hans zwart, university of twente - startseite tu …...this follows from the strong continuity and...

Elgersburg 2013: ∞ -dimensional systems

Infinite Dimensional Systems

Hans Zwart, University of Twente

– 1 –

Elgersburg 2013: ∞ -dimensional systems CONTENTS

Contents

1 Motivation 6

1.1 Potato storage . . . . . . . . . . . . . . . . . . . . . 7

1.2 UV-disinfection . . . . . . . . . . . . . . . . . . . . 10

1.3 Platoon systems . . . . . . . . . . . . . . . . . . . . 14

2 Semigroups and Generators 16

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Semigroups . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Generators . . . . . . . . . . . . . . . . . . . . . . 28

2.4 Which A generates a C0-semigroup? . . . . . . . . . 40

– 2 –


2.5 Platoon systems . . . . . . . . . . . . . . . . . . . . 42

2.6 Contraction semigroup . . . . . . . . . . . . . . . . 46

2.7 Semigroups and solutions of p.d.e.’s . . . . . . . . . . 61

2.8 Summary . . . . . . . . . . . . . . . . . . . . . . . 64

3 Inputs and Outputs 67

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 68

3.2 Inputs . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.3 Outputs . . . . . . . . . . . . . . . . . . . . . . . . 78

3.4 Feedback . . . . . . . . . . . . . . . . . . . . . . . 82

3.5 Boundary control . . . . . . . . . . . . . . . . . . . 85

3.6 Summary . . . . . . . . . . . . . . . . . . . . . . . 97

– 3 –


4 Transfer Functions 103

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 104

4.2 Exponential solutions . . . . . . . . . . . . . . . . . 106

4.3 Impulse response . . . . . . . . . . . . . . . . . . . 123

4.4 Summary . . . . . . . . . . . . . . . . . . . . . . . 135

5 Stability and Stabilizability 138

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 139

5.2 Stability notions . . . . . . . . . . . . . . . . . . . . 140

5.3 Stability of platoon models . . . . . . . . . . . . . . . 142

5.4 Exponential stability and Lyapunov equations . . . . . 144

5.5 Strong stability . . . . . . . . . . . . . . . . . . . . 153

– 4 –


5.6 Stabilizability . . . . . . . . . . . . . . . . . . . . . 162

5.7 Stabilization of the platoon model . . . . . . . . . . . 170

5.8 Transfer function and stability . . . . . . . . . . . . . 172

5.9 Summary . . . . . . . . . . . . . . . . . . . . . . . 174

6 Linear Port Hamiltonian Systems 175

6.1 A typical example . . . . . . . . . . . . . . . . . . . 176

7 Homogeneous solutions of P.H.S. 181

8 Well-posedness for PHS 192

8.1 Well-posedness for simple PHS . . . . . . . . . . . . 195

8.2 Characterization of well-posedness for PHS . . . . . . 198

– 5 –

Elgersburg 2013: ∞ -dimensional systems Motivation

1 Motivation

– 6 –


1.1 Potato storage

– 7 –


Schematic view point:

��

��

replacements

0

L

shaftbulk

Tp(x, t)Ta(x, t) v(t) x

Tc(t)

T0(t)

Φ(t)

– 8 –


The model:

dT0

dt(t) = M1α(Φ(t)) [Ta(L, t) − Tc(t)] +

M2Φ(t) [Ta(L, t) − T0(t)] +

M3 [Tout(t) − T0(t)]

∂Ta

∂t(x, t) = −v(t)

∂Ta

∂x(x, t) + M4 [Tp(x, t) − Ta(x, t)]

∂Tp

∂t(x, t) = M5Tp(x, t) + M6Ta(x, t)

Ta(0, t) = T0(t).

– 9 –


1.2 UV-disinfection

– 10 –


Schematic view point:

C1

r

r

r1 r2

L

x

C0(t)

– 11 –


The model for the velocity

∂vx

∂t(r, t) = β(t) +

1

Re

[

1

r

∂

∂r

(

r∂vx

∂r(r, t)

)]

vx(r1, t) = vx(r2, t) = 0

and for the concentration

– 12 –


∂C

∂t(x, r, t) = −vx(r, t)

∂C

∂x(x, r, t) +

1

Pe

[

1

r

∂

∂r

(

r∂C

∂r(x, r, t)

)

+∂2C

∂x2(x, r, t)

]

−Da K(r, t)C(x, r, t)

∂C

∂x(0, r, t) = 0

C(x, r1, t) = C(x, r2, t) = 0

C(0, r, t) = C0(t).

– 13 –


1.3 Platoon systems

Consider an infinite string of vehicles (a platoon)

didi+1di+2 di-1

i+1 i i-1

s i+1

s i

s i-1

LiLi+1 Li-1

The aim is to control the distance di between the cars. Every car

controls its jerk, i.e, the change of its force.

– 14 –


Standard modeling gives the following model for the i’th car

d

dt

di(t)

vi(t)

ai(t)

=

vi−1(t) − vi(t)

ai(t)

−τ−1ai(t) + τ−1ui(t)

, i ∈ Z.

where τ is a constant.

– 15 –

Elgersburg 2013: ∞ -dimensional systems Semigroups and Generators

2 Semigroups and Generators

– 16 –


2.1 Introduction

The aim of this part is to learn the concepts semigroups and

infinitesimal generator.

The reason for this is that we want to know if the partial diffe rential

equations (p.d.e.’s) or the coupled ordinary differential equations

(o.d.e.’s) have a solution.

Note there is a difference between knowing the existence of a solution

and having the form/expression of the solution.

– 17 –


Example (Transport equation)

On the spatial domain [0, 1] consider the p.d.e.

∂z

∂t(x, t) =

∂z

∂x(x, t), x ∈ [0, 1], t ≥ 0,

z(1, t) = 0

z(x, 0) = z0(x) (given).

�

– 18 –


Example (Diffusion equation)

On the spatial domain [0, 1] consider the p.d.e.

∂z

∂t(x, t) =

∂2z

∂x2(x, t), x ∈ [0, 1], t ≥ 0,

∂z

∂x(0, t) = 0

∂z

∂x(1, t) = 0

z(x, 0) = z0(x) (given).

�

– 19 –


2.2 Semigroups

Throughout this course, we assume that

• Z is a (complex) Hilbert space with inner product 〈·, ·〉.

• L(Z) denotes the set of linear and bounded operators from Z to

Z .

– 20 –


Definition

A strongly continuous semigroup is an operator-valued func tion from

[0, ∞) to L(Z) which satisfies

• T (0) = I

• T (t)T (s) = T (t + s), t, s ∈ [0, ∞)

• For all z0 ∈ Z there holds

limt↓0

T (t)z0 = z0.

�

Notation : (T (t))t≥0; Short : C0-semigroup.

– 21 –


To motivate this definition, think of

z0 7→ T (t)z0

as a solution of a time-invariant, linear differential equa tion.

• T (0) = I Trivial.

• For fixed t, T (t) is linear Is linearity of diff. eq.

• T (t)T (s) = T (t + s) Is time invariance.

• T (t)z0 → z0 if t ↓ 0 Is strong continuity.

– 22 –


Example

Let A be an n × n-matrix, then eAt is a C0-semigroup on Cn or R

n.

For instance, if

A =

1 2

0 3

,

then

eAt =

et e3t − et

0 e3t

.

�

– 23 –


Example

Let Z = L2(0, 1) and define for t ≥ 0 and x ∈ [0, 1]

(T (t)f) (x) =

f(t + x) x + t ≤ 1

0 x + t > 1

0 1x →

f

0 1x →

T (t)f

�

– 24 –


An important property of a C0-semigroup is that it is exponentially

bounded.

Lemma

There exist M ≥ 1 and ω ∈ R such that for all t ≥ 0

‖T (t)‖ ≤ Meωt.

�

– 25 –


The proof is as follows.

• supt∈[0,1] ‖T (t)‖ ≤ M .

This follows from the strong continuity and the uniform

boundedness theorem.

Note M ≥ 1.

– 26 –


• Choose t > 1 and write t = n + t0 with t0 ∈ [0, 1).

‖T (t)‖ = ‖T (n + t0)‖= ‖T (n)T (t0)‖≤ ‖T (n)‖‖T (t0)‖≤ M‖T (n)‖= M‖T (n − 1)T (1)‖≤ M‖T (1)‖‖T (n − 1)‖≤ MM‖T (n − 1)‖≤ Mn+1 = Melog Mn

≤ Melog Mt.

– 27 –


2.3 Generators

Consider the finite-dimensional semigroup

eAt =

et e3t − et

0 e3t

.

Question

What is A?

