Using saturated controls for the stabilization of the wave and KdV

Transcription

Using saturated controls for the stabilization of the
wave and KdV equations
Christophe PRIEUR
CNRS, Gipsa-lab, Grenoble, France
Groupe de Travail Contrôle,
UMPC, janvier 2016
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Given a PDE, there exists now a large variety on methods to
design linear controllers. It is well known that saturation can
reduce the performance or even destabilize the system, even for
finite-dimensional systems.
More precisely, even if
ż = Az + BKz
is asymp. stable, it may hold that
ż = Az + sat(BKz)
is not globally asymptotically stable.
It may exist new equilibrium, new limit cycles...
See e.g. [Khalil; 1996] and [Tarbouriech, et al; 2011]
Goal of this talk:
What happens for the linear wave equation?
For the nonlinear KdV equation?
In presence of bounded and unbounded control operators?
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ż = Az + BKz
ż = Az + sat(BKz)
Goal of this talk:
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ż = Az + BKz
ż = Az + sat(BKz)
Goal of this talk:
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Two objectives
Well-posedness
Stability
of the wave and KdV equations in presence of a
distributed/boundary saturating control.
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Outline
1 Well-posedness and stability of linear wave equation with a
saturated in-domain control
Lyapunov method, LaSalle invariance principle
saturated boundary control
Lyapunov method, local sector condition
3 Well-posedness and stability of nonlinear KdV equation with a
Local sector condition, contradiction argument
4 Numerical simulations on nonlinear KdV equation and
nonlinear KdV equation + sat
5 Conclusion
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Outline
5 Conclusion
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Outline
5 Conclusion
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1 – Wave equation with a bounded control operator
x =0
x =1
f (x, t)
z(x, t)
1D wave equation with bounded control operator.
Dynamics of the vibration:
ztt (x, t) = zxx (x, t) + f (x, t), ∀x ∈ (0, 1), t ≥ 0,
(1)
Boundary conditions, ∀t ≥ 0,
z(0, t) = 0 ,
z(1, t) = 0 ,
(2)
and with the following initial condition, ∀x ∈ (0, 1),
z(x, 0) = z 0 (x) ,
zt (x, 0) = z 1 (x) ,
(3)
where z 0 and z 1 stand respectively for the initial deflection and the
initial deflection speed.
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When closing the loop with a linear control
Let us define the linear control by
f (x, t) = −azt (x, t), x ∈ (0, 1), ∀t ≥ 0,
and consider
1
V1 =
2
Z
(zx2 + zt2 )dx.
Formal computation. Along the solutions to (1), (2) and (4):
R1
V̇1 = 0 (zx zxt − azt2 + zt zxx )dx
R1
= − 0 azt2 dx + [zt zx ]x=1
x=0
R1
= − 0 azt2 dx
Thus, it a > 0, V1 is a (non strict) Lyapunov function.
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(4)
Using standard technics (Lumer-Philipps thereom (for the
well-posedness) and Huang-Prüss theorem (for the exp. stability)):
Proposition
∀a > 0, ∀(z 0 , z 1 ) in H01 (0, 1) × L2 (0, 1),
∃ ! solution z: [0, ∞) → H01 (0, 1) × L2 (0, 1) to (1)-(4). Moreover,
∃ C , µ > 0, such that, for any initial condition H01 (0, 1) × L2 (0, 1),
it holds, ∀t ≥ 0,
kzkH 1 (0,1) + kzt kL2 (0,1) ≤ Ce −µt (kz 0 kH 1 (0,1) + kz 1 kL2 (0,1) ).
