Input to state stability of evolution equations
Birgit Jacob
Joint work with R. Nabiullin, J.R. Partington and F. Schwenninger
Linear systems with unbounded control operator B
ẋ(t) = Ax(t) + Bu(t),
x(0) = x0
I A generates a C0 -semigroup (etA )t≥0 on a Banach space X
I B : U → X−1 bounded
(We have: X ,→ X−1 )
Let x0 ∈ X and u ∈ L∞ (0, ∞; U). The (mild) solution
Z t
x(t) = etA x0 +
e(t−s)A Bu(s) ds
0
is in general only defined in X−1 .
(Σ)
Existence of mild solutions in X
ẋ(t) = Ax(t) + Bu(t),
B L∞ -admissible
:⇐⇒
x(t) ∈ X,
x(0) = x0
(Σ)
u ∈ L∞ , t > 0
Proposition
I B L∞ -admissible
⇐⇒
∀ t > 0 ∃ c(t) > 0
kx(t)kX ≤ ketA x0 kX + c(t)kukL∞
I B L∞ -adm. & (etA )t≥0 exp. stable
kx(t)kX ≤ Meωt kx0 k + ckukL∞
u ∈ L∞ , x0 ∈ X
⇐⇒
∃ M, c > 0, ω < 0
u ∈ L∞ , x0 ∈ X, t > 0
In the following we assume that B is L∞ -admissible.
Stability notions for inhomogeneous equations
ẋ(t) = Ax(t) + Bu(t),
x(0) = x0
(Σ)
Aim: Stability notions that consider initial conditions x0 and inputs u.
+
K = {µ ∈ C(R+
0 , R0 ) | µ strictly increasing, µ(0) = 0, unbounded}
Σ is L∞ -input-to-state stable (L∞ -ISS) :⇐⇒ ∃ β ∈ KK−1 , γ ∈ K
kx(t)k ≤ β(kx0 k, t) + γ(kukL∞ ),
u ∈ L∞ , x0 ∈ X, t > 0
Σ is L∞ -integral input-to-state stable (L∞ -iISS) :⇐⇒
∃β ∈ KK−1 , θ,µ ∈ K
Z
kx(t)k ≤ β(kx0 k, t) + θ
0
t
µ(ku(s)k)ds ,
u ∈ L∞ , x0 ∈ X, t > 0
Stability notions for inhomogeneous equations
ẋ(t) = Ax(t) + Bu(t),
x(0) = x0
(Σ)
+
K = {µ ∈ C(R+
0 , R0 ) | µ strictly increasing, µ(0) = 0, unbounded}
Σ is L∞ -input-to-state stable (L∞ -ISS) :⇐⇒ ∃ ω < 0, M ≥ 1, γ ∈ K
kx(t)k ≤ Meωt kx0 k + γ(kukL∞ ),
u ∈ L∞ , x0 ∈ X, t > 0
Σ is L∞ -integral input-to-state stable (L∞ -iISS) :⇐⇒
∃ ω < 0, M ≥ 1, θ,µ ∈ K
kx(t)k ≤ Meωt kx0 k + θ
Z
0
t
µ(ku(s)k)ds ,
u ∈ L∞ , x0 ∈ X, t > 0
L∞ -input-to-state stability
Σ is L∞ -ISS
⇐⇒
∃ ω < 0, M ≥ 1, γ ∈ K :
u ∈ L∞ , x0 ∈ X, t > 0
ωt
kx(t)k ≤ Me kx0 k + γ(kukL∞ ),
B is L∞ -adm. & (etA )t≥0 exp. stable
kx(t)kX ≤ Meωt kx0 k + ckukL∞ ,
⇐⇒ ∃ ω < 0, M ≥ 1, c > 0 :
u ∈ L∞ , x0 ∈ X, t > 0
Proposition
Σ
L∞ -ISS
⇐⇒
(etA )t≥0 exp. stable
&
B L∞ -admissible
L∞ -integral input-to-state stability
Recall
Σ is L∞ -iISS
⇐⇒ ∃ ω < 0, M ≥ 1, θ,µ ∈ K :
Z t
ωt
kx(t)k ≤ Me kx0 k + θ
µ(ku(s)k) ds
u ∈ L∞ , x0 ∈ X, t > 0
0
Proposition
Σ L∞ -iISS
=⇒
⇐⇒
Question:
(etA )t≥0 exp. stable & B L∞ -admissible
Σ L∞ -ISS
Σ L∞ -ISS
=⇒
Σ L∞ -iISS ?
Are Lp -spaces sufficient?
Σ is L∞ -iISS
⇐⇒ ∃ ω < 0, M ≥ 1, θ,µ ∈ K :
Z t
ωt
kx(t)k ≤ Me kx0 k + θ
µ(ku(s)k) ds
u ∈ L∞ , x0 ∈ X, t > 0
0
∃ ω < 0, M ≥ 1, p ≥ 1:
1/p
Z t
p
ωt
(ku(s)k) ds
kx(t)k ≤ Me kx0 k +
u ∈ L∞ , x0 ∈ X, t > 0,
0
=⇒ Σ is L∞ -iISS
Converse direction is in general false
Example: X = `2 , Aen := −2n en , U := R, B := (2n /n)n∈N .
