Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Stabilization of second order evolution equations
with feedbacks with delay
Serge NICAISE1
Université de Valenciennes et du Hainaut-Cambrésis
May 11, 2011
1
based on a joint work with J. Valein, ESAIM-COCV, 2010
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
The Problem
Let H be a real Hilbert space with inner product (·, ·)H ; and associated norm
kukH = (u, u)1/2 .
Let A : D(A) → H be a self-adjoint positive operator with a compact inverse
1
1
in H, and set V := D(A 2 ) the domain of A 2 , with i. p.
1
1
(u, v )V := (A 2 u, A 2 v )H , ∀u ∈ V ,
1
and norm kukV = kA 2 ukH , ∀u, v ∈ V .
For i = 1, 2, let Ui be a real Hilbert space (identified to its dual space) with i.
1/2
p. (·, ·)Ui and associated norm kv kUi = (v , v )Ui ,
and let Bi ∈ L(Ui , V 0 ).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Friedrichs extension
A commonly used op. A is as follows: Let V ⊂ H be another real Hilbert space
1/2
with inner product (·, ·)V ; and associated norm kukV = (u, u)V . We assume
that V is dense in H and is compactly embedded into H.
Let be given a bilinear, symmetric and continuous form
a : V × V → R such that a is coercive:
∃α > 0 : a(v , v ) ≥ αkv k2V ∀v ∈ V .
We
define
D(A) = {u ∈ V : ∃f ∈ H : a(u, v ) = (f , v ) ∀v ∈ V },
Au = f ∀u ∈ D(A).
A is called the Friedrichs extension of the triple (H, V , a). It is well known that
A is a self-adjoint positive op. with a compact inverse in H such that
1
V := D(A 2 ).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
The abstract problem
We consider the closed loop system
utt (t) + Au(t) + B1 B1∗ ut (t) + B2 B2∗ ut (t − τ ) = 0, t > 0
u(0) = u0 , ut (0) = u1 , B2∗ ut (t − τ ) = f 0 (t − τ ), 0 < t < τ.
(1)
where τ is a positive constant which represents the delay, u : [0, ∞) → H is
the state of the system.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
1st Example: bd feedbacks
The 1-d wave equation with internal bd feedbacks
 ∂2u
2
(x, t) − ∂∂xu2 (x, t)


∂t 2

∂u

 +α1 ∂t (x, t) + α2 ∂u
(x, t − τ ) = 0, 0 < x < 1, t > 0
∂t
u(0, t) = u(1, t) = 0, t > 0,



u(x, 0) = u0 (x), ut (x, 0) = u1 (x), 0 < x < 1,


ut (x, t − τ ) = f 0 (t − τ ), 0 < x < 1, 0 < t < τ.
(2)
where α1 , α2 > 0 and τ > 0.
This problem enters in the above abstract setting as shown in the next slides.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Abstract form
H = L2 (0, 1), V = H01 (0, 1); U1 = U2 = L2 (0, 1),
d2
A : H 2 (0, 1) ∩ H01 (0, 1) → H : ϕ 7→ − 2 ϕ,
dx
√
0
Bi : Ui = H → H ⊂ V : u 7→ αi u, i = 1, 2.
R1
with (u, v )H = 0 u(x)v (x) dx,
Note that Bi∗ = Bi .
Serge NICAISE
∀u, v ∈ L2 (0, 1).
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Abstract form ctd
A is a positive selfadjoint op. since it is the Friedrichs extension of
the triple (H, V , a), where
R1
a(u, v ) = 0 ux (x)vx (x) dx, ∀u, v ∈ V = H01 (0, 1).
which is coercive on V , due to Poincaré’s ≤.
A has a compact inv. since H01 (0, 1) ,→c L2 (0, 1).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
2d example: unbd feedbacks
The wave equation with boundary feedbacks
 ∂2u
2
(x, t) − ∂∂xu2 (x, t) = 0, 0 < x < 1, t > 0


∂t 2


 u(0, t) = 0, t > 0
∂u
(1, t) + α1 ∂u
(1, t) + α2 ∂u
(1, t − τ ) = 0, t > 0,
∂x
∂t
∂t



u(x,
0)
=
u
(x),
u
(x,
0)
=
u
(x), 0 < x < 1,
0
t
1


ut (ξ, t − τ ) = f 0 (t − τ ), 0 < t < τ.
(3)
where α1 , α2 > 0 and τ > 0.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
2d Example ctd
Enters in the abstract setting with
H = L2 (0, 1), V = {u ∈ H 1 (0, 1) : u(0) = 0};
A : {u ∈ H 2 (0, 1) : u(0) = ux (1) = 0} → H : ϕ 7→ −
U1 = U2 = R,
Bi : R → V 0 : k 7→
d2
ϕ,
dx 2
√
αi kδ1 , i = 1, 2.
Again A is a positive selfadjoint op. since it is the Friedrichs extension of the
triple (H, V , a), where the bil. form a is defined as before. Again A has a
compact inv. since V ,→c L2 (0, 1).
√
Note that Bi∗ : V → R : u 7→ αi u(1), i = 1, 2, because
√
hBi k, v iV 0 −V = αi kv (1) = kBi∗ v .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
3d example
The wave equation with internal unbd feedbacks
 2
2
∂u
 ∂∂tu2 − ∂∂xu2 + α1 ∂u
∂t (ξ, t)δξ + α2 ∂t (ξ, t − τ )δξ = 0, 0 < x < 1, t > 0
u(0, t) = u(1, t) = 0, t > 0

I .C .
where ξ ∈ (0, 1), α1 , α2 > 0 and τ > 0.
H = L2 (0, 1), A : H 2 (0, 1) ∩ H01 (0, 1) → H : ϕ 7→ −
d2
ϕ,
dx 2
V = H01 (0, 1); U1 = U2 = R,
√
Bi : R → V 0 : k 7→ αi kδξ , i = 1, 2.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Some instabilities
If α2 > α1 , the previous systems may be unstable, cfr. [Datko-Lagnese-Polis
1986, N.-Pignotti 2006, N.-Valein 2007]. We check this property on the next
slide for Example 2.
Hence some conditions between B1 and B2 have to be imposed to get stability.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Instabilities for Ex. 2
We look for solution u of (3) in the form
u(x, t) = e λt ϕ(x).
Hence (3) holds iff ϕ is solution of
ϕxx − λ2 ϕ = 0 in (0, 1)
ϕ(0) = 0, ϕx (1) + (α1 + α2 e −λτ )λϕ(1) = 0.
Hence for λ 6= 0, ϕ is given by
ϕ(x) = C1 sinh(λx).
Then the bc on 1 will be satisfied if
C1 λ(cosh λ + (α1 + α2 e −λτ ) sinh λ) = 0.
Since we are looking for non trivial solution, this is ⇔
cosh λ + (α1 + α2 e −λτ ) sinh λ = 0.
Serge NICAISE
(4)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Roots of (4)
Lemma
If α2 ≥ α1 ≥ 0, ∃ a discrete set T of delays τ s. t. for any τ ∈ T the system
(3) is unstable, i.e., it admits exponentially unstable solutions.
Pf. We take the delay in the form
τ = 2(2k+1)
< 1,
2n+1
with n, k ∈ N and we look for roots of (4) in the form
λ = η + ı (2n+1)π
,
2
where η ∈ R is the remaining unknown. As e −λτ = −e −ητ , (4) holds iff
F (η) := e η − e −η + (α1 − α2 e −ητ )(e η + e −η ) = 0.
But we notice that
F (0) = 2(α1 − α2 ) ≤ 0, while limη→+∞ F (η) = +∞.
Hence by the intermediate value thm, ∃η ≥ 0 s.t. F (η) = 0. The system (3) is
then unstable for the countable set of delays τ in the above form.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Well-posedness
Main idea is to introduce the auxiliary unknown:
z(ρ, t) = B2∗ ut (t − τ ρ) for ρ ∈ (0, 1) and t > 0.
Then (1) is equivalent to