– 28 –


Answer

Evaluate the derivative of the semigroup at t = 0.

Since ddt

eAt = AeAt, we have

d

dteAt |t=0= A.

�

So given the (general) C0-semigroup (T (t))t≥0, we could try to find

A by differentiating it at t = 0. However, in general (T (t))t≥0 is

only (strongly) continuous.

– 29 –


Definition

If the following limit exists,

limt↓0

T (t)z0 − z0

t,

then z0 is in the domain of A, D(A).

Furthermore, for z0 ∈ D(A), we define

Az0 = limt↓0

T (t)z0 − z0

t.

A is named the infinitesimal generator of the C0-semigroup

(T (t))t≥0. �

– 30 –


Example

Consider the shift semigroup.

(T (t)f) (x) =

f(t + x) x + t ≤ 1

0 x + t > 1

00 11x →x →

fT (t)f

– 31 –


00 11x →x →

fT (t)f

Choose x < 1, then for t ∈ (0, x), we have that

(

T (t)f − f

t

)

(x) =f(t + x) − f(x)

t.

Hence if the limit exists, then f must be differentiable, and the limit is

the derivative.

– 32 –


Now take x = 1. For any t > 0, we have that(

T (t)f − f

t

)

(1) =0 − f(1)

t.

This limit can only exist when f(1) = 0.

– 33 –


Concluding, we have that

D(A) ={

f ∈ L2(0, 1) | f is absolutely continuous,

withdf

dx∈ L2(0, 1) and f(1) = 0

}

.

Furthermore,

Af =df

dx.

�

– 34 –


Lemma

If z0 ∈ D(A), then T (t)z0 ∈ D(A), and

d

dt[T (t)z0] = AT (t)z0.

�

Hence for z0 ∈ D(A), we have that z(t) := T (t)z0 is a (classical)

solution of the abstract differential equation

z(t) = Az(t), z(0) = z0.

How does this abstract differential equation relate to a p.d .e.?

– 35 –


We make the following observations:

• The function z is at every time instant an element of a Hilbert

space, i.e., z(t) ∈ Z .

• Assume that the Hilbert space consists of functions, i.e., f ∈ Z

means a.o. f := x 7→ f(x).

For instance, Z = L2(0, 1).

– 36 –


• This implies that z(t) is for every t a function of x. We write

(z(t)) (x) = z(x, t).

• Using this last form we see that

(z(t)) (x) =∂z

∂t(x, t).

– 37 –


• For A = ddx

, the abstract differential equation

z(t) = Az(t)

becomes∂z

∂t(x, t) =

∂z

∂x(x, t).

• In the domain of A are also the boundary conditions:

z(1, t) = 0.

– 38 –


• For a given p.d.e. we (roughly) do the following

– The state z is identified, and the p.d.e. is written as

∂z

∂t= . . .

– The right-hand side defines together with the boundary

conditions A.

– 39 –


2.4 Which A generates a C0-semigroup?

So we have that every semigroup has an infinitesimal generato r, but

you would like to know which operator A generates a C0-semigroup.

The answer is given by the Hille-Yosida Theorem. This theore m we will

not treat. We focus on two special cases

• A bounded, and

• T (t) a contraction

– 40 –


If A is a bounded (linear) operator, i.e., there exists an m ≥ 0 such

that ‖Az‖ ≤ m‖z‖ for all z ∈ Z , then A generates a

C0-semigroup, given by

T (t) = eAt = I + At +A2t2

2!+

A3t3

3!+ · · ·

– 41 –


2.5 Platoon systems

For zero input our platoon system is given by

d

dt

dr(t)

vr(t)

ar(t)

=

vr−1(t) − vr(t)

ar(t)

−τ−1ar(t)

, r ∈ Z.

where τ is a constant.

As state space we choose ℓ2(Z; C3) and as state

z(t) =

· · · ,

d−1(t)

v−1(t)

a−1(t)

,

d0(t)

v0(t)

a0(t)

,

d1(t)

v1(t)

a1(t)

, · · · ,

– 42 –


For the sequence (fr)r∈Z ∈ ℓ2(Z; Cn) define the

discrete Fourier transform as

f(φ) =

∞∑

r=−∞frφ

−r, φ ∈ ∂D (unit circle)

This Fourier transform maps ℓ2(Z; Cn) onto L2(∂D; C

n).

Furthermore,

‖(fr)‖2 =

∞∑

r=−∞‖fr‖2 =

1

2π

∫ 2π

0

‖f(eθ)‖2dθ = ‖f‖2

– 43 –


For differential equations of the following type

zr(t) =N∑

l=−N

Alzr−l(t), r ∈ Z

the Fourier transform gives;

∂z

∂t(φ, t) =

[

N∑

l=−N

Alφ−l

]

z(φ, t), φ ∈ ∂D

= A(φ)z(φ, t).

Hence a parametrized finite-dimensional system.

– 44 –


It is easy to see that A(φ) =∑N

l=−N Alφ−l defines a bounded

linear operator on L2(∂D; Cn).

So A generates the C0-semigroup on L2(∂D; Cn)

(

eAt)

(θ) = eA(θ)t, θ ∈ ∂D.

Thus in the Fourier domain it is easy to find the solutions of

zr(t) =N∑

l=−N

Alzr−l(t), r ∈ Z

– 45 –


2.6 Contraction semigroup

Definition

The C0-semigroup (T (t))t≥0 is contraction semigroup if

‖T (t)z0‖ ≤ ‖z0‖ for all t ≥ 0.

�

What can we say about these semigroups?

– 46 –


0 t →

‖z0‖

‖T (t)z0‖↑

– 47 –


We known that

‖T (t)z0‖2 = 〈T (t)z0, T (t)z0〉.

For z0 ∈ D(A), we have that the derivative of T (t)z0 equals

AT (t)z0.

So if we differentiate ‖T (t)z0‖2, we find

d

dt‖T (t)z0‖2 = 〈AT (t)z0, T (t)z0〉 + 〈T (t)z0, AT (t)z0〉.

– 48 –


At time equal to zero, we find

d

dt

(

‖T (t)z0‖2)

|t=0= 〈Az0, z0〉 + 〈z0, Az0〉.

We know that at t = 0, ‖T (t)z0‖ = ‖z0‖, and so if T (t) is a

contraction semigroup, then

〈Az0, z0〉 + 〈z0, Az0〉 =d

dt‖T (t)z0‖2 |t=0≤ 0.

This holds for all z0 ∈ D(A).

– 49 –


Theorem (Lumer-Phillips)

Let A be a densely defined operator, then A generates a contraction

semigroup on Z if and only if

1. 〈Az0, z0〉 + 〈z0, Az0〉 ≤ 0 for all z0 ∈ D(A).

2. The range of A − I is the whole of Z .

�

– 50 –


Example

Consider A which is given as

Az =dz

dx, x ∈ [0, 1]

with the domain

D(A) ={

z ∈ L2(0, 1) | z is absolutely continuous,

dz

dx∈ L2(0, 1) and z(1) = 0

}

.

Let us check the properties

• A is densely defined in L2(0, 1).

– 51 –


•

〈Az, z〉+〈z, Az〉

=

∫ 1

0

dz

dx(x)z(x)dx +

∫ 1

0

z(x)dz

dx(x)dx

=

∫ 1

0

d

dx

[

z(x)z(x)]

dx

= |z(x)|2∣

∣

1

0

= 0 − |z(0)|2 ≤ 0.

– 52 –


• To see if the range of (A − I) is everything, we have for every

f ∈ L2(0, 1) to solve (A − I)z = f .

This means solving

dz

dx(x) − z(x) = f(x), x ∈ (0, 1)

with boundary condition z(1) = 0.

The solution of this differential equation with the given bo undary

value is

z(x) = −∫ 1

x

ex−ξf(ξ)dξ.

�

– 53 –


Example

Consider A given by

Az =d2z

dx2x ∈ [0, 1]

with the domain

D(A) =

{

z ∈ L2(0, 1) | z,dz

dxare absolutely continuous,

anddz

dx(0) = 0 =

dz

dx(1)

}

.

Clearly A is densely defined, and so it remains to check the other

conditions.

– 54 –


We have that

〈Az, z〉 + 〈z, Az〉

=

∫ 1

0

d2z

dx2(x)z(x) + z(x)

d2z

dx2(x)dx

=

[

dz

dx(x)z(x) + z(x)

dz

dx(x)

]1

0

− 2

∫ 1

0

∣

∣

∣

∣

dz

dx(x)

∣

∣

∣

∣

2

dx

≤ 0

for z ∈ D(A).