0
0
In the previous proposition:
stability
attractivity of the equilibrium
with an exponential speed
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Proposition
∀a > 0, ∀(z 0 , z 1 ) in H01 (0, 1) × L2 (0, 1),
0
0
stability
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Proposition
∀a > 0, ∀(z 0 , z 1 ) in H01 (0, 1) × L2 (0, 1),
0
0
stability
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When closing the loop with a saturating control
Let us consider now the nonlinear control
f (x, t) = −sat(azt (x, t)), x ∈ (0, 1), ∀t ≥ 0,
(5)
where sat is the localized saturated map:
sat(σ)
sat(σ) =
σ
σ
sign(σ)
if |σ| < 1
else
Equation (1) in closed loop with the control (5) becomes
ztt
= zxx − sat(azt )
A formal computation gives, along the solutions to (6) and (2),
Z 1
V̇1 = −
zt sat(azt )dx
0
which asks to handle the nonlinearity zt sat(azt ).
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(6)
Remark: Choice of the saturation map
[Slemrod, 1989] and [Lasiecka and Seidman, 2003] deal with
L2 saturation:
Given σ : [0, 1](→ R, sat2 (σ) is the function defined by
σ(x)
if kσkL2 (0,1) < 1
sat2 (σ)(x) =
σ(x)
else
kσk 2
L (0,1)
Here we consider localized saturation which is more physically
relevant:
σ(x)
if |σ(x)| < 1
sat(σ(x)) =
sign(σ(x)) else
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Well-posedness of this nonlinear PDE
Theorem 1 [CP, Tarbouriech, Gomes da Silva Jr; 2015]
∀a ≥ 0, for all (z 0 , z 1 ) in (H 2 (0, 1) ∩ H01 (0, 1)) × H01 (0, 1), there
exists a unique solution z: [0, ∞) → H 2 (0, 1) ∩ H01 (0, 1) to (6)
with the boundary conditions (2) and the initial condition (3).
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Consider
A1
u
v
=
uxx
v
− sat(av )
with the domain D(A1 ) = (H 2 (0, 1) ∩ H01 (0, 1)) × H01 (0, 1).
Let us use a generalization of Lumer-Phillips theorem which is the
so-called Crandall-Liggett theorem, as given in [Barbu; 1976]. See
also [Brezis; 1973] and [Miyadera; 1992].
Again two conditions
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1
A1 is dissipative, that is
u
ũ
u
ũ
< hA1
− A1
,
−
i ≤0
v
ṽ
v
ṽ
2
For all λ > 0, D(A1 ) ⊂ Ran(I − λA1 )
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First item: Easy step!
Instead
of proving
u
ũ
u
ũ
< hA1
− A1
,
−
i ≤ 0, let us
v
ṽ
v
ṽ
u
check, for all
∈ H01 (0, 1) × L2 (0, 1):
v
u
u
< hA1
,
i ≤0
v
v
To do that, using the definition of A1 , and of the scalar product in
H01 (0, 1) × L2 (0, 1), it is equal to:
R1
vx (x)ux (x)dx + 0 (uxx (x) − sat(a v (x)))v (x)dx ,
R1
R1
R1
= 0 vx (x)ux (x)dx + 0 uxx (x)v (x)dx − 0 sat(a v (x))v (x)dx
R1
= [ux (x)v (x)]x=1
x=0 − 0 sat(a v (x))v (x)dx ≤ 0
R1
0
due to the boundary and since a ≥ 0.
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Second
item
asks to deal with a nonlinear ODE. u
ũ
Let
∈ H01 (0, 1) × L2 (0, 1) we have to find
∈ D(A1 )
v
ṽ
such that
ũ
u
(I − λA1 )
=
ṽ
v
that is
ũ − λṽ = u ,
ṽ − λ(ũxx − sat(a ṽ )) = v ,
In particular, we have to find ũ such that
a
1
1
1
ũ − sat( (ũ − u)) = − v − 2 u
2
λ
λ
λ
λ
ũ(0) = ũ(1) = 0
ũxx −
holds.
Nonhomogeneous nonlinear ODE with two boundary conditions
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Lemma
If a is nonnegative and λ is positive, then there exists ũ solution to
ũxx −
1
ũ
λ2
− sat( λa (ũ − u)) = − λ1 v −
ũ(0) = ũ(1) = 0
1
u
λ2
To prove this lemma, let us introduce the following map
T : L2 (0, 1) → L2 (0, 1) ,
y 7→ z = T (y ) ,
where z = T (y ) is the unique solution to
zxx −
1
z
λ2
= − λ1 v − λ12 u + sat( λa (y − u)) ,
z(0) = z(1) = 0 .
