 ((0,T )×Ω) = hG v0 , v0 iL2 .
controllability
⇔ observability ⇔ invertibility of G
22/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
Bicharacterisitic flow: we introduce ϕ solution to
d ±
ϕs (x, ξ) = H∓|ξ|x ϕs (x, ξ) ,
ds
∗
ϕ±
0 (x, ξ) = (x, ξ) ∈ T Ω \ 0.
N.B. H∓|ξ|x is the vector field: ∓(∂ξ |ξ|x )∂x ± (∂x |ξ|x )∂ξ .
23/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
The adjoint equation is
∂t v + iλv = 0 ∈ L2 (Ω),
λ=
√
−∆,
v(0) = v0
The gives v(0) = e−itλ v0 (Fourier integral operator)
N.B. Here we ignore the eigenvalue 0
Then we compute the observation
T
T
0
0
∫ kχvk2L2 (Ω) dt = ∫ heitλ χ2 e−itλ v0 , v0 iL2 (Ω) dt
We thus have G v0 = ∫0T eitλ χ2 e−itλ v0 dt.
By the Egorov theorem eitλ χ2 e−itλ is a pseudo-differential operator of
order 0 with principal symbol σ(t, x, ξ) = χ2 ◦ ϕ−
t (x, ξ)
The Gramian is thus a pseudo-differential operator of order 0 with
symbol
T
∫ χ2 ◦ ϕ−
t (x, ξ) dt
0
24/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
The Gramian is thus a pseudo-differential operator of order 0 with
symbol
T
∫ χ2 ◦ ϕ−
t (x, ξ) dt
0
For the Gramian to be invertible one needs it to be elliptic, that is
(order 0)
T
∫ χ2 ◦ ϕ−
t (x, ξ) dt ≥ C,
∀(x, ξ) ∈ T ∗ (Ω)
0
We thus recover the GC condition of Bardos-Lebeau-Rauch.
25/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
Theorem (Dehman-Lebeau)
We have
{χ > 0} satisfies GCC
⇔
the operator G is elliptic.
Moreover in such case
1
2
the operator G is coercive and invertible;
the HUM operator G −1 can be written as G −1 = Λ + R when R is
a regularizing operator and Λ is a pseudo-differential operator of
order 0 with principal symbol
T
−1
∫ χ2 ◦ ϕ−
(x,
ξ)
dt
t
0
26/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
The proof of coercivity goes along two steps with a
compactness-uniqueness argument as in the original proof of
Bardos-Lebeau-Rauch for the controllability of the wave equation.
We acknowledge very fruitful discussions with C. Laurent on some
aspects of the proof.
27/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
First-step: from the ellipticity of G we find with the Gårding
inequality:
hG v0 , v0 iL2 ≥ Ckv0 k2L2 − C 0 kv0 k2 − 1 .
H
2
(1)
We thus have coercivity for the high-frequencies.
Second-step: We consider
N (T ) = {v0 ∈ L2 (Ω); L(v) = χv(t, x) = 0 in (0, T ) × ω},
By proving that the unit sphere in N (T ) is compact from (1) we have
Lemma
The space N (T ) is finite dimensional.
Moreover if v0 ∈ N (T ) then χv = 0, implying χ∂t v = 0. This implies
Lv0 ∈ N (T ).
28/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
Second-step (continued):
Hence, N (T ) is stable under λ and is finite dimensional.
If N (T ) 6= {0}, this implies that there exist λ and w 6= 0, such that
w ∈ N (T ),
λw = λw,
χw = 0,
We thus have −∆w = λ2 w and χw = 0. A classical unique
continuation result yields w = 0
Lemma
We have N (T ) = {0}.
29/ 36
J. Le Rousseau
coupled wave equations
Analysis of the Gramian
Second-step (continued):
We assume that coercivity does not hold:
(n)
There exists (v0 ) ⊂ L2 (Ω) such that
(n)
kv0 k = 1
(n)
We have v0
(n)
kG v0 kL2 (Ω) → 0
* v ∈ L2 (Ω).
Continuity of the half wave equation yields v ∈ N (T ), that is v = 0
(n)
In particular we have v0 → 0 in H −1 (Ω).
The high-frequency result:
hG v0 , v0 iL2 ≥ Ckv0 k2L2 − C 0 kv0 k2
1
H− 2
then implies the contradiction
0 ≥ C − C 0 × 0.
30/ 36
J. Le Rousseau
coupled wave equations
Back to systems
Again we√consider a simplified model here for the sake of exposition:
Set λ = −∆ on L2+ , projecting onto the orthogonal of the space of
constant functions.
Consider:
(
1
bu2 = 0 in (0, T ) × Ω,
(∂t − iλ)u1 − 2i
(∂t − iλ)u2 = χf
in (0, T ) × Ω.
Adjoint system:


(∂t + iλ)v1 = 0
1
(∂t + iλ)v2 + 2i
bv1 = 0


(v1 (0), v2 (0)) = (g, h) ∈ L2 (Ω; C2 )
in (0, T ) × Ω,
in (0, T ) × Ω,
The observability inequality reads
T
kgk2L2 (Ω) + khk2L2 (Ω) ≤ C ∫ kχv2 k2L2 (Ω) dt.
0
31/ 36
J. Le Rousseau
coupled wave equations
Microlocal characterization of the HUM operator
The Gramian operator is given by
T
∫ kχv2 k2L2 (Ω) dt = G (g, h), (g, h)
0
L2 (Ω;C2 )
.
If the wave system is exactly controllable then G is invertible and the
HUM operator is G −1
Theorem
There exists G ∈ Ψ0 (Ω; C2×2 ), and R an infinitely smoothing operator
on Ω such that
G = G + R,
where the principal symbol (in S 0 (T ∗ Ω, C2×2 )) of G is

2

t
t
−
−
1 T 2
1 T 2
−
−
dt
dt
χ
◦
ϕ
∫
b
◦
ϕ
dσ
χ
◦
ϕ
∫
b
◦
ϕ
dσ
∫
∫
t
t
σ
0
σ
2i 0

 4 0
0
1 T 2
− 2i
∫0T χ2 ◦ ϕ−
∫0 χ ◦ ϕ−
∫0t b ◦ ϕ−
t dt
t
σ dσ dt
32/ 36
J. Le Rousseau
coupled wave equations
Microlocal characterization of the HUM operator
Theorem (continued)
In particular, we have
det(σ0 (G)) =
t2
2
1TT 2 −
−
∫ ∫ (χ ◦ϕt1 )(χ2 ◦ϕ−
dt1 dt2 ∈ S 0 (T ∗ Ω).
t2 ) ∫ b◦ϕσ dσ
800
t1
The operator G is coercive on L2 (Ω; C2 ).
The operator G is invertible in L(L2 (Ω)). Its inverse (G )−1 , the
HUM operator, can be decomposed as (G )−1 = Λ + R where R is
smoothing and Λ ∈ Ψ0 (T ∗ Ω, C2×2 ), with principal symbol
det(σ0 (G))−1

∫0T χ2 ◦ ϕt dt
×
1 T 2
− 2i
∫0 χ ◦ ϕ−
∫0t b ◦ ϕ−
t
σ dσ dt
∫0T χ2 ◦ ϕ−
∫0t b ◦ ϕ−
t
σ dσ dt
2
t
−
1 T 2
−
∫
χ
◦
ϕ
∫
b
◦
ϕ
dσ
dt
t
0
σ
0
4
1
2i
33/ 36
J. Le Rousseau
coupled wave equations
Microlocal characterization of the HUM operator
det(σ0 (G)) =
t2
2
1TT 2
−
2
−
∫ ∫ (χ ◦ ϕ−
dt0 dt2 ∈ S 0 (T ∗ Ω).
t0 )(χ ◦ ϕt2 ) ∫ b ◦ ϕσ dσ
800
t0
T
Tω→O→ω
t2
φt1 (B)
t∗
ρ∗
t1
t0
φt0 (B)
B
O
ω
Ω
Observe that the geometric condition appears clearly in det(σ0 (G)).
J. Le Rousseau
coupled wave equations
34/ 36
Microlocal characterization of the HUM operator
Sketch of proof. The Duhamel formula gives
v1 (t) = e−itλ g,
v2 (t) = e−itλ h −
1 t −i(t−σ)λ
∫e
bv1 (σ)dσ.
2i 0
We compute
T
∫ kχv2 k2L2 (Ω) dt = (G + R)(g, h), (g, h) )L2
0
with
G=
1
4
1 T
∫0T (Bt )∗ eitλ χ2 e−itλ Bt dt + 2i
∫0 (Bt )∗ eitλ χ2 e−itλ dt
1 T itλ 2 −itλ
Bt dt
∫0T eitλ χ2 e−itλ dt
− 2i ∫0 e χ e
,
where
t
Bt := ∫ eiσλ be−iσλ dσ
0
The conclusion follows from the Egorov Theorem.
35/ 36
J. Le Rousseau
coupled wave equations
Thank you for your attention.
Further details in:
B. Dehman, J. Le Rousseau, and M. Léautaud. Controllability of two
coupled wave equations on a compact manifold, preprint 2012,
53 pages.
http://hal.archives-ouvertes.fr/hal-00686967
36/ 36
J. Le Rousseau
coupled wave equations

Documents pareils