– 55 –


To see that

(A − I)z = f

is solvable for all f ∈ L2(0, 1) we have to solve an o.d.e.

The solution is given by

z(x) = cosh(x)z(0) +

∫ x

0

sinh(x − ξ)f(ξ)dξ,

with

z(0) =−1

sinh(1)

∫ 1

0

cosh(1 − ξ)f(ξ)dξ.

Thus A generates a contraction semigroup.

– 56 –


Now we know that A generates a C0-semigroup, but can we find this

semigroup?

For this we could return to the corresponding p.d.e.

∂z

∂t(x, t) =

∂2z

∂x2(x, t), x ∈ [0, 1], t ≥ 0,

∂z

∂x(0, t) = 0

∂z

∂x(1, t) = 0

z(x, 0) = z0(x) (given).

This we can solve by the separation of variables principle. W e choose

another approach.

– 57 –


• A is a self-adjoint operator and the inverse of (A − I) is

compact.

• Hence A has an orthonormal basis of eigenfunctions.

– 58 –


• Solving Aφn = λnφn gives

φn(x) =

1 λ0 = 0√

2 cos(nπx) λn = −n2π2, n ∈ N

• Hence the solution of

z(t) = Az(t), z(0) = φn

is given by

z(t) = eλntφn

• This must be equal to T (t)φn.

– 59 –


• Since {φn, n ∈ N ∪ {0}} is an orthonormal basis, we know that

z0 =∞∑

n=0

〈z0, φn〉φn

Hence

T (t)z0 = T (t)

( ∞∑

n=0

〈z0, φn〉φn

)

=

∞∑

n=0

〈z0, φn〉T (t)φn

=∞∑

n=0

〈z0, φn〉eλntφn.

�

– 60 –


2.7 Semigroups and solutions of p.d.e.’s

We have that for any z0 ∈ D(A), the function z(t) := T (t)z0 is

the solution of

z(t) = Az(t), z(0) = z0.

How for general z0?

Note that in general Az(t) has no meaning and so has z(t).

– 61 –


To solve this question we have first to introduce the adjoint o f A.

Definition

Let A be a densely defined operator with domain D(A). The domain

of A∗, D(A∗), is defined as consisting of those w ∈ Z for which

there exists a v ∈ Z such that

〈w, Az〉 = 〈v, z〉 for all z ∈ D(A).

If w ∈ D(A∗), then A∗ is defined as

A∗w = v.

A∗ is named the adjoint of A. �

– 62 –


Lemma

Let z0 ∈ Z , and define z(t) = T (t)z0. Then for every

w ∈ D(A∗), there holds that

d

dt〈w, z(t)〉 = 〈A∗w, z(t)〉.

�

This implies that z(t) := T (t)z0 is the weak solution of

z(t) = Az(t), z(0) = z0.

– 63 –


2.8 Summary

We have seen the following

• A p.d.e. like

∂z

∂t(x, t) =

∂z

∂x(x, t)

z(1, t) = 0

will be written as

z(t) = Az(t)

where z is the state, see figure

– 64 –


– 65 –


• The state is a function of the spatial variable.

• The mapping initial state to state at time t is a strongly continuous

semigroup.

• There are conditions which tell you when the operator A

generates a strongly continuous semigroup.

• The function z(t) = T (t)z0 is always a (weak) solution of the

abstract differential equation

z(t) = Az(t), z(0) = z0.

– 66 –

Elgersburg 2013: ∞ -dimensional systems Inputs and Outputs

3 Inputs and Outputs

– 67 –


3.1 Introduction

From the previous part we know what we mean by the solution of t he

abstract differential equation

z(t) = Az(t), z(0) = z0

However, what do we mean by the solution of

z(t) = Az(t) + Bu(t), z(0) = z0,

y(t) = Cz(t) + Du(t).

To find the answer to this question is the aim of this part.

– 68 –


3.2 Inputs

In this section, we want to solve the abstract differential e quation

z(t) = Az(t) + Bu(t), z(0) = z0.

We assume that

• A generates the C0-semigroup (T (t))t≥0 on the Hilbert space

Z .

• B ∈ L(U, Z).

– 69 –


So we want to find a (candidate) solution for

z(t) = Az(t) + Bu(t), z(0) = z0.

Choose t1 > 0 and let t ∈ [0, t1].

We multiply the differential equation by T (t1 − t), and bring z to the

left-hand side

– 70 –


T (t1 − t)z(t) − T (t1 − t)Az(t) = T (t1 − t)Bu(t).

The left-hand side equals

d

dt[T (t1 − t)z(t)] = T (t1 − t)Bu(t).

Hence∫ t1

0

T (t1 − t)Bu(t)dt = [T (t1 − t)z(t)]t10

= z(t1) − T (t1)z(0).

– 71 –


Theorem

Consider the abstract differential equation

z(t) = Az(t) + Bu(t), z(0) = z0,

where A generates the C0-semigroup (T (t))t≥0 on the Hilbert

space Z , B ∈ L(U, Z), and u ∈ L1loc((0, ∞); U). Then the

(weak) solution is given by

z(t) = T (t)z0 +

∫ t

0

T (t − τ )Bu(τ )dτ.

If u is continuously differentiable and z0 ∈ D(A), then it is the

classical solution. �

– 72 –


Example

Consider the (controlled) p.d.e.

∂z

∂t(x, t) =

∂z

∂x(x, t) + u(t)

z(1, t) = 0.

We can write this as

z(t) = Az(t) + Bu(t)

with

– 73 –


Az =dz

dx,

D(A) =

{

z ∈ L2(0, 1) | dz

dx∈ L2(0, 1) and z(1) = 0

}

and

Bu = 1 · u.

– 74 –


Using the semigroup generated by A, we find that the solution of the

p.d.e. is given by

z(x, t) = z0(x+t)1[0,1](x+t)+

∫ t

0

1[0,1](x+t−τ )u(τ )dτ.

with 1[0,1](ξ) =

1 ξ ∈ [0, 1]

0 ξ 6∈ [0, 1]

�

– 75 –


Example

We consider the heated bar which is heated uniformly in the in terval

[1/2, 1].

∂z

∂t(x, t) =

∂2z

∂x2(x, t) + 1[1/2,1](x)u(t)

∂z

∂x(0, t) =

∂z

∂x(0, t) = 0.

This we can write as


with

(Bu) (x) = 1[1/2,1](x) · u

– 76 –


and A as before

Az =d2z

dx2,

with domain

D(A) =

{

z ∈ L2(0, 1) |dz

dxand

d2z

dx2∈ L2(0, 1),

dz

dx(0) = 0 =

dz

dx(1)

}

.

�

– 77 –


3.3 Outputs

Now we have solved the input problem, the understanding of th e

output equation

y(t) = Cz(t) + Du(t)

is easy.

– 78 –


If C ∈ L(Z, Y ), and D ∈ L(U, Y ), with Y a Hilbert space, then

using the solution for z(t) we find

y(t) = C

[

T (t)z0 +

∫ t

0

T (t − τ )Bu(τ )dτ

]

+ Du(t)

= CT (t)z0 +

∫ t

0

CT (t − τ )Bu(τ )dτ + Du(t).

– 79 –


Example

We take our heated bar, and we measure the (average) temperat ure in

the other half of the bar.

∂z

∂t(x, t) =

∂2z

∂x2(x, t) + 1[1

2,1](x)u(t)

∂z

∂x(0, t) =

∂z

∂x(0, t) = 0.

y(t) =

∫ 1

2

0

z(x, t)dx.

This we can written as

– 80 –



y(t) = Cz(t)

with A and B as before, and the operator C ∈ L(Z, C) given by

Cz =

∫ 1

2

0

z(x)dx.

�

– 81 –


3.4 Feedback

We have defined the open loop system, schematically given by

u y

z0

z(t) = Az(t) + Bu(t)y(t) = Cz(t)

However, for control we would like to close the loop.

– 82 –


u y

z0

z(t) = Az(t) + Bu(t)y(t) = Cz(t)

K

This gives the differential equation

z(t) = Az(t) + BKCz(t) = (A + BKC) z(t).

– 83 –


Theorem

If A generates the C0-semigroup (T (t))t≥0 on the Hilbert space Z ,

and B ∈ L(U, Z), F ∈ L(Z, U), then A + BF also generates a

C0-semigroup.

Furthermore, if we denotes this (closed-loop) semigroup by

(S(t))t≥0, then

S(t)z0 = T (t)z0 +

∫ t

0

T (t − τ )BFS(τ )z0dτ.

�

– 84 –


3.5 Boundary control

Consider the following system

∂z

∂t(x, t) =

∂z

∂x(x, t), x ∈ [0, 1]

z(1, t) = u(t).