Prove that T is well defined and apply the Schauder fixed-point
theorem (see e.g., [Coron; 2007]), to deduce that there exists y
such that T (y ) = y
ũ = y solves (7)
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(7)
Lemma
If a is nonnegative and λ is positive, then there exists ũ solution to
ũxx −
1
ũ
λ2
− sat( λa (ũ − u)) = − λ1 v −
ũ(0) = ũ(1) = 0
1
u
λ2
To prove this lemma, let us introduce the following map
T : L2 (0, 1) → L2 (0, 1) ,
y 7→ z = T (y ) ,
where z = T (y ) is the unique solution to
zxx −
1
z
λ2
= − λ1 v − λ12 u + sat( λa (y − u)) ,
z(0) = z(1) = 0 .
Prove that T is well defined and apply the Schauder fixed-point
theorem (see e.g., [Coron; 2007]), to deduce that there exists y
such that T (y ) = y
ũ = y solves (7)
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(7)
Global asymptotic stability of this nonlinear PDE
Theorem 2
∀a > 0, for all (z 0 , z 1 ) in (H 2 (0, 1) ∩ H01 (0, 1)) × H01 (0, 1), the
solution to (6) with the boundary conditions (2) and the initial
condition (3) satisfies the following stability property, ∀t ≥ 0,
kz(., t)kH 1 (0,1) + kzt (., t)kL2 (0,1) ≤ kz 0 kH 1 (0,1) + kz 1 kL2 (0,1) ,
0
0
together with the attractivity property
kz(., t)kH 1 (0,1) + kzt (., t)kL2 (0,1) → 0, as t → ∞ .
0
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Due to Theorem 1, the formal computation
Z 1
V̇1 = −
zt sat(azt )dx
0
makes sense. This is only a weak Lyapunov function V̇1 ≤ 0
(the state is (z, zt ), and there is no −z 2 ).
To be able to apply LaSalle’s Invariance Principle, we have to
check that the trajectories are precompact
(see e.g. [Dafermos, Slemrod; 1973], [d’Andréa-Novel et al; 1994]).
It comes from:
Lemma
The canonical embedding from D(A1 ), equipped with the graph
norm, into H01 (0, 1) × L2 (0, 1) is compact.
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Sketch of the proof of
The canonical embedding from D(A1 ), equipped with the graph
norm, into H01 (0, 1) × L2 (0, 1) is compact.
un
Consider a sequence
in D(A1 ), which is bounded with
vn n∈N
the graph norm, that is ∃M > 0, ∀n ∈ N,
2
un 2 un 2
un ,
+ A1
:= vn vn vn D(A1 )
Z 1
2
2
=
(un0 + |vn |2 + vn0 0
2
+ un00 − asat(vn ) )dx < M
R1
From that, we deduce that 0 (|vn |2 + |vn0 |2 )dx and
R1 0 2
00 2
0 (|un | + |un | )dx are bounded.
Thus there exists a subsequence which converges in
H01 (0, 1) × L2 (0, 1).
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Using the dissipativity
of A1 , and previous lemma the trajectory
z(., t)
is precompact in H01 (0, 1) × L2 (0, 1).
zt (., t)
z(., 0)
Moreover the ω-limit set ω
⊂ D(A1 ), is not empty
zt (., 0)
and invariant with respect to the nonlinear semigroup T (t) (see
[Slemrod; 1989]).
We
invariance principle to show that
now use LaSalle’s
z(., 0)
ω
= {0}.
zt (., 0)
Therefore the convergence property holds.
For linear PDE, we have exponential convergence
(see Proposition on Slide 8).
Do we have exp. stability for the nonlinear PDE?