utt (t) + Au(t) + B1 B1∗ ut (t) + B2 z(1, t) = 0, t > 0



∂z

+ ∂ρ
= 0, t > 0, 0 < ρ < 1
τ ∂z

∂t
u(0) = u0 , ut (0) = u1 ,



z(ρ, 0) = f 0 (−τ ρ), 0 < ρ < 1


z(0, t) = B2∗ ut (t), t > 0.
(5)
Note that in (5) the second identity is a transport equation in z.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Reduction of order
By a standard reduction of order, we set
U = (u, ut , z)> and problem (5) can be rewritten as
Ut = AU
U(0) = (u0 , u1 , f 0 (−τ ·)),
(6)
where formally the operator A is defined by


 
v
u
∗
A  v  =  −Au − B1 B1 v − B2 z(1)  .
∂z
− τ1 ∂ρ
z
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Well-posedness ctd
Denote by H the Hilbert space H = V × H × L2 ((0, 1), U2 ).
Then A is well defined if
D(A) := {(u, v , z) ∈ V × V × H 1 ((0, 1), U2 ); z(0) = B2∗ v ,
Au + B1 B1∗ v + B2 z(1) ∈ H}.
In that case A maps D(A) into H.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Well-posedness ctd
Let us now suppose that
∃0 < α ≤ 1, ∀u ∈ V , kB2∗ uk2U2 ≤ αkB1∗ uk2U1 .
(7)
We fix a positive real number ξ such that
1≤ξ≤
2
− 1.
α
(8)
We now introduce the following inner product on H:

 

Z 1
u
ũ
1
1
h v  ,  ṽ i= (A 2 u, A 2 ũ)H +(v , ṽ )H + τ ξ
(z(ρ), z̃(ρ))U2 dρ.
0
z
z̃
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Illustration
For the three examples: (7) ⇔ α2 ≤ α1 .
Indeed for ex 1, we have
kBi∗ uk2Ui = αi kuk2H .
Hence (7) ⇔
∃0 < α ≤ 1, ∀u ∈ V , α2 kuk2H ≤ αα1 kuk2H
m
∃0 < α ≤ 1, α2 ≤ αα1
m
α2 ≤ α1 .
Idem for ex 2 (resp. 3), since here we have
kBi∗ uk2Ui = αi |u(1)|2 (resp. kBi∗ uk2Ui = αi |u(ξ)|2 ).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Well-posedness
We show that A generates a C0 semigroup on H, by showing that A is
dissipative in H for the above inner product and that λI − A is surjective for
any λ > 0. By Lumer-Phillips Theorem ⇒
Theorem
Under the assumption (7), for an initial datum U0 ∈ H, there exists a unique
solution U ∈ C ([0, +∞), H) to system (6). Moreover, if U0 ∈ D(A), then
U ∈ C ([0, +∞), D(A)) ∩ C 1 ([0, +∞), H).
Remark
D(A) is dense in H, cfr. [Pazy 1983].
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Dissipativeness
Take U = (u, v , z)> ∈ D(A). Then
*
hAU, UiH =
1
 
+
v
u
∗
 −Au − B1 B1 v − B2 z(1)  ,  v 
∂z
− τ1 ∂ρ
z
H
1
= (A 2 v , A 2 u)H − (Au + B1 B1∗ v + B2 z(1), v )H
Z 1
∂z
−ξ
( (ρ), z(ρ))U2 dρ.
∂ρ
0
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Dissip. ctd
Since Au + B1 B1∗ v + B2 z(1) ∈ H, we obtain by duality
1
1
hAU, UiH = A 2 v , A 2 u − hAu, v iV 0 , V − hB1 B1∗ v , v iV 0 , V
H
Z 1
∂z
− hB2 z(1), v iV 0 , V −ξ
( (ρ), z(ρ))U2 dρ
∂ρ
0
Z 1
∂z
= −kB1∗ v k2U1 − (z(1), B2∗ v )U2 −ξ
( (ρ), z(ρ))U2 dρ.
∂ρ
0
R
R 1 ∂z
1
d
But 0 ( ∂ρ (ρ), z(ρ))U2 dρ = 21 0 dρ
k z(ρ)k2U2 dρ
2
∗
1
= 2 (kz(1)kU2 − kB2 v k2U2 )
⇓
hAU, UiH = −kB1∗ v k2U1 − (z(1), B2∗ v )U2 −
Serge NICAISE
ξ
ξ
kz(1)k2U2 + kB2∗ v k2U2 .
2
2
(9)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Dissip. end
Therefore, by Cauchy-Schwarz’s and Young’s ≤, we find
hAU, UiH = −kB1∗ v k2U1 − (z(1), B2∗ v )U2 − ξ2 kz(1)k2U2 + ξ2 kB2∗ v k2U2
≤ −kB1∗ v k2U1 + ( 12 − ξ2 )kz(1)k2U2 + ( 21 + ξ2 )kB2∗ v k2U2 .
By (7), we obtain
hAU, UiH ≤ (
Since condition (8) ⇒
α
2
ξα
1
ξ
α
+
− 1)kB1∗ v k2U1 + ( − )kz(1)k2U2 .
2
2
2
2
+
ξα
2
− 1 ≤ 0 and
1
2
−
ξ
2
≤ 0 we have shown that
hAU, UiH ≤ 0
and then the dissipativeness of A.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Maximality
We show that I − A is surjective: For (f , g , h)T ∈ H, we look for
U = (u, v , z)> ∈ D(A) solution of

 