We cannot write this in the form z(t) = Az(t) + Bu(t), with

B ∈ L(Z, C).

– 85 –


∂z

∂t(x, t) =

∂z

∂x(x, t), x ∈ [0, 1]

z(1, t) = u(t).

We perform the following trick.

Define v(x, t) = z(x, t) − u(t).

Then v satisfies the following partial differential equation

∂v

∂t(x, t) =

∂v

∂x(x, t) − u(t), x ∈ [0, 1]

v(1, t) = 0.

This we can write as v(t) = Av(t) + Bu(t), for u = u.

– 86 –


Definition

The system

z(t) = Az(t), z(0) = z0

Bz(t) = u(t)

is a boundary control system if

• A is a linear operator from D(A) ⊂ Z to Z , B is a linear

operator from D(B) ⊂ Z to U , and D(A) ⊂ D(B).

• The operator A defined as Az = Az with domain

D(A) = D(A) ∩ ker B generates a C0-semigroup on Z .

• The range of B equals U .

�

– 87 –


• Since B is surjective, we can find a B such that for all u ∈ U

– Bu ∈ D(B) and

– BBu = u.

• Define v(t) = z(t) − Bu(t). Then

– If z ∈ D(A), then v ∈ D(A), and

– since Bv = 0, we have that v ∈ D(A).

• v satisfies the following abstract differential equation

v(t) = Az(t) − Bu(t)

= Av(t) + ABu(t) − Bu(t)

= Av(t) + ABu(t) − Bu(t).

– 88 –


So if z(t) is the classical solution of

z(t) = Az(t), z(0) = z0

Bz(t) = u(t),

then v(t) = z(t) − Bu(t) is a classical solution of

v(t) = Av(t) + ABu(t) − Bu(t)

v(0) = z0 − Bu(0).

The reserve direction also holds.

– 89 –


For this last differential equation:

v(t) = Av(t) + ABu(t) − Bu(t)

v(0) = z0 − Bu(0)

we know the solution

v(t) = T (t)v(0) +

∫ t

0

T (t − τ ) [ABu(τ ) − Bu(τ )] dτ.

– 90 –


Example

We apply this to the transport equation with boundary contro l

∂z

∂t(x, t) =

∂z

∂x(x, t), x ∈ [0, 1]

z(1, t) = u(t).

Here we have that

Az =dz

dx, D(A) = {z ∈ L2(0, 1) | dz

dx∈ L2(0, 1)}.

Bz = z(1), D(B) = D(A).

Let us check the assumptions.

– 91 –


Az =dz

dx, D(A) = {z ∈ L2(0, 1) | dz

dx∈ L2(0, 1)}.

Bz = z(1), D(B) = D(A).

• Both operators are linear.

• A on the domain D(A) ∩ ker B generates a C0-semigroup.

• The range of B is C = U .

• Choose B = 1, then BBu = u.

– 92 –


Since AB = 0, we find

z(t) − 1 · u(t) = v(t) = T (t)v0 −∫ t

0

T (t − τ )1u(τ )dτ.

Using the expression of the (shift) semigroup, we find that th e solution

of the boundary controlled p.d.e. is

z(x, t) − u(t) = v0(x + t)1[0,1](x + t) −∫ t

0

1[0,1](x + t − τ )u(τ )dτ.

This we can simplify.

– 93 –


For t > 1 this expression becomes

z(x, t) − u(t) = v0(x + t)1[0,1](x + t) −∫ t

0

1[0,1](x + t − τ )u(τ )dτ

= 0 − [u(τ )]tx+t−1

= −u(t) + u(x + t − 1).

– 94 –


For t ∈ [0, 1] we have to distinguish two cases:

• x ∈ [0, 1 − t]:

z(x, t) − u(t) = v0(x + t)1[0,1](x + t) −∫ t

0

1[0,1](x + t − τ )u(τ )dτ

= z0(x + t) − u(0) − [u(τ )]t0

= z0(x + t) − u(t).

• x ∈ [1 − t, 1]:

z(x, t) − u(t) = 0 − [u(τ )]tx+t−1

= −u(t) + u(x + t − 1).

– 95 –


This we can summarize as follows

z(x, t) =

z0(x + t) x + t < 1

u(x + t − 1) x + t > 1

The solution for t < 1 is depicted for all x ∈ [0, 1].

1100 xx

z0-part u-part

Hence for the control transport equation we have found a (wea k)

solution for all L2-input functions. �

– 96 –


3.6 Summary

We have seen the following

• A controlled p.d.e. like

∂z

∂t(x, t) =

∂2z

∂x2(x, t) + 1[1/2,1](x)u(t)

∂z

∂x(0, t) =

∂z

∂x(0, t) = 0

y(t) =

∫ 1

2

0

z(x, t)dx

can be written as

– 97 –




with B, C , and D bounded operators.

• We know what we mean by the solution of this (controlled)

abstract differential equation, and we have a formula for it .

– 98 –


• If the control appears at the boundary, like

∂z

∂t(x, t) =

∂z

∂x(x, t)

z(1, t) = u(t)

we can write this in the abstract form

z(t) = Az(t)

Bz(t) = u(t)

• We know that we have a solution (for smooth inputs) of such a

boundary control system.

– 99 –


With this knowledge let us revisit the UV-reactor model disc ussed in

the first part.

– 100 –


Schematic view point of UV-reactor:

C1

r

r

r1 r2

L

x

C0(t)

– 101 –


∂C

∂t(x, r, t) = −vx(r, t)

∂C

∂x(x, r, t) +

1

Pe

[

1

r

∂

∂r

(

r∂C

∂r(x, r, t)

)

+∂2C

∂x2(x, r, t)

]

−Da K(r, t) C(x, r, t)

∂C

∂x(0, r, t) = 0

C(x, r1, t) = C(x, r2, t) = 0

C(0, r, t) = C0(t).

– 102 –

Elgersburg 2013: ∞ -dimensional systems Transfer Functions

4 Transfer Functions

– 103 –


4.1 Introduction

The aim of this part is to define transfer function for systems

described by partial differential equations.

We derive these transfer functions via a very simple calcula tion.

Furthermore, the relation with the impulse response is give n.

– 104 –


Consider the simple ordinary differential equation

y(t) + 5y(t) = 3u(t)

The transfer function of this system is given by

G(s) =3

s + 5

How do you come to this?

• Laplace transform, or

• Exponential solutions.

– 105 –


4.2 Exponential solutions

One way for obtaining the transfer function of

y(t) + 5y(t) = 3u(t)

is to take u(t) = est, s ∈ C, and to try to find a solution of the same

format, i.e., y(t) = αest.

Substituting this in the differential equation, gives

– 106 –


sαest + 5αest = 3est.

Since est is non-zero, we may divide by it, and we find

sα + 5α = 3.

If s 6= −5, this is solvable;

α =3

s + 5.

– 107 –


Definition

Given an (abstract) differential equation in the variables

(u(t), z(t), y(t)), where u(t), z(t), and y(t) take their values in

the (Hilbert) spaces U , Z , and Y , respectively.

Let s ∈ C. If for every u0 ∈ U , there exists a unique solution of the

form (u0est, z0e

st, y0est), and the mapping u0 7→ y0 is linear and

bounded, then this mapping is called the transfer function at s, and

will be denoted by G(s). �

– 108 –






Let s ∈ C, and u0 ∈ U . We try to find a solution of the form

(u(t), z(t), y(t)) = (u0est, z0e

st, y0est).

Substituting, this in the abstract differential equation g ives

sz0est = Az0e

st + Bu0est

y0est = Cz0e

st + Du0est.

– 109 –


(sI − A)z0 = Bu0

y0 = Cz0 + Du0.

If sI − A is (boundedly) invertible, then we find

y0 = C(sI − A)−1Bu0 + Du0.

This clearly defines a bounded linear mapping from u0 to y0, and so

the transfer function at s is given by

G(s) = C(sI − A)−1B + D.

This holds for all s ∈ ρ(A).

– 110 –


Example

We take our heated bar. We heat it uniformly at one half, and we

measure (half) the average temperature in the other half.

∂z

∂t(x, t) =

∂2z

∂x2(x, t) + 1[1

2,1](x)u(t)

∂z

∂x(0, t) =

∂z

∂x(1, t) = 0

y(t) =

∫ 1

2

0

z(x, t)dx.

We obtain the transfer function, by two approaches.

