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Byrnes-Isidori form for infinite dimensional
systems
Carsten Trunk
joint work with A. Ilchmann (Ilmenau) and B. Jacob(Wuppertal)
TU Ilmenau
19. July 2011
Assumptions
z(t) = Az(t) + u(t)b, z(0) = z0,
y(t) = (z(t), c).
Assumptions
z(t) = Az(t) + u(t)b, z(0) = z0,
y(t) = (z(t), c).
1 (H, ( . , . )) Hilbert space,
Assumptions
z(t) = Az(t) + u(t)b, z(0) = z0,
y(t) = (z(t), c).
1 (H, ( . , . )) Hilbert space,
2 A : dom A → H generator of a C0-semigroup,
Assumptions
z(t) = Az(t) + u(t)b, z(0) = z0,
y(t) = (z(t), c).
1 (H, ( . , . )) Hilbert space,
2 A : dom A → H generator of a C0-semigroup,
3 u ∈ L1loc(R
+, R),
Assumptions
z(t) = Az(t) + u(t)b, z(0) = z0,
y(t) = (z(t), c).
1 (H, ( . , . )) Hilbert space,
2 A : dom A → H generator of a C0-semigroup,
3 u ∈ L1loc(R
+, R),
4 Relative degree r :
Assumptions
z(t) = Az(t) + u(t)b, z(0) = z0,
y(t) = (z(t), c).
1 (H, ( . , . )) Hilbert space,
2 A : dom A → H generator of a C0-semigroup,
3 u ∈ L1loc(R
+, R),
4 Relative degree r :
b ∈ dom Ar and c ∈ dom A∗r
and
(Ar−1b, c) 6= 0 and for j = 0, 1, . . . , r − 2 we have (Ajb, c) = 0.
Decomposition of H
Lemma
Assume we have relative degree r . Then
H = ls{c}.
+ ls{A∗c}.
+ · · ·.
+ ls{A∗r−1c}
.
+ H0, (1)
whereH0 := {b}⊥ ∩ {Ab}⊥ ∩ · · · ∩ {Ar−1b}⊥. (2)
Decomposition of H
Theorem
With respect to H = ls{c}.
+ ls{A∗c}.
+ · · ·.
+ ls{A∗r−1c}
.
+ H0:
x = (P0x)c + (P1x)A∗c + · · · + (P r−1x)A∗r−1c + PH0
x ,
where P0x , . . . ,P r−1 : H → R (except for PH0):
Decomposition of H
Theorem
With respect to H = ls{c}.
+ ls{A∗c}.
+ · · ·.
+ ls{A∗r−1c}
.
+ H0:
x = (P0x)c + (P1x)A∗c + · · · + (P r−1x)A∗r−1c + PH0
x ,
where P0x , . . . ,P r−1 : H → R (except for PH0):j = m + 2, . . . , r
Pmj x :=
(A∗
j−1
c , Ar−(m+1)b)
(c , Ar−1b)−
j−1∑
k=m+2
Pmk A∗
j−1
c
(x , Ar−jb
)
(c , Ar−1b),
Pmx :=
(x , Ar−(m+1)b
)
(c , Ar−1b)−
r∑
j=m+2
Pmj x , m = 0, . . . , r − 1,
Decomposition of H
Theorem
With respect to H = ls{c}.
+ ls{A∗c}.
+ · · ·.
+ ls{A∗r−1c}
.
+ H0:
x = (P0x)c + (P1x)A∗c + · · · + (P r−1x)A∗r−1c + PH0
x ,
where P0x , . . . ,P r−1 : H → R (except for PH0):j = m + 2, . . . , r
Pmj x :=
(A∗
j−1
c , Ar−(m+1)b)
(c , Ar−1b)−
j−1∑
k=m+2
Pmk A∗
j−1
c
(x , Ar−jb
)
(c , Ar−1b),
Pmx :=
(x , Ar−(m+1)b
)
(c , Ar−1b)−
r∑
j=m+2
Pmj x , m = 0, . . . , r − 1,
PH0x := x −r−1∑
j=0
(P jx)A∗j
c .