u
f
(I − A)  v  =  g 
z
h
or equivalently

 u−v =f
v + Au + B1 B1∗ v + B2 z(1) = g

∂z
= h.
z + τ1 ∂ρ
Serge NICAISE
(10)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Maximality ctd
Suppose that we have found u with the appropriate regularity. Then, we have
v = −f + u ∈ V .
We can then determine z: indeed z satisfies the diff. eq.
∂z
= h and the boundary condition z(0) = B2∗ v = −B2∗ f + B2∗ u.
z + τ1 ∂ρ
Therefore z is explicitly given by
Z ρ
e τ σ h(σ)dσ.
(11)
z(ρ) = B2∗ ue −τ ρ − B2∗ fe −τ ρ + τ e −τ ρ
0
This means that once u is found with the appropriate properties, we can find z
and u. In particular, we have
Z 1
z(1) = B2∗ ue −τ − B2∗ fe −τ + τ e −τ
e τ σ h(σ)dσ = B2∗ ue −τ + z 0 ,
(12)
0
z 0 = −B2∗ fe −τ + τ e −τ
R1
0
e τ σ h(σ)dσ ∈ U2 depends only on f , h.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Maximality ctd
It remains to find u. By (10), eliminating v = u − f , u must satisfy
u + Au + B1 B1∗ u + B2 z(1) = g + B1 B1∗ f + f ,
and thus by (12),
u + Au + B1 B1∗ u + e −τ B2 B2∗ u = g + B1 B1∗ f + f − B2 z 0 =: q,
where q ∈ V 0 . We take the duality brackets h·, ·iV 0 , V with φ ∈ V and get the
problem
u + Au + B1 B1∗ u + e −τ B2 B2∗ u, φ V 0 , V = hq, φiV 0 , V .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Maximality ctd
we have
Since u ∈ V ⊂ H
u + Au + B1 B1∗ u + e −τ B2 B2∗ u, φ V 0 , V =
hu, φiV 0 , V + hAu, φiV 0 , V + hB1 B1∗ u, φiV 0 , V + e −τ hB2 B2∗ u, φiV 0 , V
1
1
= (u, φ)H + (A 2 u, A 2 φ)H + (B1∗ u, B1∗ φ)U1 + e −τ (B2∗ u, B2∗ φ)U2 .
Consequently, we arrive at the problem: Find u ∈ V sol. of
b(u, φ) = hq, φiV 0 , V ,
∀φ ∈ V ,
(13)
where
b(u, φ)
1
1
=
(u, φ)H + (A 2 u, A 2 φ)H + (B1∗ u, B1∗ φ)U1
+
e −τ (B2∗ u, B2∗ φ)U2 .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Maximality ctd
The bil. form b is continuous and coercive on V . Indeed, we have
1
1
|b(u, φ)| ≤ kukH kφkH + kA 2 ukH kA 2 φkH
+kB1∗ ukU1 kB1∗ φkU1 + e −τ kB2∗ v kU2 kB2∗ φkU2
≤ C kukV kφkV + kukV kφkV
+kB1∗ k2L(V , U1 ) kukV kφkV + e −τ kB2∗ k2L(V , U2 ) kukV kφkV
≤ C kukV kφkV ,
and for φ = u ∈ V
1
1
b(u, u) = kuk2H + (A 2 u, A 2 u)H + kB1∗ uk2U1 + e −τ kB2∗ uk2U2
1
≥ kA 2 uk2H = kuk2V .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Maximality end
Lax-Milgram’s lemma ⇒ problem (13) has a unique solution u ∈ V .
Then we take v = u − f that belongs to V , z defined by (11) that belongs to
H 1 ((0, 1), U2 ). With these choices, we see that (10) holds.
Therefore by construction Au + B1 B1∗ v + B2 z(1) = g + f − u. Since this rhs
belongs to H, the triple (u, v , z)> belongs to D(A).
In summary, we have found (u, v , z)> ∈ D(A) satisfying (10).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Back to (1)
For initial data (u0 , u1 , f 0 (−τ ·))> in D(A), we easily show that the solution
(u(t), v (t), z(t))T = T (t)(u0 , u1 , f 0 (−τ ·))T , where T (t) is the semigroup
generated by A, is indeed solution of (1) in the sense that
v = ut ,
and
z(ρ, t) = B2∗ ut (t − τ ρ),
and u satisfies (1).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
The energy
We restrict the ass. (7) to obtain the decay of the energy:
∃0 < α<1, ∀u ∈ V , kB2∗ uk2U2 ≤ αkB1∗ uk2U1
(14)
We define the energy as
E (t) :=
1
2
Z
1
kA 2 uk2H + kut k2H + τ ξ
1
0
kB2∗ ut (t − τ ρ)k2U2 dρ ,
where ξ is a positive constant satisfying 1 < ξ <
2
α
(15)
− 1.
Ex. (14) ⇔ α2 < α1 .
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
3.1. Energy decay
Proposition
If (14), then for any regular sol. of (1), the energy is non increasing and
E 0 (t) ∼ − kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 .
(16)
Remark
Integrating the expression (16) between 0 and T , we obtain
Z T
1
kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 dt ≤
(E (0) − E (T ))
C
1
0
1
≤
E (0).
C1
This estimate implies that B1∗ ut (·) ∈ L2 ((0, T ), U1 ) and
B2∗ ut (· − τ ) ∈ L2 ((0, T ), USerge
2 ). NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
Deriving (15), we obtain
1
1
E 0 (t)=(A 2 u, A 2 ut )H + (ut , utt )H
Z 1
+τ ξ
(B2∗ ut (t − τ ρ), B2∗ utt (t − τ ρ))U2 dρ
0
= hAu, ut iV 0 ,V − (ut , Au + B1 B1∗ ut + B2 B2∗ ut (t − τ ))H
Z 1
+ξτ
(B2∗ ut (t − τ ρ), B2∗ utt (t − τ ρ))U2 dρ.
0
Hence we see that
E 0 (t) = hAU(t), U(t)iH , where U(t) = (u(t), ut (t), B2∗ ut (t − τ ·)).
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
By the id. (9) we have
E 0 (t) = −kB1∗ ut (t)k2U1 − (B2∗ ut (t − τ ), B2∗ ut (t))U2
− ξ2 kB2∗ ut (t − τ )k2U2 + ξ2 kB2∗ ut (t)k2U2 .
Cauchy-Schwarz’s inequality ⇒
kB2∗ ut (t)k2U2 + 1−ξ
kB2∗ ut (t − τ )k2U2 ,
E 0 (t) ≤ −kB1∗ ut k2U1 + 1+ξ
2
2
ξ−1
1+ξ
0
∗
2
∗
2
E (t) ≥ −kB1 ut kU1 + 2 kB2 ut (t)kU2 − 2 kB2∗ ut (t − τ )k2U2 . Therefore, by
(14), these estimates lead to E 0 (t) ≤ −C1 kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2
0
∗
2
∗
2
E (t) ≥ −C2 kB1 ut kU1 + kB2 ut (t − τ )kU2 with
C1 = min{1 − (ξ+1)α
, ξ−1
} > 0, C2 = max{1, ξ+1
}>0
2
2
2
due to the choice of ξ.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
3.2. Energy decay to 0
Proposition
Assume that (14) holds. Then, for all initial data in H, limt→∞ E (t) = 0 iff
∀(non zero) eigenvector ϕ ∈ D(A) : B1∗ ϕ 6= 0.
(17)
Proof: ⇐ We closely follow [Tucsnak-Weiss 03].
⇒ We use a contradiction argument. Remark
This NSC is the same than without delay; therefore, (1) with delay is st. stable
(i.e. the energy tends to zero) iff the system without delay (i.e. for B2 = 0) is
st. stable.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Illustration
Ex. 1
Ex. 2
always
Ex. 3
√
ϕk = √2 sin(kπ·)∀k ∈ N∗ : (17) ⇔ sin(kπ·) 6= 0∀k ∈ N∗ : always true.
ϕk = 2 sin((k − 12 )π·)∀k ∈ N∗ : (17) ⇔ sin((k − 12 )π) 6= 0∀k ∈ N∗ :
true. √
ϕk = 2 sin(kπ·)∀k ∈ N∗ :
(17) ⇔ sin(kπξ) 6= 0∀k ∈ N∗ ⇔ ξ 6∈ Q.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Conservative syst.
The stability of (1) is based on some observability estimates for the associated
conservative system: We split up u sol of (1) in the form
u = φ + ψ,
where φ is solution of the problem without damping
φtt (t) + Aφ(t) = 0
φ(0) = u0 , φt (0) = u1 .
and ψ satisfies
ψtt (t) + Aψ(t) = −B1 B1∗ ut (t) − B2 B2∗ ut (t − τ )
ψ(0) = 0, ψt (0) = 0.
Serge NICAISE
(18)
(19)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Cons. syst. ctd
By setting B = (B1 B2 ) ∈ L(U, V 0 ) where U = U1 × U2 , ψ is solution of
ψtt (t) + Aψ(t) = Bv (t)
v (t) = (−B1∗ ut (t), −B2∗ ut (t − τ ))> .
(20)
ψ(0) = 0, ψt (0) = 0,
