Byrnes-Isidori form for infinite dimensional systemsfa/cdps/talks/Trunk.pdf · Byrnes-Isidori form...

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.