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Using the dissipativity
of A1 , and previous lemma the trajectory
z(., t)
is precompact in H01 (0, 1) × L2 (0, 1).
zt (., t)
z(., 0)
Moreover the ω-limit set ω
⊂ D(A1 ), is not empty
zt (., 0)
and invariant with respect to the nonlinear semigroup T (t) (see
[Slemrod; 1989]).
We
invariance principle to show that
now use LaSalle’s
z(., 0)
ω
= {0}.
zt (., 0)
Therefore the convergence property holds.
For linear PDE, we have exponential convergence
(see Proposition on Slide 8).
Do we have exp. stability for the nonlinear PDE?
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2 – Wave equation with a boundary control
g (t)
x =0
x =1
z(x, t)
1D wave equation with a boundary control.
Dynamics:
ztt (x, t) = zxx (x, t), ∀x ∈ (0, 1), t ≥ 0,
(8)
Boundary conditions, ∀t ≥ 0,
z(0, t) = 0 ,
zx (1, t) = g (t) ,
(9)
and with the same initial condition, ∀x ∈ (0, 1),
z(x, 0) = z 0 (x) ,
zt (x, 0) = z 1 (x) .
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(10)
When closing the loop with a linear boundary control
g (t) = −bzt (1, t), x ∈ (0, 1), ∀t ≥ 0
(11)
and consider
1
V2 =
2
Z
(e
µx
2
(zt + zx ) dx +
Z
(e −µx (zt − zx )2 dx,
V̇2 = −µV2 + 12 e µ (1 − b)2 − e −µ (1 + b)2 zt2 (1, t)
Assuming b > 0 and letting µ > 0 such that
e µ (1 − b)2 ≤ e −µ (1 + b)2 , it holds V̇2 ≤ −µV2 and thus V2 is a
strict Lyapunov function and thus (8)-(11) is exponentially stable.
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When closing the loop with a linear boundary control
g (t) = −bzt (1, t), x ∈ (0, 1), ∀t ≥ 0
(11)
and consider
1
V2 =
2
Z
(e
µx
2
(zt + zx ) dx +
Z
(e −µx (zt − zx )2 dx,
V̇2 = −µV2 + 12 e µ (1 − b)2 − e −µ (1 + b)2 zt2 (1, t)
Assuming b > 0 and letting µ > 0 such that
e µ (1 − b)2 ≤ e −µ (1 + b)2 , it holds V̇2 ≤ −µV2 and thus V2 is a
strict Lyapunov function and thus (8)-(11) is exponentially stable.
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When closing the loop with a saturating control
Let us consider now the nonlinear control
g (t) = −sat(bzt (1, t)), ∀t ≥ 0. The boundary conditions become:
z(0, t) = 0 ,
zx (1, t) = −sat(bzt (1, t)) .
(12)
Theorem 3
∀b > 0, for all (z 0 , z 1 ) in {(u, v ), (u, v ) ∈
1 (0, 1), u (1) + sat(bv (1)) = 0, u(0) = 0}, the
H 2 (0, 1) × H(0)
x
solution to (8) with the boundary conditions (12) and the initial
condition (3) satisfies the following stability property, ∀t ≥ 0,
kz(., t)kH 1
(0)
(0,1)
+ kzt (., t)kL2 (0,1) ≤ kz 0 kH 1
(0)
(0,1)
+ kz 1 kL2 (0,1) ,
together with the attractivity property
kz(., t)kH 1
(0)
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(0,1)
+ kzt (., t)kL2 (0,1) → 0, as t → ∞ .
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To prove the well-posedness of the Cauchy problem we prove that
A2 defined by
u
v
A2
=
v
u 00
with the domain D(A2 ) = {(u, v ), (u, v ) ∈
1 (0, 1), u 0 (1) + sat(bv (1)) = 0, u(0) = 0} is a
H 2 (0, 1) × H(0)
semigroup of contraction.
The global stability property comes directly from the dissipativity
of A2 .
The global attractivity property comes from the following lemma:
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To prove the well-posedness of the Cauchy problem we prove that
A2 defined by
u
v
A2
=
v
u 00
with the domain D(A2 ) = {(u, v ), (u, v ) ∈
1 (0, 1), u 0 (1) + sat(bv (1)) = 0, u(0) = 0} is a
H 2 (0, 1) × H(0)
semigroup of contraction.