Lemma
[Ammari-Tucsnak 01]If B satisfies: ∃β > 0 :
λ ∈ {µ ∈ C |<µ = β } → λB ∗ (λ2 I + A)−1 B ∈ L(U) is bd,
then ψ satisfies
Z
Z T X
k(Bi∗ ψ)0 k2Ui dt≤ Ce 2βT
0
i=1,2
T
(21)
(kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 )dt. (22)
0
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
As (20) is time reversible,
we can extend v by zero outside [0, T ] (set ṽ this
ytt (t) + Ay (t) = B ṽ (t), t ∈ R,
extension) and solve
y (0) = 0, yt (0) = 0.
In that way Rψ = y on (0, T ).
Let ŷ (λ) = R e λt y (t) dt, be the Laplace transform of y that is well defined for
λ = β + ıη for any β > 0, η ∈ R.
Now we see that (22) holds if
ke −β· B ∗ y k2H 1 (R;U) ≤ C kṽ k2L2 (R;U) .
By the Plancherel identity, this is ⇔
k(β + ıη)B ∗ ŷ k2 1
≤ C kṽˆ k2 2
.
H (R;U)
L (R;U)
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Stabilization of second order evolution equations with feedback
Outline of the talk
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Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf end
But we see that
λ2 ŷ (λ) + Aŷ (λ) = B ṽˆ (λ).
Therefore
ŷ (λ) = (λ2 + A)−1 B ṽˆ (λ),
and consequently
λB ∗ ŷ (λ) = [λB ∗ (λ2 + A)−1 B]ṽˆ (λ).
The conclusion follows from the assumption (21).
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Cons. syst. ctd
Lemma
Suppose that the assumption (21) is satisfied. Then the solutions
RT P
∗ 0 2
u of (1) and φ of (18) satisfy 0
i=1,2 k(Bi φ) kUi dt ≤
R
T
Ce 2βT 0 (kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 )dt,
with C > 0 independent of T .
RT P
RT P
2
∗ 0 2
∗
Pf. 0
i=1,2 kBi ut kUi dt
i=1,2 k(Bi φ) kUi dt ≤ 0
RT P
∗
0 2
+ 0
i=1,2 k(Bi ψ) kUi dt.
The conclusion follows from (22) and the assump. (7).
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Exp. stab.
Theorem
Assume that (14) and (21) are satisfied. If ∃T > 0 and a constant C > 0 ind.
of τ s. t. the observability estimate
Z T
1
kA 2 u0 k2H + ku1 k2H ≤ C
kB1∗ φt (t)k2U1 dt
(23)
0
holds, where φ is solution of (18), then the system (1) is exp. stable. in energy
space: there exist C > 0 independent of τ and ν > 0 such that, for all initial
data in H,
E (t) ≤ CE (0)e −νt ∀t > 0.
(24)
Remark
Notice that the SC (23) is the same than the case without delay
[Ammari-Tucsnak 01]. Therefore, if (21) holds, then (1) is exp. stable if the
dissipative system without Serge
delayNICAISE
(i.e. withStabilization
B2 = 0) ofis second
exp. order
stable.
evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
Let u be a solution of (1) with initial data in D(A).
Without loss of generality, we can always assume that (23) holds with T > τ
and C independent of τ .
By Rk 2, we obtain
RT
E (0) − E (T ) ≥ C 0 (kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 )dt
R
T
≥ C2 0 (kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 )dt
RT
C
+ 2 0 kB2∗ ut (t − τ )k2U2 dt.
Lemma 2 ⇒ ∃C1 > 0 (dep. on T but ind. of τ ) s.t.
RT
RT
E (0) − E (T ) ≥ C1 ( 0 k(B1∗ φ)0 (t)k2U1 dt + 0 kB2∗ ut (t − τ )k2U2 dt).
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
By assumption (23), we obtain
1
E (0) − E (T ) ≥ C2 (kA 2 u0 k2H + ku1 k2H +
T
Z
kB2∗ ut (t − τ )k2U2 dt),
0
with C2 > 0 independent of τ . As T > τ , by change of variables:
R T −τ
RT
kB2∗ ut (t − τ )k2U2 dt= −τ kB2∗ ut (t)k2U2 dt
0
R0
R1
≥ −τ kB2∗ ut (t)k2U2 dt = τ 0 kB2∗ ut (−τ ρ)k2U2 dρ.
The two previous inequalities directly imply that
Z
1
E (0) − E (T ) ≥ C3 kA 2 u0 k2H + ku1 k2H + ξτ
1
kB2∗ ut (−τ ρ)k2U2 dρ ,
0
with C3 > 0 dep. on T but ind. of τ .
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
By the def. of E (0), this is ⇔
E (0) − E (T ) ≥ C3 E (0).
⇓ (energy is decreasing)
E (0) − E (T ) ≥ C3 E (T ).
This leads to
E (T ) ≤ γE (0), where γ =
1
< 1.
1 + C3
Applying this argument on [(m − 1)T , mT ], for m = 1, 2, ... (which is valid
because the system is invariant by translation in time), we will get
E (mT ) ≤ γE ((m − 1)T ) ≤ · · · ≤ γ m E (0).
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Stabilization of second order evolution equations with feedback
Outline of the talk
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Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf end
Therefore, we have
E (mT ) ≤ e −νmT E (0), m = 1, 2, · · ·
with ν = T1 ln γ1 > 0 which depends on T and thus on τ (because T > τ ). For
an arbitrary positive t, there exists m ∈ N∗ such that (m − 1)T < t ≤ mT and
by the non-increasing property of the energy, we conclude
E (t) ≤ E ((m − 1)T ) ≤ e −ν(m−1)T E (0) ≤
1 −νt
e
E (0).
γ
Hence the energy decays exponentially with the decay rate ν.
By density of D(A) into H, we deduce that (24) holds for any initial data in
H.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Preliminary res.
The proof of the polynomial stability result requires the next technical Lemma
proved by [Ammari-Tucsnak 00].
Lemma
Let (εk )k be a sequence of positive real numbers s. t. ∃C > 0 and µ > −1:
εk+1 ≤ εk − C ε2+µ
k+1
∀k ≥ 0.
(25)
Then ∃M > 0 (depending on µ and C ) such that
εk ≤
M
1
∀k ≥ 0.
(26)
(1 + k) 1+µ
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
Consider the mapping
F (x) = x + Cx 2+µ , ∀x > 0. ((25) ⇔ F (εk+1 ) ≤ εk , ∀k ≥ 0)
Then F 0 (x) = 1 + C (2 + µ)x 1+µ ≥ 0∀x > 0 ⇒ F is increasing.
1
Write for shortness ν := 1+µ
.
ν
As for k ≥ 1, (1 + k) ∼ k ν and since (26) trivially holds for k = 0, it then
suffices to show
εk ≤ Mk −ν ∀k ≥ 1.
(27)
We show (27) by induction on k. Again it trivially holds for k small, hence it
suffices to show that the induction procedure for k ≥ k0 with k0 :
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Stabilization of second order evolution equations with feedback
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ind. proc.
Assume that (27) for k ≥ k0 . Then by (25) we get
F (εk+1 ) ≤ kMν .
If one can show that
M
M
≤ F(
)
kν
(k + 1)ν
∀k ≥ k0 ,
(28)
then we will obtain
M
F (εk+1 ) ≤ F ( (k+1)
ν ),
and since F is increasing, we deduce (27) for k + 1.
Hence it remains to show (28), that is ⇔
M
M
M 2+µ
≤ (k+1)
ν + C
kν
(k+1)(2+µ)ν
m
1≤
k
k+1
ν
ν
+ CM 1+µ (k+1)k(2+µ)ν .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf of (28)
Introduce
themapping
ν
ν
x
g (x) = x+1
+ CM 1+µ (x+1)x(2+µ)ν .
Then direct calculations yield
ν−1
ν−1
x
1
g 0 (x) = ν x+1
+ CM 1+µ ν (x+1)x(2+µ)ν+1 (x − 1 − µ). Hence for
(x+1)2
x ≥ 1 + µ, g is increasing and therefore
g (x) ≥ g (1 + µ), ∀x ≥ 1 + µ.
