BSc Thesis Applied Mathematics
A study on backstepping boundary
control for two classes of linear
port-Hamiltonian systems
E.M. van der Meer
Supervisor: prof. dr. H.J. Zwart
January 22, 2020
Preface
This paper has been written as a Bachelor thesis for the study Applied Mathematics at the Uni-versity of Twente. I’ve enjoyed studying Applied Mathematics at the UniUni-versity of Twente im-mensely and I am proud of this final product.
First, I would like to thank prof. dr. H.J. Zwart for his help in finding a suitable bachelor project and for supervising it. This project fit my preferences beautifully and was challenging and rewarding to do.
E.M. van der Meer
January 22, 2020
Abstract
For two classes of linear port-Hamiltonian systems with constant parameters backstep-ping boundary control is investigated. An exponentially stable port-Hamiltonian system with homogeneous boundary conditions is for both classes the target system and the goal is to use linear multiplicative coordinate transformations. For the first class a coordinate transformation based on a multiplicative operator suffices and the condition for its exis-tence is algebraic. Using this a boundary controller is constructed. For the second class it shown that a multiplication mapping does not suffice.
Keywords: Backstepping, Boundary control, Linear port-Hamiltonian systems, Multiplica-tive coordinate transformation.
1 Introduction
Many physical systems described by partial differential equations (PDE’s) are controlled through their boundaries. To stabilize these systems, backstepping boundary control has been success-fully applied to, among other examples, the wave equation, the slender Timoshenko beam and the linearized Saint-Venant–Exner model [1], [4], [5], [7]. The construction of a stabilizing con-troller has been done by mapping the controlled system to an exponentially stable target sys-tem. To achieve this mapping from the PDE into a target PDE, an invertible Volterra integral coordinate transformation has to be found. The finding of such a Voltera integral mapping has been proven to be quite difficult and is usually done on specific cases for different classes of PDE’s.
In the study of physical systems the theory of port-Hamiltonian systems has been devel-oped and due to the advantages of this framework research has focused on the control of port-Hamiltonian systems [9]. A backstepping approach has been used to develop a boundary con-troller for a linear port-Hamiltonian system with constant parameters [6]. Again an exponen-tially stable target system is used but a linear multiplicative coordinate transformation suffices to achieve an invertible mapping. The condition for the existence of this transformation is an algebraic equation depending on the boundaries. Hence this transformation is simpler than the Volterra integral mapping and still allows the computation of the desired boundary con-troller.
class of port-Hamiltonian systems it is deduced that a linear multiplicative coordinate trans-formation cannot lead to the desired result. Instead, a short study on the mapping shows that a Volterra integral coordinate transformation is likely to be necessary.
This paper is organized as follows. In Chapter 2 the first class of linear port-Hamiltonian systems is introduced, after which in Section 2.1 the backstepping boundary controller is pre-sented. In Chapter 3 the second class of linear port-Hamiltonian systems is introduced, after which in Section 3.1 it is shown that a multiplication mapping cannot give the desired result. In Section 3.2 the form of the mapping is further studied. Finally, in Chapter 4 are some con-cluding remarks and future directions.
2 First class of port-Hamiltonian systems
LetMn(R) denote the space of realn×nmatrices and letMn(C) denote the space of complex
n×n matrices. The first class of port-Hamiltonian systems under study is described by the following partial differential equation (PDE):
∂xe(t,ζ)
∂t =P1 ∂
∂ζ(xe(t,ζ))+P0xe(t,ζ), (1)
ζ∈(0, 1), whereP1∈Mn(R) is non-singular and symmetric,P0∈Mn(R) andxetakes values in
Rn. The variabletrepresents time. Furthermore,P
1andP0are simultaneously diagonalizable. There are homogeneous and controlled boundary conditions. That is, there are matrices of appropriate sizes such that:
u(t)=WfB,1
· e
x(t, 1)
e
x(t, 0)
¸
(2)
and
0=WfB,2
· e
x(t, 1)
e
x(t, 0)
¸
. (3)
Here (2) represents the controlled boundary conditions, thusu(t) is the control, while (3) rep-resents the homogeneous boundary conditions.
Define the state spaceX asX =L2((0, 1);Rn) with the standard inner product〈xe1,xe2〉and
norm||xe||
2
= 〈xe,xe〉. The Sobolev space of orderk is denoted by H
k((0, 1),Rn), which is the
subspace of the state spaceXwhere the elementsxeand their derivatives up to orderkhave a
finite norm.
Associated with the homogenous PDE we define the operatorExe=P1ddζ(xe) with domain
D(E)=
½ e
x∈H1((0, 1),Rn)|
· e
x(1)
e
x(0)
¸
∈kerWfB ¾
, (4)
whereWfB=
· f
WB,1
f
WB,2
¸
.