Byrnes-Isidori form
Set dom A := Rr × (H0 ∩ dom A) and define A in R
r × H0
A =
0 1 · · · 0 0
0 0. . . 0 0
...... 1 0
P0A∗r
c P1A∗r
c · · · P r−1A∗r
c SR 0 · · · 0 Q
Byrnes-Isidori form
Set dom A := Rr × (H0 ∩ dom A) and define A in R
r × H0
A =
0 1 · · · 0 0
0 0. . . 0 0
...... 1 0
P0A∗r
c P1A∗r
c · · · P r−1A∗r
c SR 0 · · · 0 Q
Rα =α
(c ,Ar−1b)P⊥Arb, where P⊥ : orthogonal Proj. on H0,
Byrnes-Isidori form
Set dom A := Rr × (H0 ∩ dom A) and define A in R
r × H0
A =
0 1 · · · 0 0
0 0. . . 0 0
...... 1 0
P0A∗r
c P1A∗r
c · · · P r−1A∗r
c SR 0 · · · 0 Q
Rα =α
(c ,Ar−1b)P⊥Arb, where P⊥ : orthogonal Proj. on H0,
Sη =(PH0
A∗r
c , η)
Byrnes-Isidori form
Set dom A := Rr × (H0 ∩ dom A) and define A in R
r × H0
A =
0 1 · · · 0 0
0 0. . . 0 0
...... 1 0
P0A∗r
c P1A∗r
c · · · P r−1A∗r
c SR 0 · · · 0 Q
Rα =α
(c ,Ar−1b)P⊥Arb, where P⊥ : orthogonal Proj. on H0,
Sη =(PH0
A∗r
c , η)
Qη = P⊥Aη for η ∈ H0 ∩ dom A.
Byrnes-Isidori form
Set dom A := Rr × (H0 ∩ dom A) and define A in R
r × H0
A =
0 1 · · · 0 0
0 0. . . 0 0
...... 1 0
P0A∗r
c P1A∗r
c · · · P r−1A∗r
c SR 0 · · · 0 Q
Theorem
U : H → Rr × H0, x 7→ Ux :=
P0x...
P r−1xPH0
x
.
Then we haveAU∗ = U∗A.
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use:
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z =
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z = (U−1)∗Az + (U−1)∗bu =
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z = (U−1)∗Az + (U−1)∗bu = Ax +
0...0
(Ar−1b, c)0
u,
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z = (U−1)∗Az + (U−1)∗bu = Ax +
0...0
(Ar−1b, c)0
u,
y = (z , c)
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z = (U−1)∗Az + (U−1)∗bu = Ax +
0...0
(Ar−1b, c)0
u,
y = (z , c) = (U∗x , c) =
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z = (U−1)∗Az + (U−1)∗bu = Ax +
0...0
(Ar−1b, c)0
u,
y = (z , c) = (U∗x , c) = (x ,Uc) =
Rewrite the system
z(t) = Az(t) + u(t)b, y(t) = (z(t), c).Set
x := (U−1)∗z use: AU∗ = U∗A ⇔ (U−1)∗A = A(U−1)∗.
We have
x = (U−1)∗z = (U−1)∗Az + (U−1)∗bu = Ax +
0...0
(Ar−1b, c)0
u,
y = (z , c) = (U∗x , c) = (x ,Uc) =
x ,
10...0
Rr×H0
.
Rewrite the system in Byrnes-Isidori form
Theorem
z = Az + ub, y = (z , c).
Rewrite the system in Byrnes-Isidori form
Theorem
z = Az + ub, y = (z , c).