As for M , g (1 + µ) ≥ 1, we conclude that
g (k) ≥ 1∀k ≥ k0 ≥ 1 + µ.
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Stabilization of second order evolution equations with feedback
Outline of the talk
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Interp. ≤
Lemma
For (u0 , u1 ) ∈ D(A) × V , we have
ku0 km+11
≤
ku1 kHm+1
C ku0 km
D(A) ku0 k
D(A
D(A 2 )
≤
C ku1 km
1
D(A 2 )
1−m
2
,
)
ku1 kD(A− m2 ) ,
where C > 0.
1
Pf If we denote by {λk }k≥1 the eigenvalues of A 2 counted with their
multiplicity, and {ϕk }k≥1 the orthonormal eigenvectors associated with the
eigenvalue λP
k . Then we have the equivalence
kuk2D(As ) ∼ k≥1 |uk |2 λ4s
∀s ∈ R,
k
P
when u = k≥1 uk ϕk .
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Hence
kuk2
1
D(A 2 )
∼
X
|uk |2 λ2k =
k≥1
X
2
2
2β
|uk | p |uk | q λ2α
k λk ,
k≥1
with p, q > 1 s.t. p1 + q1 = 1 and α + β = 1.
Hölder’s inequality ⇒
1
1 P
P
p
q
kuk2 1 .
|uk |2 λ2αp
|uk |2 λ2βq
. Therefore
k
k
k≥1
k≥1
D(A 2 )
P
m+1 P
m+1
2p
2q
2 2αp
2 2βq
kukm+11 .
.
k≥1 |uk | λk
k≥1 |uk | λk
D(A 2 )
The first estimate follows by choosing p =
β=
1−m
q
since
m+1
p
= m,
m+1
q
m+1
,
m
q = m + 1, α =
2
p
and
= 1, 2αp = 4, βq = 1 − m and finally α + β = 1.
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Stabilization of second order evolution equations with feedback
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Similarly
kuk2H
∼
X
|uk |2 =
k≥1
X
2
2
−α
|uk | p |uk | q λα
k λk ,
k≥1
with p, q > 1 s.t. p1 + q1 = 1 and any α ∈ R.
Hölder’s inequality ⇒
P
1 P
1
p
q
kuk2 1 .
|uk |2 λαp
|uk |2 λ−αq
. Therefore
k
k
k≥1
k≥1
D(A 2 )
P
m+1 P
m+1
2p
2q
2 αp
2 −αq
kukm+1
.
.
H
k≥1 |uk | λk
k≥1 |uk | λk
The 2d estimate follows by choosing p =
m+1
p
= m,
m+1
q
m+1
,
m
q = m + 1, α =
2
p
since
= 1, αp = 2, −αq = −2m.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pol. stab.
(u0 , u1 , f 0 (−τ ·)) ∈ D(A) 6⇒ u0 ∈ D(A), we can not use standard interpolation
inequalities from Le 5. Therefore we need to make the following hypo.:
∃m, C > 0∀(u0 , u1 , z) ∈ D(A) :
ku0 km+1
≤ C k(u0 , u1 , z)km
D(A) ku0 k
V
Serge NICAISE
D(A
1−m
2
.
(29)
)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pol. stab. ctd
Theorem
Let u sol. of (1) with (u0 , u1 , f 0 (−τ ·)) ∈ D(A). Assume that (14), (21) and
(29) are verified for some m > 0. If ∃ a time T > 0 and C > 0 ind. of τ s. t.
Z T
k(B1∗ φ)0 (t)k2U1 dt
(30)
ku0 k2 1−m + ku1 k2 − m2 ≤ C
D(A
2
D(A
)
)
0
holds where φ is sol. of (18), then the energy decays polynomially, i.e., ∃C > 0
depending on m and τ s. t.
E (t) ≤
C
1
(1 + t) m
k(u0 , u1 , f 0 (−τ ·))k2D(A) , ∀t > 0.
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(31)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
As the hypothesis (21) is satisfied for B = (B1 B2 ), U = U1 × U2 , by using
Lemma
2, we obtain
RT
∗
2
∗
2
−2βT
(kB
u
(ku0 k2 1−m + ku1 k2 − m ).
t (t)kU1 + kB2 ut (t − τ )kU2 )dt ≥ Ce
1
0
D(A
2
)
D(A
2
)
Now fix T ∗ large enough s.t. T ∗ ≥ max(T , τ ). Then by Rk 2, we obtain
R T∗
E (0) − E (T ∗ ) ≥ C 0 (kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 )dt
R T∗
≥ C2 0 (kB1∗ ut (t)k2U1 + kB2∗ ut (t − τ )k2U2 )dt
R
T∗
+ C2 0 kB2∗ ut (t − τ )k2U2 dt
R1
∗
≥ Ce −2βT (ku0 k2 1−m + ku1 k2 − m + τ 0 kB2∗ ut (−τ ρ)k2U2 dρ),
D(A
2
)
D(A
2
)
by change of variable (because T ∗ > τ ).
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Therefore ∃K1 > 0 depending on T ∗ and τ s. t.
E (T ∗ ) ≤ E (0)
R1
−K1 (ku0 k2 1−m + ku1 k2 − m + τ ξ 0 kB2∗ ut (−τ ρ)k2U2 dρ),
D(A
2
)
D(A
2
)
for some K1 > 0 independent of T ∗ and τ . As B2∗ ut (−τ ρ) = f 0 (−τ ρ), for all
ρ ∈ (0, 1), this is equivalent to
E (T ∗ ) ≤ E (0)
−K1 (ku0 k
(32)
2
D(A
1−m
2
+
)
ku1 k2 − m2
D(A
)
+ τ ξkf
0
(−τ ·)k2L2 (U2 ) ),
where k · kL2 (U2 ) := k · kL2 ((0, 1), U2 ) .
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Outline of the talk
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Now, by (29), Le 5 and a convexity inequality, we have (recalling that
U0 = (u0 , u1 , f 0 (−τ ·)))
m+1
k(u0 , u1 )km+1
+ ku1 km+1
)
V ×H ≤C (ku0 kV
H
m
≤C (kU0 kD(A) ku0 k 1−m + ku1 km 1 ku1 kD(A− m2 ) )
≤C kU0 km
D(A) (ku0 k
D(A
2
1−m
D(A 2
)
)
D(A 2 )
+ ku1 kD(A− m2 ) ).
This implies
m
k(u0 , u1 )km+1
V ×H ≤C Ẽ (0) (ku0 k
D(A
1−m
2
)
+ ku1 kD(A− m2 ) ),
(33)
where we have introduced the modified energy
Ẽ (t) =
1
1
kU(t)k2D(A) = (kU(t)k2H + kAU(t)k2H ).
2
2
As in Proposition 1, this energy Ẽ is decaying.
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Outline of the talk
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Combining the estimates (32) and (33), we obtain
E (T ∗ ) ≤ E (0) − K2 (
k(u0 , u1 )k2m+2
V ×H
Ẽ (0)m
+ ξτ kf 0 (−τ ·)k2L2 (U2 ) ),
(34)
for some K2 > 0 depending on T ∗ and τ . Using the trivial est.
(ξτ )m+1 kf 0 (−τ ·)k2m+2
= ξτ kf 0 (−τ ·)k2L2 (U2 )
L2 (U )
2
0
2
m
·(ξτ )m kf 0 (−τ ·)k2m
L2 (U2 ) ≤ τ ξkf (−τ ·)kL2 (U2 ) Ẽ (0) ,
the above inequality (34) becomes
m+1
k(u0 , u1 )k2m+2
kf 0 (−τ ·)k2m+2
V ×H + (ξτ )
L2 (U2 )
)
E (T ∗ )≤E (0) − K2 (
m
Ẽ (0)
m+1
E (0)
≤E (0) − K 0
,
Ẽ (0)m
0
∗
with K > 0 depending on T and τ .
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Stabilization of second order evolution equations with feedback
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Since the energy of our system is decaying, we have proved that
E (T ∗ ) ≤ E (0) − K 0
E (T ∗ )m+1
.
Ẽ (0)m
(35)
The estimate (35) being valid on the intervals [kT ∗ , (k + 1)T ∗ ], for any k ≥ 0,
we have
E ((k + 1)T ∗ )m+1
E ((k + 1)T ∗ ) ≤ E (kT ∗ ) − K 0
.
(36)
Ẽ (kT ∗ )m
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The energy
Exp. stab.
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Some Examples
Pf end
Setting εk =
E (kT ∗ )
,
Ẽ (0)
and dividing (36) by Ẽ (0), we obtain
εk+1 ≤ εk − K 0 εm+1
k+1 ,
(37)
because Ẽ (kT ∗ ) ≤ Ẽ (0). By Lemma 4 with µ = m − 1 > −1 (as m > 0),
there exists a constant M 0 > 0 s. t.
1
εk ≤ M 0 (1 + k)− m ∀k ≥ 0,
m
1
E (kT ∗ ) ≤ M 0 (1 + k)− m Ẽ (0).
This estimate and again the decay of the energy lead to the estimate (31),
1
where C = M 0 (1 + T ∗ ) m .
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
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Stabilization of second order evolution equations with feedback
Outline of the talk
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ingham’s ≤
Assume that the eigenvalues λk , k ∈ N∗ are simple and that the standard gap