– 111 –


Method 1. :

The p.d.e. can be written as


y(t) = Cz(t)

with Cz =∫ 1

2

0z(x)dx, Bu = 1[1

2,1](x)u, and Az = d2z

dx2with

domain

D(A) =

{

z ∈ L2(0, 1) | dz

dx,d2z

dx2∈ L2(0, 1),

dz

dx(0) =

dz

dx(1) = 0

}

.

– 112 –


Now we know that

G(s) = C(sI − A)−1B.

Using the fact that A has an orthonormal basis of eigenfunctions, i.e.,

Aφn = λnφn, we find that

(sI − A)−1z =

∞∑

n=0

1

s − λn

〈z, φn〉φn.

Hence

(sI − A)−1Bu =∞∑

n=0

1

s − λn

〈Bu, φn〉φn

and

C(sI − A)−1Bu =∞∑

n=0

1

s − λn

〈Bu, φn〉Cφn.

– 113 –


Thus

G(s) =∞∑

n=0

1

s − λn

〈B, φn〉Cφn.

We know that λn = −n2π2, n=0,1,2,. . . , and

φn(x) =

1 n = 0√

2 cos(nπx) n ≥ 1.

Using this we find that

– 114 –


G(s) =∞∑

n=0

1

s − λn

〈B, φn〉Cφn

=1

s

∫ 1

1/2

1 · 1dx

∫ 1/2

0

1 · 1dx +

∞∑

n=1

1

s + n2π2

∫ 1

1/2

1 ·√

2 cos(nπx)dx ·∫ 1/2

0

1 ·√

2 cos(nπx)dx.

Evaluating the integrals gives

G(s) =1

4s− 2

∞∑

n=1

sin(nπ 12)2

n2π2(s + n2π2).

– 115 –


Method 2 :

We try to find an exponential solution of the p.d.e. This gives the

following equation

sz0(x)est =d2z0

dx2(x)est + 1[1

2,1](x)u0e

st

dz0

dx(0)est =

dz0

dx(1)est = 0

y0est =

∫ 1

2

0

z0(x)estdx.

– 116 –


Hence

sz0(x) =d2z0

dx2(x) + 1[1

2,1](x)u0

dz0

dx(0) =

dz0

dx(1) = 0

y0 =

∫ 1

2

0

z0(x)dx.

The first two lines represent an o.d.e. with boundary conditi ons.

– 117 –


The solution of

sz0(x) =d2z0

dx2(x) + 1[1

2,1](x)u0

dz0

dx(0) =

dz0

dx(1) = 0

is given as

z0(x) = cosh(√

sx)z0(0) −1

√s

∫ x

0

sinh(√

s(x − ξ))1[1/2,1](ξ)u0dξ

with

z0(0) =sinh(

√s/2)u0

s sinh(√

s)=

u0

2s cosh(√

s/2).

– 118 –


Using this we find that

y0 =

∫ 1

2

0

z0(x)dx

=sinh(

√s/2)u0

2s√

s cosh(√

s/2).

Hence the transfer function is given by

G(s) =tanh(

√s/2)

2s√

s.

– 119 –


The transfer function is unique, and so we find that

tanh(√

s/2)

2s√

s= G(s) =

1

4s− 2

∞∑

n=1

sin(nπ 12)2

n2π2(s + n2π2).

�

– 120 –


Example

Consider the system with boundary control and observation

∂z

∂t(x, t) =

∂z

∂x(x, t)

z(1, t) = u(t)

y(t) = z(0, t).

Substituting exponential functions for all signals, gives

sz0(x)est =dz0

dx(x)est

z0(1)est = u0e

st

y0est = z0(0)e

st.

Thus

– 121 –


sz0(x) =dz0

dx(x)

z0(1) = u0

y0 = z0(0).

This is an ordinary differential equation with given (end) c ondition, u0

and unknown (initial) condition, y0.

The solution equals z0(x) = es(x−1)u0. Thus y0 = e−su0.

The transfer function equals

G(s) = e−s s ∈ C.

�

– 122 –


4.3 Impulse response

Definition

Let the (abstract) differential equation have the transfer function

G(s). If this transfer function exists on some right-half plane a nd

possesses an inverse (one-sided) Laplace transform, then w e call the

inverse Laplace transform, the impulse response. �

– 123 –


Theorem





The impulse response of this differential equation equals

h(t) = CT (t)B + Dδ(t), t ≥ 0−.

�

– 124 –


Question

If we take the Laplace transform of the impulse response do we get

the transfer function? �

– 125 –


Answer

No.

Example

Let Z = ℓ2(Z), and consider the following differential equation

dz

dt(n, t) = z(n + 1, t), n ∈ Z \ {0}

dz

dt(0, t) = z(1, t) + u(t)

y(t) = z(1, t).

– 126 –


Substituting the exponential functions, leads to

sz0(n) = z0(n + 1), n ∈ Z \ {0}sz0(0) = z0(1) + u0

y0 = z0(1).

From the first equation, we obtain

z0(n) =

sn−1z0(1) n ≥ 1

snz0(0) n ≤ 0.

– 127 –


Since we must have that z0 ∈ ℓ2(Z), we find that

z0(n) =

sn−1z0(1) = 0 n ≥ 1 and |s| ≥ 1

snz0(0) = 0 n ≤ 0 and |s| ≤ 1.

Using the equation

sz0(0) = z0(1) + u0,

we find for |s| > 1

z0(n) =

0 n ≥ 1

sn−1u0 n ≤ 0

– 128 –


For |s| < 1

z0(n) =

−sn−1u0 n ≥ 1

0 n ≤ 0

So we have that

y0 = z0(1) =

0 |s| > 1

−u0 |s| < 1.

The transfer function is given by

– 129 –


1

iC

Yellow means G(s) = −1. White means G(s) = 0.

– 130 –


The inverse Laplace transform of G(s) gives h(t) = 0.

If we take of this the Laplace transform, then it is obviously zero

everywhere, and thus not equals to G(s) for all s. �

Remark

The system can be written in the standard format

z(t) = Az(t) + Bu(t), y(t) = Cz(t) + Du(t) with all

operators bounded. �

– 131 –


Theorem

Consider the system


y(t) = Cz(t) + Du(t),

where B, C , and D are bounded operators, and A generates the

C0-semigroup (T (t))t≥0. If this semigroup satisfies

‖T (t)‖ ≤ Meωt, for all t > 0,

then the transfer function and the Laplace transform of the i mpulse

response exist in {s ∈ C | Re(s) > ω}. Furthermore, they are

equal on this set. �

– 132 –


Example (revisited)

In our previous example we have that the semigroup is bounded by et,

and so the theorem tells us that on the right-half plane

{s ∈ C | Re(s) > 1} the transfer function and the Laplace

transform of the impulse response are the same.

Since h(t) = 0 and since G(s) = 0 on this right-half plane, we are

in agreement with the theorem.

We also see that (in general) a larger half-plane is not possi ble. �

– 133 –


Theorem

If the transfer function G(s) is a meromorphic function on C, then

the meromorphic continuation of the Laplace transform of th e impulse

response is equal to G(s) on C. �

– 134 –


4.4 Summary

In this part we have seen the following

• Transfer functions are easy to find by using exponential solu tions.

• You don’t have to write a p.d.e. as an abstract differential e quation

in order to obtain the transfer function.

• It is not hard to find the transfer function of a system with

boundary control and observation .

• The Laplace transform of the impulse response only equals th e

transfer function on some right-half plane.

• For scalar transfer function the design rules of Bode and Nyq uist

still apply.

– 135 –


Here we show the Nyquist plot of e−st.

1−1

iC

Re →

Im ↑

– 136 –


The (magnitude) Bode plot of

G(s) = −2∞∑

n=1

sin(nπ 12)2

n2π2(s + n2π2)

−1

−1−4 0 1 2 3log(ω) →

log

|G(i

ω)|

– 137 –

Elgersburg 2013: ∞ -dimensional systems Stability and Stabilizability

5 Stability and Stabilizability

– 138 –


5.1 Introduction

In this part we want to define stability and show how these noti ons

relate to similar notions for ordinary differential equati ons.

Furthermore, we show how we can stabilize systems.

– 139 –


5.2 Stability notions

There are several stability notions for the abstract differ ential equation

z(t) = Az(t), z(0) = z0.

We assume that A is the infinitesimal generator of a strongly

continuous semigroup (T (t))t≥0 on the Hilbert space Z .

– 140 –


Definition

The semigroup (T (t))t≥0 is exponentially stable if there exists

M, ω > 0 such that

‖T (t)‖ ≤ Me−ωt, t ≥ 0.

�

Definition

The semigroup (T (t))t≥0 is strongly stable if for all z0 ∈ Z

limt→∞

T (t)z0 = 0.