Assumption 2.1. We assume that for the operator E the following hold:
1. The matrixWfB is an n×2n matrix of full rank.
2. Forxe0∈D(E)we have〈Exe0,xe0〉 ≤0.
2.1 The Backstepping Controller
Now a stabilizing controller for the port-Hamiltonian system (1) will be constructed. For this the port-Hamiltonian system will be written into a different form, after which it is mapped onto an exponentially stable target system. Then a coordinate transformation will be derived based on a multiplicative operator.
Lemma 2.1. The linear port-Hamiltonian system (1) can be rewritten into the form
∂x(t,ζ) ∂t =Λ
∂
∂ζ(x(t,ζ))+M x(t,ζ), (5)
ζ∈(0, 1), whereΛ=diag(λi)∈Mn(R)with non-zero elements on the diagonal and
M=diag(µi)∈Mn(C). The boundary conditions are
u(t)=WB,1
·
x(t, 1) x(t, 0)
¸
(6)
and
0=WB,2
·
x(t, 1) x(t, 0)
¸
. (7)
Proof. SinceP1is a real, symmetric, nonsingular matrix, there exists a matrixJ ∈Mn(C) such
that
Λ=J−1P1J=JTP1J (8)
whereΛis diagonal with non-zero elements on the diagonal [8]. The columns of Jconsists of the eigenvectors ofP1and the eigenvalues ofP1form the diagonal ofΛ[2]. SinceP1is a real symmetric matrix, all eigenvalues are real, thusΛ= diag (λi)∈Mn(R) [8]. Notice that the matrix
Jneed not be unique, for example this matrix is not unique whenP1equals the identity matrix. It is given thatP1andP0are simultaneously diagonalizable. Thus the matrixJ can be cho-sen such that
M=J−1P0J (9)
is also diagonal. LetM= diag (µi)∈Mn(C).
Definingx=J−1xeand using (8) and (9) we can rewrite (1) to (5). The boundary conditions
also become of the desired form withWB,1=WfB,1
·
J 0 0 J
¸
andWB,2=WfB,2
·
J 0 0 J
¸
.
Theorem 2.1. Beside the notation as introduced above, we define for c∈R,c>0
Υ(ζ)=
e
¡c+µ1
λ1
¢
ζ . . . 0
..
. . .. ...
0 . . . e
¡c+µn
λn
¢
ζ
. (10)
If there exists an invertible, diagonal matrix A∈Mn(C)such that or all p,q∈Rnsatisfying
·
p q
¸
we have that
·
J AΥ(1)J−1p J A J−1q
¸
∈ker
· f
WB,1
f
WB,2
¸
, (12)
then the boundary controller
u(t)= −WfB,1
·J(AΥ(1)
−I)J−1
e
x(t, 1) J(A−I)J−1xe(t, 0)
¸
(13)
stabilizes the port-Hamiltonian system (1). Furthermore, there exists a (positive) constant m0 such that solutions of the closed loop system satisfy
||xe(t)|| ≤m0e
−c t
||xe(0)||. (14)
Proof. In order to prove that the given boundary controller exponentially stabilizes the port-Hamiltonian system (1), it will be mapped onto the following target system
∂ez(t,ζ)
∂t =P1 ∂
∂ζ(ez(t,ζ))−c Ize(t,ζ), (15)
wherec∈R,c>0 and with homogeneous boundary conditions
0=
· f
WB,1
f
WB,2
¸ · e
z(t, 1)
e
z(t, 0)
¸
. (16)
Using the assumptions, it can be proven that the target system is exponentially stable, which has been done in Appendix A.1. Furthermore, for smooth initial conditions
||ze(t)|| ≤e
−c t
||ez(0)||. (17)
Defining z=J−1
e
z and using (8) the target system can equivalently be formulated as the diagonal target system
∂z ∂t =Λ
∂
∂ζ(z)−c I z, (18)
with homogeneous boundary conditions
0=
·W
B,1 WB,2
¸ ·z(t, 1)
z(t, 0)
¸
. (19)
Now the rewritten port-Hamiltonian system (5) will be mapped to the rewritten target sys-tem (18). Consider the coordinate transformation
z(t,ζ)=x(t,ζ)+Q(x(t,ζ)), (20)