condition holds:
∃γ0 > 0, ∀k ≥ 1, λk+1 − λk ≥ γ0 .
(38)
Theorem (Ingham 1936)
If the sequence (λn )n≥1 satisfies (38). Then for all sequence (an )n∈Z∗ (where
Z∗ = Z \ {0}), and setting λn = −λn , for all n < 0, the function
X
f (t) =
an e iλn t ,
n∈Z∗
satisfies the estimate
Z T
0
|f (t)|2 dt ∼
X
n∈Z∗
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|an |2 , for any T >
2π
.
γ0
(39)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
2d Ingham’s ≤
Now, let U be a separable Hilbert space (in the sequel, U will be U1 ). Then we
obtain the inequality of Ingham’s type in U:
Proposition
If we have the standard gap condition (38), then for all sequence (an )n in U,
the function
X
u(t) =
an e iλn t
n∈Z∗
satisfies the estimates
T
Z
ku(t)k2U dt ∼
0
for T >
X
kan k2U ,
(40)
n∈Z∗
2π
.
γ0
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
As U is a separable Hilbert space, there exists a Hilbert basis (ψk )k≥1 of U.
Therefore,
P ank ∈ U can be written as
an = +∞
k=1 an ψk .
K
X
(K )
We truncate an as follows: for K ∈ N∗ , let an =
ank ψk and set
k=1
K
X
X k iλ t
uK (t) =
(
an e kn )ψk . Since (ψk )k≥1 is a Hilbert basis, we have by
k=1 n∈Z∗
Fubini’s theorem
kuK (t)k2U
2
K X
X
k iλkn t =
an e
.
∗
k=1 n∈Z
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Thus, by applying Ingham’s inequality (39), we have
2
Z T
K Z T X
X
k iλkn t 2
an e
kuK (t)kU dt =
dt
0
0 ∗
n∈Z
k=1
∼
K X
X
(ank )2 ∼
k=1 n∈Z∗
Therefore
T
Z
kuK (t)k2U dt ∼
0
(K )
As uK → u and an
K
XX
(ank )2 .
n∈Z∗ k=1
X
(K ) 2
an .
n∈Z∗
U
→ an when K → +∞, we obtain the result.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Checking the 1st OE
Proposition
If (38) holds. Then (23) holds, namely
RT
1
kA 2 u0 k2H + ku1 k2H ≤ C 0 kB1∗ φt (t)k2U1 dt
iff
∃α > 0, ∀k ≥ 1, kB1∗ ϕk kU1 ≥ α.
(41)
√
Ex. 1 ϕ√
2 sin(kπ·)∀k ∈ N∗ :
k =
(41) ⇔ k 2 sin(kπ·)kL2 (0,1) ≥ α > 0∀k ∈ N∗ : always true since
√
k 2 sin(kπ·)kL2 (0,1) = 1.
√
Ex. 2 ϕk = 2 sin((k − 12 )π·)∀k ∈ N∗ :
(41) ⇔ | sin((k − 21 )π)| ≥ α > 0∀k ∈ N∗ : always true since
| sin((k − 12 )π)| = 1.
Ex. 3
Not exp. stable for any ξ ∈ (0, 1) because
6 ∃α > 0 : | sin(kπξ)| ≥ α
∀k.
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Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
(41)⇒(23):
Writing P
P
u0 = k≥1 ak ϕk , u1 = k≥1 bk ϕk
where (λk ak )k , (bk )k ∈ l 2 (N∗ ), then the solution φ of system without damping
(18) is given by P
φ(·, t) = k≥1 ak cos(λk t) + λbkk sin(λk t) ϕk .
ConsequentlyP
(B1∗ φ)0 (t) = k≥1 (−ak λk sin(λk t) + bk cos(λk t)) B1∗ ϕk .
This is equivalent to
P
(B1∗ φ)0 (t) = k∈Z∗ αk e iλk t ,
where
αk = 21 (bk + iak λk )B1∗ ϕk , α−k = 12 (bk − iak λk )B1∗ ϕk .
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Stabilization of second order evolution equations with feedback
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Integrating the square of the norm of this identity between 0 and T > 0 and
using Ingham’s inequality (40) in U1 , for T large enough, we get
RT
P
k(B1∗ φ)0 (t)k2U1 dt ≥ C k∈Z∗ kαk k2U1
0
0P
2
2
2
∗
2
≥C
k∈N∗ (|bk | + |λk | |ak | )kB1 ϕk kU1 .
Hence using (41), we get
Z T
X
k(B1∗ φ)0 (t)k2U1 dt ≥ C 0 α
(|bk |2 + |λk |2 |ak |2 )
0
k∈N∗
≥
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C 00 (kA 2 u0 k2H + ku1 k2H ).
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Necessity
(41)⇐(23):
Let k ∈ N be fixed. Take u0 = ak ϕk and u1 = bk ϕk . Then the solution φ of
system (18) is given by
φ(·, t) = (ak cos(λk t) + λbkk sin(λk t))ϕk ,
and then
(B1∗ φ)0 (t) = (ak λk cos(λk t) + bk sin(λk t))B1∗ ϕk .
Applying
again Ingham’s inequality, we get for T large enough
RT
∗
0
k(B
φ)
(t)k2U1 dt ∼ kB1∗ ϕk k2U1 (|ak |2 |λk |2 + |bk |2 ). But (23) means here that
1
0
RT
1
2
2
|ak | |λk | + |bk |2 = kA 2 u0 k2H + ku1 k2H ≤ C 0 kB1∗ φt (t)k2U1 dt.
The conclusion follows from these two estimates.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Checking the 2d OE
Ingham’s ≤⇒
Proposition
Assume that the standard gap condition (38) holds. Then (30) holds, namely
RT
ku0 k2 1−m + ku1 k2 − m ≤ C 0 k(B1∗ φ)0 (t)k2U1 dt iff
D(A
2
)
D(A
2
)
∃α > 0, ∀k ≥ 1, kB1∗ ϕk kU1 ≥
Serge NICAISE
α
.
λm
k
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf
The proof is similar to the one of Proposition 4 because
X 1
(ak2 λ2k + bk2 )
λ2m
k
=
k≥1
X 2 2(1−m)
(ak λk
+ bk2 λ−2m
)
k
k≥1
∼
ku0 k2
D(A
Serge NICAISE
1−m
2
)
+ ku1 k2
D(A
−m
2
)
.
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Illustration
Denote by S the set of all irrational real numbers ρ ∈ (0, 1) such that if
[0, a1 , ..., an , ...] is the expansion of ρ as a continued fraction, then the
sequence (an )n is bounded. It is well known that S is uncountable and that its
Lebesgue measure is zero. In particular, by the Euler-Lagrange theorem, S
contains all irrational quadratic numbers (i.e. the roots of a second order
equation with rational coefficients). Roughly speaking, the set S contains all
irrational numbers which are badly approximated by rational numbers. More
precisely the next result holds cfr [Lang 1966]:
Proposition
ρ ∈ S iff ∃C > 0 s. t. minn∈Z |qρ − n| ≥
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C
q
∀q ≥ 1.
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Illustration ctd
Lemma
If ξ ∈ S, then ∃α > 0 s. t. |sin(kπξ)| ≥
α
k
∀k ≥ 1.
Pf. Prop 6 ⇒
minn∈Z |qπξ − nπ| ≥ Cqπ ∀q ≥ 1.
For a fixed q, denote by nq one index such that
|qπξ − nq π| = minn∈Z |qπξ − nπ|.
Since this minimum is ≤ π2 and the sin fct is ↑ on [0, π2 ], we get
| sin(qπξ)| = | sin(qπξ − nq π)| = sin |qπξ − nq π| ≥ sin Cqπ .
As sinx x is ↓ on [0, π2 ], we get
0
| sin(qπξ)| ≥ Cq , ∀q ≥ 4C .
This proves the lemma, since for 1 ≤ q ≤ 4C , this estimate is trivial.
Ex. 3
If ξ ∈ S, then (30) holds with m = 1, since
∃α > 0 : | sin(kπξ)| ≥
α
k
∀k ≥ 1.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Outline of the talk
1
The Problem
2
Well-posedness
3
The energy
4
Exp. stab.
5
Pol. stab.
6
Ingham’s ineq.
7
Some Examples
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Some Examples
Ex 1: Bounded distributed dampings in 1d,
Ex 2: Bounded distributed dampings in Rn ,
Ex 3: Non bounded distributed dampings in 1d.
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Stabilization of second order evolution equations with feedback
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 1: Distributed dampings