The global stability property comes directly from the dissipativity
of A2 .
The global attractivity property comes from the following lemma:
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Lemma (semi-global exponential stability)
For all r > 0, there exists µ > 0 such that, for all initial condition
satisfying
kz 000 k2L2 (0,1) + kz 1 k2H 1 (0,1) ≤ r 2 ,
(13)
(0)
it holds
V̇2 ≤ −µV2
along the solution to (8) with the boundary conditions (12).
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Sketch of the proof of this lemma
First note that by dissipativity of A2 , it holds that
z(.,
t)
t 7→ A2 zt (., t) is a non-increasing function. Thus, for all t ≥ 0,
z(., 0) A
|zt (1, t)| ≤ 2 zt (., 0) .
(14)
Now for all initial conditions satisfying (13), there exists c 6= b
such that, for all t ≥ 0,
(b − c)|zt (1, t)| ≤ 1
and thus the following local sector condition holds:
sat(bσ)
bσ
Letting σ = zt (1, t), it holds
(sat(bσ) − bσ)(sat(bσ) − (b − c)σ) ≤ 0
b
|b−c|
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We come back to the Lyapunov function candidate V2 . Given
b > 0, using the previous inequality, we compute
2
−µ
2
V̇2 = −µV2 + e µ (σ − sat(bσ))
µ −−µe 2(σ + sat(bσ))
µ
−µ
>
σ
e −e
− b (b − c)
−e − e
+ b + b(b − c)
≤ −µV2 + sat(bσ)
µ
−µ
µ
−µ
−e − e
+ b + b(b − c)
−1 + e − e
σ
× sat(bσ)
≤ −µV2
with a suitable choice of constant values µ and c.
The semi-global exponential stability follows.
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2
−µ
2
V̇2 = −µV2 + e µ (σ − sat(bσ))
µ
−µ
>
σ
e −e
− b (b − c)
−e − e
+ b + b(b − c)
µ
−µ
µ
−µ
−e − e
+ b + b(b − c)
−1 + e − e
σ
× sat(bσ)
≤ −µV2
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2
−µ
2
V̇2 = −µV2 + e µ (σ − sat(bσ))
µ
−µ
>
σ
e −e
− b (b − c)
−e − e
+ b + b(b − c)
µ
−µ
µ
−µ
−e − e
+ b + b(b − c)
−1 + e − e
σ
× sat(bσ)
≤ −µV2
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3 – KdV equation
Let us consider the following nonlinear PDE


 zt + zx + zxxx + zzx + f = 0, x ∈ [0, L], t ≥ 0,
z(0, t) = z(L, t) = zx (L, t) = 0, t ≥ 0,


z(x, 0) = z 0 (x), x ∈ [0, L],
where z stands for the state and f for the control.
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(KdV)
As noted in [Rosier; 1997], if f = 0 and
( r
)
k 2 + kl + l 2 .
∗
L ∈ 2π
k, l ∈ N ,
3
then, there exist solutions of the linearized version of (KdV) for
which the L2 (0, L)-energy does not decay to zero. Consider, for
instance, L = 2π and z 0 = 1 − cos(x) for all x ∈ [0, L].
See [Coron, Crépeau; 2004], [Cerpa; 2007] and [Cerpa, Crépeau;
2009] for controllability properties with critical lengths.