⇔
x =
0 1 · · · 0 0
0 0. . . 0 0
...... 1 0
P0A∗r
c P1A∗r
c · · · P r−1A∗r
c SR 0 · · · 0 Q
x +
0...0
(Ar−1b, c)0
u,
y =
x ,
10...0
Rr×H0
.
Relative degree 1 and Funnel
(b, c) 6= 0.
Relative degree 1 and Funnel
(b, c) 6= 0.
Lemma
H = sp{c}.
+ {b}⊥.
Rewrite in Byrnes-Isidori form
Relative degree 1 and Funnel
(b, c) 6= 0.
Lemma
H = sp{c}.
+ {b}⊥.
Rewrite in Byrnes-Isidori form
(z1(t)z2(t)
)=
[A1 A2
A3 A4
](z1(t)z2(t)
)+
(u(t)(b, c)
0
)
Relative degree 1 and Funnel
(b, c) 6= 0.
Lemma
H = sp{c}.
+ {b}⊥.
Rewrite in Byrnes-Isidori form
(z1(t)z2(t)
)=
[A1 A2
A3 A4
](z1(t)z2(t)
)+
(u(t)(b, c)
0
)
(z1(0)z2(0)
)=
(z01
z02
)
Relative degree 1 and Funnel
(b, c) 6= 0.
Lemma
H = sp{c}.
+ {b}⊥.
Rewrite in Byrnes-Isidori form
(z1(t)z2(t)
)=
[A1 A2
A3 A4
](z1(t)z2(t)
)+
(u(t)(b, c)
0
)
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Funnel
Let ϕ smooth, ϕ(0) = 0, ϕ(t) > 0, lim inft→∞ ϕ(t) > 0.
Fϕ := graph(t 7→
{z ∈ R
∣∣ ϕ(t)|z | < 1})
.
Funnel
Let ϕ smooth, ϕ(0) = 0, ϕ(t) > 0, lim inft→∞ ϕ(t) > 0.
Fϕ := graph(t 7→
{z ∈ R
∣∣ ϕ(t)|z | < 1})
.
1/ϕ(t)
Funnel
Let ϕ smooth, ϕ(0) = 0, ϕ(t) > 0, lim inft→∞ ϕ(t) > 0.
Fϕ := graph(t 7→
{z ∈ R
∣∣ ϕ(t)|z | < 1})
.
1/ϕ(t)
For a cont. differentiable bounded reference yref let
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Theorem
Assume A4 is the generator of an exponentially stable semigroup
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Theorem
Assume A4 is the generator of an exponentially stable semigroupand yref ∈ W 1,∞(R+).
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Theorem
Assume A4 is the generator of an exponentially stable semigroupand yref ∈ W 1,∞(R+). The above system with Funnel control hasa global solution.
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Theorem
Assume A4 is the generator of an exponentially stable semigroupand yref ∈ W 1,∞(R+). The above system with Funnel control hasa global solution. Each global solution z and u, k are bounded.
Funnel II
u(t) = −k(t) sgn (b, c) e(t), k(t) = 11−ϕ(t)|e(t)| , e(t) = y(t) − yref(t).
(z1(t)z2(t)
)=
[A1 A2
A3 A4
] (z1(t)z2(t)
)+
(u(t)(b, c)
0
),
(z1(0)z2(0)
)=
(z01
z02
)
y(t) = z1(t)
Theorem
Assume A4 is the generator of an exponentially stable semigroupand yref ∈ W 1,∞(R+). The above system with Funnel control hasa global solution. Each global solution z and u, k are bounded.For all t > 0 we have
e(t) ∈ Fϕ.
Example: Heat equation
x = 0 x = 1
z(t) =d2z(t)
dx2+ u(t)b, z(0) = z0,
dz(t)
dx(0) =
dz(t)
dx(0) = 0,
y(t) = (z(t), c)L2(0,1).
c(x) ≡ 1, b smooth with compact support in (0, 1). (b, c) 6= 0, i.e.relative degree 1.
Thank you.
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