∂2u
∂t 2
∂2u
∂x 2
+ α1 ∂u
(x, t)χ|I1
∂t
t − τ )χ|I2 = 0 in (0, 1) × (0, ∞)
u(0, t) = u(1, t) = 0 t > 0
u(x, 0) = u0 (x), ∂u
(x, 0) = u1 (x) in (0, 1)
∂t
0
∂u
(x,
t
−
τ
)
=
f
(x,
t − τ ) in I2 × (0, τ ),
∂t
−
+α2 ∂u
(x,
∂t
(42)
where χ|I = characteristic fct of I .
We assume that 0 < α2 < α1 , τ > 0 and
I2 ⊂ I1 ⊂ [0, 1],
as well as
∃δ ∈ [0, 1], > 0 : [δ, δ + ] ⊂ I1 .
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Outline of the talk
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 1 Wellposedness
H = L2 (0, 1), V = H01 (0, 1), D(A) = H01 (0, 1) ∩ H 2 (0, 1),
d2
A : D(A) → H : ϕ 7→ − dx
2ϕ
√
Ui = L2 (Ii ), Bi : Ui → H ⊂ V 0 : k 7→ αi k̃χ|Ii ,
where k̃ = extension of k by zero outside Ii .
√
Hence Bi∗ (ϕ) = αi ϕ|Ii ⇒ Bi Bi∗ (ϕ) = αi ϕχ|Ii ∀ϕ ∈ V ,
R
kBi∗ ϕk2Ui = αi I |ϕ|2 dx.
i
Therefore, we notice
that (7) is equivalent
to
R
R
∃0 < α ≤ 1, α2 I2 |ϕ|2 dx ≤ αα1 I1 |ϕ|2 dx, and consequently, this system is
well posed for α2 ≤ α1 by Theorem 1, and the energy is decreasing for α2 < α1
by Proposition 1.
Thm 3 and Prop 4 ⇒ problem (42) is exponentially stable.
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Stabilization of second order evolution equations with feedback
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 1 Obs. est.
√
The eigenvectors of the operator A are ϕk (x) = 2 sin(kπx) of eigenvalue
2
(kπ) , k ≥ 1 of multiplicity 1. Hence, the standard gap condition (38) is
verified.
Lemma
Under the previous hypo., (41) holds.
Pf. From the definition of B1∗ we have
R
kB1∗ ϕk k2U1 = α1 I1 |ϕk (x)|2 dx
h
R δ+
≥ 2α1 δ |sin(kπx)|2 dx = α1 x −
1
≥ α1 − kπ
≥ α1 2 ,
sin(2kπx)
2kπ
iδ+
δ
2
for k ≥ E( π
) + 1 =: k , where E(x) is the entire part of x. This leads to the
conclusion because for k ∈ {1, ..., k }, we have
R δ+
|sin(kπx)|2 dx > 0.
δ
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 1 AT’s cond.
Lemma
The operators A and B = (B1 B2 ) ∈ L(U, V 0 ) where U = U1 × U2 satisfy the
assumption (21).
Pf. Let i ∈ {1, 2} and ϕ ∈ L2 (Ii ). It can be easily checked that
v = (λ2 + A)−1 Bi ϕ ∈ H01 (0, 1) satisfies
2
√
λ2 v − ddx v2 = αi ϕ̃χ|Ii .
Writing
∞
X
√
α
v=
ck ϕk , ck are given by ck = λ2 +λi 2 (ϕ̃, ϕk ) ,
k
k=1
Hence
kλv k2L2 (0, 1) =
∞ X
λ 2
2
λ2 +λ2 αi |(ϕ̃, ϕk )| .
k=1
k
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf ctd
Now, if λ = β + iy , with β > 0, y ∈ R, then ∃C (β) > 0 (depending only on β):
2
2
λ 2
λ2 +λ2 = (β 2 −y 2β+λ+y
2 )2 +4β 2 y 2 ≤ C (β).
k
k
Therefore, we have
∞
X
kλv k2L2 (0, 1) ≤ C (β)αi
|(ϕ̃, ϕk )|2 ≤ C (β)αi kϕk2L2 (Ii ) ,
k=1
which leads to
kλBj∗ v k2L2 (Ij ) ≤ αj kλv k2L2 (0, 1) ≤ C (β)αi αj kϕk2L2 (Ii ) , ∀j ∈ {1, 2} .
Consequently, the operator λ → λBj∗ (λ2 I + A)−1 Bi is bounded on Cβ and the
lemma is proved.
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Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 2: Distributed dampings
Let Ω ⊂ Rn , n ≥ 1, with a C 2 bdy Γ. We assume that Γ = ΓD ∪ ΓN , with
Γ̄D ∩ Γ̄N = ∅ and ΓD 6= ∅. ∃x0 ∈ Rn is s. t. (x − x0 ) · ν(x) ≤ 0, ∀x ∈ ΓD .
Let O2 ⊂ O1 ⊂ Ω s. t. ΓN ⊂ ∂O1 .