�

– 141 –


5.3 Stability of platoon models

Consider the the platoon model

z(t) = Az(t), z(0) = z0 ∈ Z

where Z = L2(∂D; C), and(

Az(t))

(φ) = A(φ)z(φ, t),

φ ∈ ∂D.

Theorem

Assume that A is a continuous function of φ. The above system is

exponentially stable if and only if there exists a δ > 0 such that for all

φ ∈ ∂D the eigenvalues of A(φ) have real part less or equal to −δ.

– 142 –


Example

Consider our platoon model

d

dt

dr(t)

vr(t)

ar(t)

=

vr−1(t) − vr(t)

ar(t)

−τ−1ar(t)

, r ∈ Z.

For this model A is given by

A(φ) =

0 φ−1 − 1 0

0 0 1

0 0 −τ−1

, φ ∈ ∂D.

It is clear that this is not exponentially stable (also not st rongly

stable). �

– 143 –


5.4 Exponential stability and Lyapunov equations

The following theorem relates exponential stability with t he solution of

a Lyapunov equation.

Theorem

The following are equivalent.

• (T (t))t≥0 is exponentially stable.

• For all z0 ∈ Z we have that T (t)z0 ∈ L2((0, ∞); Z).

• There exists a positive L ∈ L(Z) satisfying the following

Lyapunov equation

〈Az, Lz〉 + 〈Lz, Az〉 = −〈z, z〉 z ∈ D(A).

�

– 144 –


Some remarks concerning the Lyapunov equation.

Lemma

For a positive operator L ∈ L(Z) the following are equivalent:

• L satisfies

〈Az, Lz〉 + 〈Lz, Az〉 = −〈z, z〉 ∀z ∈ D(A).

• L satisfies

〈Az1, Lz2〉+〈Lz1, Az2〉 = −〈z1, z2〉 ∀z1, z2 ∈ D(A).

• L maps the domain of A into the domain of A∗, and L satisfies

A∗L + LA = −I on D(A).

– 145 –


Proof of the equivalence between the last two items.

Assume that z0 ∈ L2((0, ∞); Z) for all z0 ∈ Z . For z1, z2 ∈ Z ,

define

〈z1, Lz2〉 =

∫ ∞

0

〈T (t)z1, T (t)z2〉dt.

This operator is well-defined, self-adjoint, and positive. Furthermore,

for z1, z2 ∈ D(A)

〈Az1, Lz2〉 + 〈z1, LAz2〉

=

∫ ∞

0

〈T (t)Az1, T (t)z2〉dt +

∫ ∞

0

〈T (t)z1, T (t)Az2〉dt.

– 146 –


Hence

〈Az1, Lz2〉 + 〈z1, LAz2〉

=

∫ ∞

0

〈T (t)Az1, T (t)z2〉dt +

∫ ∞

0

〈T (t)z1, T (t)Az2〉dt

=

∫ ∞

0

d

dt[〈T (t)z1, T (t)z2〉] dt

= 0 − 〈z1, z2〉.

– 147 –


If the Lyapunov equation has a positive solution, then

∫ t1

0

‖T (t)z0‖2dt

=

∫ t1

0

〈T (t)z0, T (t)z0〉dt =

∫ t1

0

〈T (t)z0, IT (t)z0〉dt

= −∫ t1

0

〈AT (t)z0, LT (t)z0〉dt

−∫ t1

0

〈T (t)z0, LAT (t)z0〉dt

= −∫ t1

0

d

dt[〈T (t)z0, LT (t)z0〉] dt

= −〈T (t1)z0, LT (t1)z0〉 + 〈z0, Lz0〉.

– 148 –


Hence for all t1 > 0∫ t1

0

‖T (t)z0‖2dt ≤ ‖L‖‖z0‖2

and so T (t)z0 ∈ L2((0, ∞); Z).

Thus we have proved the equivalence between the last two item s.

– 149 –


Proof of the equivalence between the first two items.

If T (t)z0 ∈ L2((0, ∞), Z) for all z0, then

• ‖T (t)‖ ≤ M for all t > 0

• The following estimate holds

t‖T (t)z0‖2 =

∫ t

0

‖T (t)z0‖2dt

=

∫ t

0

‖T (t − τ )T (τ )z0‖2dτ

≤ M2

∫ t

0

‖T (τ )z0‖2dτ ≤ K‖z0‖2.

– 150 –


Thus

‖T (t)z0‖2 ≤K

t‖z0‖2.

Now using the semigroup property, we find that (T (t))t≥0 is

exponentially stable. �

– 151 –


There are examples of unstable semigroups for which the infin itesimal

generator A has no spectrum in the set {s ∈ C | Re(s) ≥ 0}.

Hence, in general, we cannot conclude stability by only look ing at the

spectrum of A. However,

Theorem

The semigroup (T (t))t≥0 is exponentially stable if and only if

sup{s∈C|Re(s)>0}

‖(sI − A)−1‖ < ∞.

�

– 152 –


5.5 Strong stability

We may also conclude exponential stability when there exist s a

positive L such that

A∗L + LA ≤ −εI, ε > 0.

Question

Is the semigroup strongly stable when there exists an L ∈ L(Z),

L > 0 such that

A∗L + LA < 0 on D(A)?

Answer

No.

– 153 –


Example

Take the Hilbert space

Z =

{

f : [0, ∞) 7→ C |∫ ∞

0

|f(x)|2[

e−x + 1]

dx < ∞}

.

It inner product is given by:

〈f, g〉 =

∫ ∞

0

f(x)g(x)[

e−x + 1]

dx.

Note that Z “is” L2(0, ∞).

– 154 –


As semigroup, we choose the shift:

(T (t)f) (x) =

f(x − t) x > t

0 x ∈ [0, t)

= f(x − t)1[0,∞](x − t).

– 155 –


There holds

‖T (t)f‖2 =

∫ ∞

0

|f(x − t)1[0,∞](x − t)|2[

e−x + 1]

dx

=

∫ ∞

0

|f(ξ)|2[

e−(ξ+t) + 1]

dξ

≥∫ ∞

0

|f(ξ)|2dξ

≥ 1

2

∫ ∞

0

|f(ξ)|2[

e−ξ + 1]

dξ

=1

2‖f‖2.

Hence (T (t))t≥0 is not strongly stable.

– 156 –


The infinitesimal generator is given by

Af = −df

dx

with domain

D(A) = {f ∈ Z | f is absolutely continuous

df

dx∈ Z, and f(0) = 0

}

.

Next we evaluate

〈Az, z〉 + 〈z, Az〉for z ∈ D(A) and z 6= 0.

– 157 –


For A = − ddx

with boundary condition z(0) = 0, we find

〈Az, z〉 + 〈z, Az〉

=

∫ ∞

0

(−1)dz

dx(x)z(x)

[

e−x + 1]

dx +

∫ ∞

0

z(x)(−1)dz

dx(x)

[

e−x + 1]

dx

= −∫ ∞

0

d

dx

(

|z(x)|2) [

e−x + 1]

dx.

– 158 –


Hence

〈Az, z〉 + 〈z, Az〉= −

[

|z(x)|2[

e−x + 1]]∞

0+

∫ ∞

0

|z(x)|2[

−e−x]

dx

= 0 −∫ ∞

0

|z(x)|2e−xdx

< 0.

– 159 –


Concluding, we have constructed an infinitesimal generator A for

which the semigroup is not strongly stable, but the Lyapunov

inequality

〈Az, z〉 + 〈z, Az〉 < 0, z ∈ D(A), z 6= 0

holds. Note that L = I . �

Remark

If the trajectories, t 7→ T (t)z0, are pre-compact, then this Lyapunov

inequality implies strong stability. �

– 160 –


Theorem

Let (T (t))t≥0 is a strongly continuous semigroup on the Hilbert

space Z . If

• The semigroup is uniformly bounded, i.e., ‖T (t)‖ ≤ M

• A has no eigenvalues on the imaginary axis

• The spectrum on the imaginary axis is countable,

then (T (t))t≥0 is strongly stable.

– 161 –


5.6 Stabilizability

In this section we study the question whether there exist a fe edback,

u = Fz which stabilizes the system

z(t) = Az(t) + Bu(t).

That is, the operator A + BF generates an exponentially stable

semigroup.

– 162 –


Theorem

Consider z(t) = Az(t) + Bu(t), with B ∈ L(Cm, Z). The

following are equivalent.

• There exists an F ∈ L(Z, Cm) such that A + BF generates

an exponentially stable semigroup on Z .

• Z can be decomposed as Z = Zu ⊕ Zs with

– dim(Zu) < ∞.