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Saturating control for KdV
Given a > 0, let us consider the KdV equation controlled by a
saturated distributed control as follows


zt + zx + zxxx + zzx + sat(az) = 0,



 z(0, t) = z(L, t) = 0,
(KdVsat)

zx (L, t) = 0,



 z(x, 0) = z 0 (x).
where sat is the same saturation as before:
sat(σ)
sat(σ) =
σ
σ
sign(σ)
if |σ| < 1
else
In a work in progress with S. Marx, E. Cerpa, and V. Andrieu, we
consider also the other saturation sat2 :
Given σ : [0, L] → R, sat2 (σ) is the function defined by
(
σ(x)
if kσkL2 (0,L) < 1
σ(x)
sat2 (σ)(x) =
.
else
kσk
L2 (0,L)
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Theorem 4 [Marx, Cerpa, CP, Andrieu, 2016]: Well posedness
For any initial condition z 0 ∈ L2 (0, L), there exists a unique
solution z ∈ C ([0, T ]; L2 (0, L)) ∩ L2 (0, T ; H 1 (0, L)) to (KdVsat).
Sketch of proof
First we prove the existence of solution on [0, T 0 ] for T 0 small, as
done in [Rosier, Zhang; 2006] and [Chapouly; 2009].
Then we remove the smallness assumption on T 0 by proving an a
priori estimate.
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Sketch of proof
priori estimate.
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Sketch of proof
priori estimate.
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Theorem 5: Global asymptotic stability.
There exist µ and a class K function α : [0, ∞) → [0, ∞) such that
for any z 0 ∈ L2 (0, 1), any solution z to (KdVsat) satisfies, for all
t ≥ 0,
kz(t)kL2 (0,1) ≤ α(kz 0 kL2 (0,1) )e −µt .
In the previous result, we have stability and attractivity properties.
To prove this result, we follow the approach of [Rosier, Zhang;
2006] and [Cerpa; 2014].
To be more specific:
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Theorem 5: Global asymptotic stability.
There exist µ and a class K function α : [0, ∞) → [0, ∞) such that
for any z 0 ∈ L2 (0, 1), any solution z to (KdVsat) satisfies, for all
t ≥ 0,
kz(t)kL2 (0,1) ≤ α(kz 0 kL2 (0,1) )e −µt .
In the previous result, we have stability and attractivity properties.
To prove this result, we follow the approach of [Rosier, Zhang;
2006] and [Cerpa; 2014].
To be more specific:
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Lemma: semi-global exponential stability
Given r > 0, there exist positive values C and µ such that for all
initial condition z 0 satisfying kz 0 kL2 (0,L) ≤ r , it holds, for all t ≥ 0,
kz(t)kL2 (0,L) ≤ C kz 0 kL2 (0,L) e −µt .
To do that, the key result is:
∀T > 0, r > 0, ∃ C > 0 such that for any solution z to (KdVsat)
starting from z 0 ∈ L2 (0, L) with kz 0 kL2 (0,L) ≤ r , it holds
kz 0 k2L2 (0,L) ≤ C
Z
T
|zx (t, 0)|2 dt + 2
0
Z
0
T
Z
L
sat(az)zdtdx .
0
(15)
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C. Prieur
Paris, January 2016
Assume (15) for the time being. Then with
RT
kz(., T )k2L2 (0,L) = kz 0 k2L2 (0,L) − 0 |zx (0, t)|2 dt
RT RL
−2 0 0 sat(az)zdxdt
we get
kz(., kT )k2L2 (0,L) ≤ γ k kz 0 k2L2 (0,L)
∀k ≥ 0
where γ ∈ (0, 1). From the dissipativity property, we have
kz(., t)kL2 (0,L) ≤ kz(., kT )kL2 (0,L) for kT ≤ t ≤ (k + 1)T . Thus
we obtain, for all t ≥ 0,
kz(., t)k2L2 (0,L) ≤
log γ
1 0
kz kL2 (0,L) e T t
γ
We conclude the proof of the semi-global exponential stability.
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C. Prieur
Paris, January 2016
Assume (15) for the time being. Then with
RT
kz(., T )k2L2 (0,L) = kz 0 k2L2 (0,L) − 0 |zx (0, t)|2 dt
RT RL
−2 0 0 sat(az)zdxdt
we get
kz(., kT )k2L2 (0,L) ≤ γ k kz 0 k2L2 (0,L)
∀k ≥ 0
where γ ∈ (0, 1). From the dissipativity property, we have
kz(., t)kL2 (0,L) ≤ kz(., kT )kL2 (0,L) for kT ≤ t ≤ (k + 1)T . Thus
we obtain, for all t ≥ 0,
kz(., t)k2L2 (0,L) ≤
log γ
1 0
kz kL2 (0,L) e T t
γ
We conclude the proof of the semi-global exponential stability.