Orlicz Spaces
Definition:
Φ : (0, ∞) → (0, ∞) is a Young-Function
Z s
Φ(s) =
φ(τ )dτ
with
if
0
I φ : (0, ∞) → (0, ∞) right-continuous & increasing,
I lims→0+ φ(s) = 0, lims→∞ φ(s) = ∞
Example:
es − s − 1, sp for p ∈ (1, ∞)
Definition (Orlicz Space)
k·kE
(0,t)
EΦ (0, t) := L∞ (0, t) Φ
with
Z t
kukEΦ (0,t) = inf{k > 0 |
Φ(k−1 ku(s)k)ds ≤ 1}
0
Orlicz Spaces
k·kE
(0,t)
with
EΦ (0, t) := L∞ (0, t) Φ
Z t
kukEΦ (0,t) = inf{k > 0 |
Φ(k−1 ku(s)k)ds ≤ 1}
0
I EΦ (0, t) is a Banach space.
I L∞ (0, t) ,→ EΦ (0, t) ,→ L1 (0, t).
I u ∈ L1 (0, t)
∃ Young-Function Φ : u ∈ EΦ (0, t).
Z t
Φ(k−1 ku(s)k)ds ≤ 1 ⇐⇒
ku(s)kp ds ≤ kp
0
0
Z t
p
1/p
kukEΦ (0,t) = ( ku(s)k ds) ,
EΦ (0, t) = Lp (0, t)
Let Φ(s) = sp :
=⇒
⇒
Z
t
0
L∞ -iISS = EΦ -ISS
Recall
Σ is L∞ -iISS
⇐⇒
∃ ω < 0, M ≥ 1, θ, µ ∈ K:
Z t
ωt
kx(t)k ≤ Me kx0 k + θ
µ(ku(s)k) ds , u ∈ L∞ , x0 ∈ X, t > 0
0
Definition
Σ is EΦ -ISS
⇐⇒
∃ ω < 0, M ≥ 1, γ ∈ K :
kx(t)k ≤ Meωt kx0 k + γ(kukEΦ ) ∀u ∈ EΦ , x0 ∈ X, t > 0
Theorem
Σ
L∞ -iISS
(J., Nabiullin, Partington, Schwenninger ’16)
⇐⇒
∃ Young-Function Φ : Σ EΦ -ISS
Continuous mild solutions
Recall:
B is L∞ -admissible
⇐⇒
∀ t > 0 ∃ c(t) > 0
kx(t)kX ≤ ketA x0 kX + c(t)kukL∞
Mild solution: x(t) = etA x0 +
Z
∀u ∈ L∞ , x0 ∈ X
t
e(t−s)A Bu(s) ds
0
Proposition
(J., Nabiullin, Partington, Schwenninger ’16)
I Σ has cont. mild solutions in X
I Σ L∞ -iISS
=⇒
=⇒
B is L∞ -adm.
Σ has continuous mild solutions in X
Open Problem of G. Weiss ’89
Does L∞ -adm. of B imply the continuity of all mild solutions in X?
What do we know so far?
Let (etA )t≥0 be exp. stable
L∞ -iISS
EΦ -ISS
L∞ -ISS
mild soln cont.
L∞ -adm.
Questions:
Σ L∞ -ISS
=⇒ Σ L∞ -iISS?
B L∞ -adm. =⇒ Σ has continuous mild solutions?
Parabolic diagonal systems
ẋ(t) = Ax(t) + Bu(t),
Definition:
x(0) = x0
Σ is called a parabolic, diagonal system
(Σ)
if
Aen = λn en
D(A) = {x : (λn xn ) ∈ `2 (N)}
λn
(en )n∈N ONB of X,
B ∈ L(C, X−1 )
Then APgenerates an exp. stable, analytic
C0 -semigroup and
−1 x ) ∈ `2 (N)
X−1 =
x
e
|
(λ
n n
n n n
P |bn |2
Remark: B ∈ L(C, X−1 ) =⇒ B = (bn ) ∈ X−1 ⇐⇒ n |λ
2 < ∞.
n|
Parabolic diagonal systems: Main result
Theorem
(J., Nabiullin, Partington, Schwenninger ’16)
Σ diagonal, parabolic system
=⇒
Σ L∞ -iISS
Remark
Let Σ be a diagonal, parabolic system. Then
(i) Σ is L∞ -admissible
(ii) Σ is L∞ -iISS
(iii) Σ is L∞ -ISS
(iv) Mild solutions are continuous.
Remark: Result also hold for:
(etA )t≥0 exp. stable & analyic on a Hilbert space,
(−A)1/2 & (−A∗ )−1/2 L2 -admissible and U = Cn
(see talk of Felix Schwenninger)
Sketch of the proof: w.l.o.g. X = `2
2
Z
2
Z t
tX
λn (t−s)
e(t−s)A Bu(s)ds = e
b
e
u(s)ds
k k
2 0 k
0
`
`2
Z
2
X
t
=
|bn |2 eλn s u(s) ds ≤ · · ·
n∈N
0
Z tX
|bn |2 Re λn s
≤
e
|u(s)|2 ds ≤ · · ·
|Re
λ
|
n
0 n∈N
Z tX
2
|bn | Re λn s
≤
e
|u(s)|2 ds
|Re
λ
|
n
0 n∈N
|
{z
}
=:f (s)
We have f ∈ L1 (0, ∞)
Thus there exits a Young-Function Ψ : f ∈ EΨ (0, t).
Further we use Orlicz space theory to obtain the result.
Positive systems
Theorem
(J., Wintermayr ’17)
Let
I X be a Banach lattice
I (etA )t≥0 be a positive semigroup
I B ∈ L(Cn , X−1 ) positive
Then:
I B L∞ -admissible
I X reflexive
⇒
⇒
Σ
Mild solutions are continuous in X
L∞ -iISS
Thanks for your attention!