– For all t > 0 we have that T (t)Zu ⊂ Zu and T (t)Zs ⊂ Zs.

– T (t)|Zs is exponentially stable.

– The system restricted to Zu is controllable.

– 163 –


1

iC

Red denotes the spectrum.

– 164 –


1

iC

Zu

Zs

– 165 –


Example

We consider the heated bar. We heat it uniformly at one half, a nd we

measure (half) the average temperature in the other half.

∂z

∂t(x, t) =

∂2z

∂x2(x, t) + 1[1

2,1](x)u(t)

∂z

∂x(0, t) =

∂z

∂x(1, t) = 0

y(t) =

∫ 1

2

0

z(x, t)dx.

We know that the corresponding A has the eigenvalues

– 166 –


λn = −n2π2, n = 0, 1, 2, . . ., and the eigenfunctions

φn(x) =

1 n = 0√

2 cos(nπx) n = 1, 2, . . .

This A has an unstable eigenvalue, λ0 = 0. Now we define

Zu = span{φ0}

Zs = spann=1,2,...{φn} = Z⊥u .

Let us check the conditions.

– 167 –


• dim(Zu) = 1 < ∞.

• T (t)Zu ⊂ Zu and T (t)Zs ⊂ Zs.

•

T (t)|Zs =

∞∑

n=1

e−n2π2t〈·, φn〉φn.

This is exponentially stable.

• A restricted to Zu is Au = [0], and the projection of B onto Zu

is Bu = 〈B, φ0〉 = [1/2].

The pair (Au, Bu) is controllable.

Hence the system is stabilizable.

– 168 –


You can stabilize it by stabilizing the finite-dimensional p art, and not

effecting the stable part.

A feedback that works is

Fz = −k〈z, φ0〉φ0,

with k > 0. �

– 169 –


5.7 Stabilization of the platoon model

After Fourier transform our platoon model becomes

∂z

∂t(φ, t) = A(φ)z(φ, t) + B(φ)u(φ, t),

with φ ∈ ∂D, and

A(φ) =

0 φ−1 − 1 0

0 0 1

0 0 −τ−1

, B(φ) =

0

0

τ−1

.

Note that U = L2(∂D). So infinite-dimensional .

– 170 –


Suppose that we apply a state feedback of the type

u(φ, t) = F (φ)z(φ, t) with

F (φ) =(

f1(φ) f2(φ) f3(φ))

Then A(φ) + B(φ)F (φ) equals

0 φ−1 − 1 0

0 0 1

τ−1f1(φ) τ−1f2(φ) τ−1f3(φ) − τ−1

.

Exponentially stable?

– 171 –


5.8 Transfer function and stability

One can check stability by looking at the transfer function. We can do

that for the state-space system.


y(t) = Cz(t) + Du(t).

– 172 –


Theorem

Consider the (standard) state space system with B, C , and D

bounded operators. Let the input and output space be

finite-dimensional.

Assume further that there exists a F ∈ L(Z, U) and a

K ∈ L(Y, Z) such that A + BF and A + KC generate

exponentially stable semigroups.

The semigroup generated by A is exponentially stable if and only if

the transfer function G(s) = C(sI − A)−1B + D is analytic and

bounded in {s ∈ C | Re(s) > 0}. �

– 173 –


5.9 Summary

• We have introduce different notions of stability for a

C0-semigroup.

• For exponential stability there are nice equivalent condit ions.

• For strong stability there are nice sufficient conditions, b ut they

are not sufficient.

• If the input is finite-dimensional, then there is a complete

characterizing of stabilizability.

– 174 –

Elgersburg 2013: ∞ -dimensional systems Linear Port Hamiltonian Systems

6 Linear Port Hamiltonian Systems

Warning: x will denote the state from the state space X , and ζ

denotes the spatial variable.

– 175 –


6.1 A typical example

V (a)

I(a)

V (b)

I(b)a b

Consider the transmission line on the spatial interval [a, b]

∂Q

∂t(ζ, t) = − ∂

∂ζ

φ(ζ, t)

L(ζ)

∂φ

∂t(ζ, t) = −

∂

∂ζ

Q(ζ, t)

C(ζ).

We write x1 = Q (charge) and x2 = φ (flux), and we find that

– 176 –


∂

∂t

x1

x2

(ζ, t) =

0 −1

−1 0

∂

∂ζ

1C(ζ)

x1(ζ, t)

1L(ζ)

x2(ζ, t)

=

0 −1

−1 0

∂

∂ζ

1C(ζ)

0

0 1L(ζ)

x1(ζ, t)

x2(ζ, t)

= P1

∂

∂ζ(Hx) (ζ, t)

So we have, J = P1∂∂ζ

(skew-symmetric) and the Hamiltonian

(energy) equals H = 12

∫ b

axTHxdζ (positive).

To use this structure we differentiate the Hamiltonian (ene rgy) along

trajectories.

– 177 –


dH

dt(t) =

1

2

∫ b

a

∂x

∂t(ζ, t)TH(ζ)x(ζ, t)dz+

1

2

∫ b

a

x(ζ, t)TH(ζ)∂x

∂t(ζ, t)dz

=1

2

∫ b

a

(

P1

∂

∂ζ(Hx) (ζ, t)

)T

H(ζ)x(ζ, t)dζ+

1

2

∫ b

a

x(ζ, t)TH(ζ, t)(P1

∂

∂ζ(Hx) (ζ, t)dζ

=1

2

∫ b

a

∂

∂ζ

[

(Hx)T (ζ, t)P1 (Hx) (ζ, t)]

dζ

=1

2

[

(Hx)T (ζ, t)P1 (Hx) (ζ, t)]b

a,

where we have used the symmetry of P1 and H.

– 178 –


So we have that the time-change of Hamiltonian satisfies

dH

dt(t) =

1

2

[

(Hx)T (ζ, t)P1 (Hx) (ζ, t)]b

a. (1)

That is the change of internal energy goes via the boundary (p orts).

Note the we only used that P1 and H are symmetric. We did not need

the specific form of P1 or H.

The balance equation (1) also holds for the system

∂x

∂t(ζ, t) = P1

∂Hx

∂ζ(ζ, t) + P0 [Hx] (ζ, t) (2)

with P0 anti-symmetric, i.e., P T0 = −P0.

Many (hyperbolic) systems can be written in this format.

– 179 –


Example

Consider transmission lines in a network

VV

IITrans. line I

Trans. line II

Trans. line III

Trans. line IVKK

In the coupling parts K , we have that Kirchhoff laws holds. Hence

charge flowing out of the transmission line I, enters II and II I, etc.

The P1 of the big system is the diagonal matrix, build from the

uncoupled P1’s (which are all the same). The H of the coupled

system is the diagonal matrix of the uncoupled H’s.

The coupling is written down as boundary conditions of the pde.

– 180 –

Elgersburg 2013: ∞ -dimensional systems Homogeneous solutions of P.H.S.

7 Homogeneous solutions of P.H.S.

– 181 –


Consider the p.d.e. of our Port-Hamiltonian system

∂x

∂t(ζ, t) = P1

∂

∂ζ[H(ζ)x(ζ, t)] + P0 [H(ζ)x(ζ, t)] .

with energy (Hamiltonian)

E(t) =1

2

∫ b

a

x(ζ, t)TH(ζ)x(ζ, t)dζ.

We assume that

• P1 is real and symmetric, i.e., P T1 = P1,

• P0 is real and anti-symmetric, i.e., P T0 = −P0,

• H(ζ) is a positive (real) symmetric matrix, uniformly satisfyin g

0 < mI ≤ H(ζ) ≤ MI , and continuously differentiable.

– 182 –


The energy

E(t) =1

2

∫ b

a

x(ζ, t)TH(ζ)x(ζ, t)dζ.

can be seen as a weighted L2-norm.

– 183 –


Associated to this p.d.e. we define the state space

X = L2((a, b); Cn)

with inner product

〈f, g〉 =1

2

∫ b

a

f(ζ)TH(ζ)g(ζ)dζ.

and the differential operator

Ax := P1

d

dζ[Hx] + P0 [Hx]

with domain

D(A) = {x ∈ X | x is absolutely continuous, anddx

dζ∈ X}.

Thus the squared norm of x equals the energy of this state.

– 184 –


The following balance equation holds

Lemma

On the space X , the operator Ax = P1ddζ

[Hx] + P0 [Hx]

satisfies

〈Ax, x〉 + 〈x, Ax〉

=1

2

[

(Hx)T (b)P1(Hx) (b) − (Hx)T (a)P1(Hx) (a)]

for x ∈ D(A). �

– 185 –


Proof (for P0 = 0) Using the definition of the inner product, we find

〈Ax, x〉 + 〈x, Ax〉

=1

2

∫ b

a

[

P1

∂

∂ζ(Hx) (ζ)

]T

H(ζ)x(ζ)dζ +

1

2

∫ b

a

x(ζ)TH(ζ)

[

P1

∂

∂ζ(Hx) (ζ)

]

dζ.