∂2u
− ∆u + α1 ∂u
(x, t)χ|O1


∂t
∂t 2



+α2 ∂u
(x,
t
−
τ
)χ
in Ω × (0, ∞),
|O2 = 0
∂t
(43)
u(x, t) = 0 on ΓD × (0, ∞),

∂u


(x,
t)
=
0
on
Γ
×
(0,
∞),
N

∂ν

I .C .,
0 < α2 < α1 , τ > 0.
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Stabilization of second order evolution equations with feedback
Outline of the talk
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The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 2 Wellposedness
H = L2 (Ω), V = HΓ1D (Ω), A : D(A) → H : ϕ 7→ −∆ϕ
√
Ui = L2 (Oi ), Bi : Ui → H ⊂ V 0 : k 7→
αi k̃χ|0i .
R
√
Bi∗ (ϕ) = αi ϕ|Oi . ⇒ kBi∗ ϕk2Ui = αi O |ϕ|2 dx,
i
and therefore (7)R is equivalent to R
2
∃0 < α ≤ 1, α2 O2 |ϕ| dx ≤ αα1 O1 |ϕ|2 dx.
Consequently, (43) is well posed for α2 ≤ α1 by Thm 1, and the energy is
decreasing for α2 < α1 by Prop 1.
The obs. est. (23) was proved in [Lasiecka-Triggiani-Yao 99] and (21) is
proved like in Le 9 ⇒ the pb is exponentially stable.
Remark
Generalization of [N.-Pignotti 2006] where O2 = O1 .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 3: Non bd int feedback
We come back to the Example 3 from section 1. As seen before we cannot
expect to have exponential decay.
To get the pol. decay (see Thm 6), we need to check that (14), (21) and (29)
and (30) are verified for some m ∈ N∗ .
(14) ok if α2 < α1 .
(30) with m = 1 ok if ξ ∈ S (see Le 8).
Hence if we check (21) and (29), the energy will decay as t −1 .
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 3: Checking (21)
For i ∈ {1, 2} and k ∈ Ui = R. It can be easily checked that
v = (λ2 + A)−1 Bi k ∈ H01 (0, 1) satisfies
2
λ2 v − ddx v2 = 0 in (0, 1) \ {ξ}, vx (ξ+) − vx (ξ−) = k. Hence
direct calculations yield
v (x) = c1 sinh(λx) if x < ξ, v (x) = d1 sinh(λ(1 − x)) if x > ξ,
with c1 = k sinh(λ(1−ξ))
, d1 = k sinh(λξ)
. Hence
sinh λ
sinh λ
sinh(λ(1−ξ))
λv (ξ) = k sinh(λξ)sinh
.
λ
For
λ
=
β
+
iy
,
with
β
>
0
and
y
∈
R,
∃C
(β) > 0 (depending only on β) s. t.
sinh(λξ) sinh(λ(1−ξ)) ≤
C
(β).
This
follows
from
the fact that
sinh λ
| sinh λ|2 = sinh2 β cos2 y + cosh2 β sin2 y
= − cos2 y + cosh2 β ≥ cosh2 β − 1 > 0.
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Ex 3: Checking (29)
2
For (u, v , z) ∈ D(A), there exists
R 1 f ∈ H = L (0, 1) s.t. √
∗
Au + B1 B1 v + B2 z(1) = f ⇔ 0 ux wx dx + (α1 v (ξ) + α2 z(1))w (ξ) =
R1
fw dx ∀w ∈ H01 (0, 1).
0
Hence by Green’s formula we find
√
−uxx = f in (0, 1) \ {ξ} and vx (ξ+) − vx (ξ−) = −(α1 v (ξ) + α2 z(1)).
We then introduce the Hilbert space X defined by
X = {u ∈ H01 (0, 1) : uxx ∈ L2 (0, ξ) and uxx ∈ L2 (ξ, 1)}, with norm
kuk2X = kuk2H 1 (0,1) + kuxx k2L2 (0,ξ) + kuxx k2L2 (ξ,1) .
0
From the previous consideration
k(u, v , z)kD(A) & kukX .
Hence (29) holds with m = 1 if
kuk2V . kukX kukH , ∀u ∈ X .
Serge NICAISE
(44)
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf of (44)
Let u ∈ X arbitrary. Fix a cut-off function η ∈ D(R) s. t.
η = 1 on (ξ − , ξ + ), η = 0 on [0, ] ∪ [1 − , 1], with .
1. (1 − η)u ∈ D(A) and therefore
k(1 − η)uk2V . k(1 − η)ukD(A) k(1 − η)ukH . kukX kukH .
2. Set v (y ) = η(y − ξ)ũ(y − ξ), ∀y ∈ R.
Then we introduce the following extension of v1 := v|R+ (resp. v2 := v|R− ) as
follows:
Ev1 (y ) = v1 (y ), ∀y > 0, Ev1 (y ) = ν0 v1 (−y ) + ν1 v1 (−y ), ∀y < 0
Ev2 (y ) = v2 (y ), ∀y < 0, Ev2 (y ) = ν0 v (−y ) + ν1 v (−y ), ∀y > 0
with ν0 and ν1 s. t.
ν0 + ν1 = 1,
ν0 + 2ν1 = −1
Serge NICAISE
Stabilization of second order evolution equations with feedback
Outline of the talk
The Problem
Well-posedness
The energy
Exp. stab.
Pol. stab.
Ingham’s ineq.
Some Examples
Pf of (44) ctd
In that way: Evi ∈ D(Ã), i = 1, 2, where
D(Ã) = H01 (−1, 1) ∩ H 2 (−1, 1). Therefore Le 5 ⇒
kEvi k2H 1 (−1,1) . kEvi kD(Ã) kEvi kL2 (−1,1) .
In particular we get
kv1 k2H 1 (0,1)
.
kEv1 kD(Ã) kEv1 kL2 (−1,1)
.
(kv1 kH 1 (0,1) + kv1xx kL2 (0,1) )kv1 kL2 (0,1) ,
since |Ev1 |`,(−1,0) . |v1 |`,(0,1) . In the same manner, we have
kv2 k2H 1 (−1,0) . (kv2 kH 1 (−1,0) + kv2xx kL2 (−1,0) )kv2 kL2 (−1,0) .
These two estimates prove (44).
Serge NICAISE
Stabilization of second order evolution equations with feedback