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C. Prieur
Paris, January 2016
Proof of the blue equation (15). By contradiction !
Assume that there exists a sequence of solution z n to (KdVsat) with
kz n (., 0)kL2 (0,L) ≤ r
(16)
and such that
kz n k2L2 (0,T ;L2 (0,L))
= +∞.
RT RL
|zxn (0, t)|2 dt + 2 0 0 sat(az n (x, t))z n (x, t)dtdx
0
(17)
By dissipativity property, ∃ β > 0,
Z T
sup kz n (., t)kL2 (0,L) ≤ r ,
sup
|z n (x, t)|2 dt ≤ β.
(18)
lim R
n→+∞ T
t∈[0,T ]
x∈[0,L]
0
n
o
Now let us define Ωi := t ∈ [0, T ], supx∈[0,L] |z(x, t)| > i ⊂ [0, T ].
We have
Z T
Z
n
2
β≥
sup |z (x, t)| dt ≥
sup |z n (x, t)|2 dt ≥ i 2 ν(Ωi ),
0
x∈[0,L]
Ωi x∈[0,L]
Therefore, weobtain forthe complementary set
ν(Ωci ) ≥ max T − iβ2 , 0 .
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C. Prieur
Paris, January 2016
Proof of the blue equation (15). By contradiction !
Assume that there exists a sequence of solution z n to (KdVsat) with
kz n (., 0)kL2 (0,L) ≤ r
(16)
and such that
kz n k2L2 (0,T ;L2 (0,L))
= +∞.
RT RL
|zxn (0, t)|2 dt + 2 0 0 sat(az n (x, t))z n (x, t)dtdx
0
(17)
By dissipativity property, ∃ β > 0,
Z T
sup kz n (., t)kL2 (0,L) ≤ r ,
sup
|z n (x, t)|2 dt ≤ β.
(18)
lim R
n→+∞ T
t∈[0,T ]
x∈[0,L]
0
n
o
Now let us define Ωi := t ∈ [0, T ], supx∈[0,L] |z(x, t)| > i ⊂ [0, T ].
We have
Z T
Z
n
2
β≥
sup |z (x, t)| dt ≥
sup |z n (x, t)|2 dt ≥ i 2 ν(Ωi ),
0
x∈[0,L]
Ωi x∈[0,L]
Therefore, weobtain forthe complementary set
ν(Ωci ) ≥ max T − iβ2 , 0 .
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C. Prieur
Paris, January 2016
Moreover, using again the local sector condition, we have, for a
suitable positive function k and ∀i ∈ N,
Z
T
Z
L
n
n
sat(az )z dtdx
0
Z
L
Z
n
=
0
Z
Z
0+
Ωci
C. Prieur
L
....
Ωci
0
Z
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Z
sat(az )z dtdx +
Ωi
≥
n
0
L
ak(i)(z n )2 dtdx.
0
Paris, January 2016
(19)
n
Let λn := kz n kL2 (0,T ;L2 (0,L)) and v n (x, t) = z (x,t)
λn . Due to (16),
up to extracting a subsequence, we may assume that λn → λ ≥ 0.
Due to (17) and (19), we have, for all i ∈ N
Z
0
T
|vxn (0, t)|2 dt
Z
Z
+2
Ωci
L
ak(i)(v n )2 dtdx → 0
(20)
0
Using Aubin-Lions lemma in [Simon, 1987]: {v n }n∈N converges
strongly in L2 (0, T ; L2 (0, L)). Thus, with (20), we have, for all
i ∈N
vx (0, t) = 0, ∀t ∈ (0, T ) and v (x, t) = 0, ∀x ∈ [0, L], ∀t ∈ Ωci .