Using the fact that P1, H(ζ) are real symmetric, we write the last

– 186 –


expression as

1

2

∫ b

a

[

d

dζ(Hx) (ζ)

]T

P1H(ζ)x(ζ)+

[H(ζ)x(ζ)]T[

P1

d

dζ(Hx) (ζ)

]

dζ

=1

2

∫ b

a

d

dζ

[

(Hx)T (ζ)P1(Hx) (ζ)]

dζ

=1

2

[


.

�

– 187 –


Theorem

Consider the operator A which is defined as

Ax = Ax = P1

∂

∂ζ(Hx) + P0 (Hx)

on the domain D(A) = D(A) ∩ ker B, where

Bx = M

(Hx)(b)

(Hx)(a)

.

If M is n × 2n-matrix of full rank, then A generates a contraction

semigroup on X if and only if

〈Ax, x〉 + 〈x, Ax〉 ≤ 0 for all x ∈ D(A).

�

– 188 –


Hence by posing n boundary conditions, we only have to check the

inequality.

Since we have the power balance this can be done quickly.

If H is non-constant, it can be very hard (impossible) to obtain t he

expression for the C0-semigroup.

– 189 –


Example

Consider the transmission line.

Ax =

0 −1

−1 0

d

dζ

1C(ζ)

0

0 1L(ζ)

x1(ζ)

x2(ζ)

and

〈Ax, x〉 + 〈x, A〉

=1

2

[


= V (a)I(a) − V (b)I(b),

since x1/C = Q/C = V and x2/L = φ/L = I .

– 190 –


If we choose V (a) = 0 and V (b) = RI(b), R ≥ 0, then

M =

0 0 1 0

1 −R 0 0

.

This matrix has rank n = 2.

Furthermore, using these boundary conditions, we have that

V (a)I(a) − V (b)I(b) ≤ 0.

Hence the operator A generates a contraction semigroup on Z . �

– 191 –

Elgersburg 2013: ∞ -dimensional systems Well-posedness for PHS

8 Well-posedness for PHS

For t ≥ 0 and ζ ∈ [a, b] we consider the system:

∂x

∂t(ζ, t) = P1

∂

∂z(Hx) (ζ, t), x(0, z) = x0(ζ) (3)

0 = M11 (Hx) (b, t) + M12 (Hx) (a, t) (4)

u(t) = M21 (Hx) (b, t) + M22 (Hx) (a, t) (5)

y(t) = C1 (Hx) (b, t) + C2 (Hx) (a, t). (6)

We assume that x takes values in Cn, P T

1 = P1, det(P1) 6= 0,

H > 0, and that rank[

M11 M12

M21 M22

C1 C2

]

= n + rank [ C1 C2 ].

– 192 –


Definition

Consider the system (3)–(6). This system is well-posed if there exists

a tf > 0 and mf such that the following holds:

1. The homogeneous p.d.e., i.e., u ≡ 0 has for any initial condition,

x0, a unique (weak) solution.

2. The following inequality holds for all smooth initial con ditions, and

all smooth inputs

‖x(tf)‖2+

∫ tf

0

‖y(t)‖2dt ≤ mf

[

‖x0‖2 +

∫ tf

0

‖u(t)‖2dt

]

,

where ‖ · ‖2 is the energy/Hamiltonian: 12

∫ b

axTHxdζ.

�

– 193 –


Remark

• The semigroup need not to be a contraction.

• So well-posedness gives you that you have a unique solution f or

every square integrable input function and every initial co ndition.

Furthermore, the output signal is always square integrable .

– 194 –


8.1 Well-posedness for simple PHS

Consider the (very) simple port-Hamiltonian system

∂x

∂t(ζ, t) = v

∂x

∂ζ(ζ, t)

vx(b, t) = u(t)

y(t) = vx(a, t),

with v a positive constant.

The norm (energy/Hamiltonian) is given by 12

∫ b

av‖f(ζ)‖2dζ.

– 195 –


We know that for smooth inputs and smooth initial conditions , there

exists a solution. For this (classical) solution the balanc e equation

reads as

d

dt‖x(t)‖2 =

1

2

[

|vx(ζ, t)|2]b

a

=1

2

[

|u(t)|2 − |y(t)|2]

Hence, we have that for all tf > 0

‖x(tf)‖2 +

∫ tf

0

‖y(t)‖2dt =

[

‖x0‖2 +

∫ tf

0

‖u(t)‖2dt

]

.

Since the smooth inputs and smooth initial conditions are de nse in X

and L2(0, tf), respectively, we find that this system is well-posed.

– 196 –


A similar result holds for the other (very) simple port-Hami ltonian

system

∂x

∂t(ζ, t) = −v

∂x

∂ζ(ζ, t)

vx(a, t) = u(t)

y(t) = vx(b, t),

with v a positive constant.

The energy is the same as for the previous p.d.e.

– 197 –


8.2 Characterization of well-posedness for PHS

Theorem

Under the conditions we have imposed, the following holds:

• If condition 1 holds, then automatically condition 2 holds.

• That is: If the homogeneous p.d.e., i.e., u ≡ 0, has a weak

solution, then the system (3)–(6) is well-posed.

• There is a matrix condition for checking condition 1.

• Same theorem holds for∂x∂t

(ζ, t) = P1∂∂ζ

(Hx) (ζ, t) + P0 (Hx) (ζ, t).

– 198 –


We want to give an idea of the proof. Therefor we do it for our

transmission line system with the following choice of bound ary input

and output

u(t) =

V (a, t)

V (b, t)

, y(t) =

I(a, t)

I(b, t)

. (7)

Now we preform the following steps:

1. Since V = Q/C = x1/C and I = φ/L = x2/L, we have

that

[M21, M22]=

0 0 1C

0

1C

0 0 0

, [C1, C2]=

0 0 0 1L

0 1L

0 0

.

– 199 –


2. Write the system in the ”characteristic”. That is diagona lize

P1H =(

0 −1

C−1

L0

)

. Hence

P1H=

1 1

−√

LC

√

LC

√

1LC

0

0 −√

1LC

1 −√

CL

1√

CL

1

2.

In the new variables ( characteristics )

ξ1

ξ2

:=

1 −√

CL

1√

CL

x1

x2

the transmission line becomes

∂ξ1

∂t=

√

1

LC

∂ξ1

∂ζ,

∂ξ2

∂t= −

√

1

LC

∂ξ2

∂ζ.

– 200 –


The “perfect” input and output for this system is

us(t) =

ξ1(b, t)

ξ2(a, t)

, ys(t) =

ξ1(a, t)

ξ2(b, t)

. (8)

3. Write our input-output pair in this input-output pair.

u(t) =

0 12C

12C

0

us(t) +

12C

0

0 12C

ys(t)

= Kus(t) + Qys(t), (9)

– 201 –


y(t) =

0 1

2√

LC

−1

2√

LC0

us(t) +

−1

2√

LC0

0 1

2√

LC

ys(t)

= O1us(t) + O2ys(t). (10)

– 202 –


We regard the system with input/output u, y as a feedback of the

system with input/output us, ys, i.e.,

−u(t)

us(t)

y(t)

ys(t)Gs(s)

Q

K−1 O2

O1

Now Gs has feed-through zero, and so any feedback is allowed.

Only condition: K−1 exists.

– 203 –


References

1. R.F. Curtain and H. Zwart, An Introduction to Infinite-Dimensional

Linear Systems Theory , Springer-Verlag, New York, 1995.

2. Y. Le Gorrec, H. Zwart, and B. Maschke, “Dirac structures a nd

boundary control systems associated with skew-symmetric

differential operators,” SIAM J. Control and Optim. , vol. 44, no. 2,

pp. 1864–1892, 2005.

3. J.A. Villegas, A Port-Hamiltonian Approach to Distributed

Parameter Systems , Ph.D. thesis, Department of Applied

Mathematics, University of Twente, Enschede, The Netherla nds,

May 2007, available at

http://www.eng.ox.ac.uk/control/peopleAJV.shtml.

– 204 –


4. B. Jacob and H. Zwart, Linear Port-Hamiltonian Systems on

Infinite-Dimensional Spaces , Operator Theory: Advances and

Applications, vol. 223, Birkh auser, 2012.

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hans zwart, university of twente - startseite tu …...this follows from the strong continuity and...

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