S
c
We know that ν
i∈N Ωi = T . We get a contradiction with
kv kL2 (0,T ;L2 (0,L)) = 1.
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C. Prieur
Paris, January 2016
4 – Numerical simulations
To discretize (KdVsat), we follow the approach of [Pazoto,
Sepúlveda, Vera Villagrán; 2010], and solve, at each time-step, a
fixed point problem.
(No proof of convergence of the numerical scheme)
Consider the first critical length L = 2π and the initial condition
z 0 (x) = 1 − cos(x). The energy is constant along the linearized
KdV equation.
Consider the control f (x, t) = sat(a(x)z(x, t)) where
a(x) = 1[ 1 L, 2 L] (x).
3
3
(Our proof remains valid even for more general controls)
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C. Prieur
Paris, January 2016
KdV equation.
a(x) = 1[ 1 L, 2 L] (x).
3
3
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C. Prieur
Paris, January 2016
KdV equation.
a(x) = 1[ 1 L, 2 L] (x).
3
3
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C. Prieur
Paris, January 2016
Figure: Solution to (KdV) with
linear control
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Figure: Solution to (KdVsat)
with saturated control
C. Prieur
Paris, January 2016
Figure: Control f = sat(az)
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Figure: Time evolution of the
energy function kzk2L2 (0,L) for
(KdV) (· · · ) and for (KdVsat) (–)
C. Prieur
Paris, January 2016
5 – Conclusion, under investigation and open questions
Conclusion
Well-posedness and global asymptotic stability for the
nonlinear PDEs:
ztt = zxx − sat(azt )
ztt = zxx
z(0, t) = z(1, t) = 0
z(0, t) = 0, zx (1, t) = −sat(bz(1, t))
and also
zt + zx + zxxx + zzx + sat(a(x)z) = 0
z(0, t) = z(L, t) = zx (L, t) = 0
distributed/localized (localized/L2 ) saturating control
Lyapunov function has been used for the wave equation
non-constructive proof for the KdV equation
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C. Prieur
Paris, January 2016
5 – Conclusion, under investigation and open questions
Conclusion
Well-posedness and global asymptotic stability for the
nonlinear PDEs:
ztt = zxx − sat(azt )
ztt = zxx
z(0, t) = z(1, t) = 0
z(0, t) = 0, zx (1, t) = −sat(bz(1, t))
and also
(
zt + zx + zxxx + zzx + sat(2) (a(x)z) = 0
z(0, t) = z(L, t) = zx (L, t) = 0
distributed/localized (localized/L2 ) saturating control
Lyapunov function has been used for the wave equation
non-constructive proof for the KdV equation
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C. Prieur
Paris, January 2016
Under actual investigation
Under actual investigation #1
Boundary saturating control for KdV?
zt + zx + zxxx + zzx = 0
z(0, t) = sat(g (t)), z(L, t) = zx (L, t) = 0
In [Marx, Cerpa; 2014] an output feedback control has been
computed for the linearized KdV
zt + zx + zxxx = 0
y (t) = zxx (L, t) output
z(0, t) = F(y ), z(L, t) = zx (L, t) = 0
Could we compute a saturating output feedback control?
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C. Prieur
Paris, January 2016
Under actual investigation
Boundary saturating control for KdV?
zt + zx + zxxx + zzx = 0
z(0, t) = sat(g (t)), z(L, t) = zx (L, t) = 0
In [Marx, Cerpa; 2014] an output feedback control has been
computed for the linearized KdV
zt + zx + zxxx = 0
y (t) = zxx (L, t) output
z(0, t) = F(y ), z(L, t) = zx (L, t) = 0
Could we compute a saturating output feedback control?
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C. Prieur
Paris, January 2016
Open questions
Output feedback controller for KdV? Saturating controller?
Exponential convergence for the nonlinear dynamics?
Rapid stabilization, as in [Cerpa, Crépeau; 2009]?
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C. Prieur
Paris, January 2016

Using saturated controls for the stabilization of the wave and KdV

Transcription

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