whereQ is a bounded linear mapping fromL2((0, 1);Rn) toL2((0, 1);Rn) independent oft. By taking the partial derivative with respect to time one obtains
∂z ∂t =
∂x ∂t +Q
µ∂x
∂t
¶
=Λ∂x
∂ζ+M x+Q
µ
Λ∂x ∂ζ
¶
+Q(M x). (21)
On the other hand
∂z ∂t =Λ
∂z
∂ζ−c z=Λ ∂x ∂ζ+Λ
∂Q(x)
Thus the following equation should be satisfied
Λ∂Q(x) ∂ζ =Q
µ
Λ∂x
∂ζ+(c+M)x
¶
+(c+M)x. (23)
In order to solve the equation above, we first try to solve it for specialxof the formx(ζ)=
esζvwiths∈C,v∈Cn. Thus one obtains
Λ∂Q(esζv) ∂ζ =Q
µ
(sΛ+c+M)esζv
¶
+(c+M)esζv. (24)
Consider the particularvgiven byv1=
£
1 0 ... 0¤T
, then the equation above becomes
Λ∂Q(e
sζv1)
∂ζ =Q
µ
(sλ1+c+µ1)esζv1
¶
+(c+µ1)esζv1. (25)
To computeQ(esζv1) we write
Q(esζv1)=
qs,1(ζ)
ps,1(ζ)
.. . ps,n−1(ζ)
. (26)
Using this notation the differential equation (25) gives for the top element ofQ(esζv1):
λ1∂qs,1(ζ)
∂ζ =(sλ1+c+µ1)qs,1(ζ)+(c+µ1)e
sζ, (27)
which is equivalent to
∂qs,1(ζ)
∂ζ =
µ
s+c+µ1
λ1
¶
qs,1(ζ)+
c+µ1
λ1 e
sζ, (28)
and thus implies that
qs,1(ζ)=αs,1e
³
s+c+λ1µ1
´
ζ
−esζ, (29)
withαs,1∈C.
Equation (25) implies that the second element ofQ(esζv1) satisfies
λ2∂ps,1(ζ)
∂ζ =(sλ1+c+µ1)ps,1(ζ) (30)
which admits as solutionps,1=0. By (25) the following elements ofQ(esζv1) admit the same solution, thus
Q(esζv1)=
qs,1(ζ)
0 .. . 0
We can argue similarly for the other basis vectors ofRn. Thus, using the linearity ofQ, we obtain
Q(esζv)=
qs,1(ζ) 0 . . . 0
0 qs,2(ζ) . . . 0
..
. . .. . .. ... 0 . . . 0 qs,n(ζ)
v (32)
with qs,i(ζ) =αs,ie
¡
s+c+µi
λi
¢
ζ
−esζ for i =1, ...,n. Since Q is a bounded linear mapping from L2((0, 1);Rn) toL2((0, 1);Rn), it is necessary that the mapping does not depend ons. Thus we impose thatαs,i≡αi∈Cand it follows that
Q(esζv)=
α1e
¡c+µ1 λ1
¢
ζ
−1 . . . 0 ..
. . .. ...
0 . . . αne
¡c+µn
λn ¢ ζ −1
esζv. (33)
So forx(ζ)=esζvwe see that
Q(x(ζ))=diag³αie
¡c+µi
λi
¢
ζ
−1´x(ζ). (34)
SinceQis linear and the subspace spanned by exponential functions lies dense inL2((0, 1);Rn), Q must be a multiplicative operator. This has been proven in Appendix A.2. Thus the coordi-nate transformation becomes
z(t,ζ)=Y(ζ)x(t,ζ), (35)
with
Y(ζ)=
α1e
¡c+µ1 λ1
¢
ζ . . . 0
..
. . .. ...
0 . . . αne
¡c+µn
λn ¢ ζ . (36)
It remains to show thatY(ζ) is invertible and that the boundary conditions can be satisfied. For this we writeY as
Y(ζ)=AΥ(ζ) (37)
withA=diag (αi) and
Υ(ζ)=
e
¡c+µ1
λ1
¢
ζ . . . 0
..
. . .. ...
0 . . . e
¡c+µn
λn ¢ ζ . (38)
NowY is invertible if and only if all alpha’s are nonzero or equivalently A is invertible. The boundary conditions of the target system are given by
0=
· f
WB,1
f
WB,2
Recalling thatze=J z,z=AΥxandx=J
−1
e
x, we can rewrite the above as
0=
· f
WB,1
f
WB,2
¸ ·
J AΥ(1)J−1xe(1) J A J−1xe(0)
¸
. (40)
By the assumption in Theorem 2.1, we can transform, boundedly and invertible, our system into the target system. From the first line of (40) we have that
0=WfB,1 ·
J AΥ(1)J−1xe(t, 1)
J A J−1xe(t, 0)
¸
(41)
= WfB,1
· e
x(t, 1) x(t, 0)
¸
+WfB,1
·J
(AΥ(1)−I)J−1
e
x(t, 1) J(A−I)J−1xe(t, 0)
¸
(42)
= u(t)+WfB,1
·
J(AΥ(1)−I)J−1xe(t, 1) J(A−I)J−1
e
x(t, 0)
¸
(43)
from which the stabilizing input follows. Since only invertible state transformations were used between the port-Hamiltonian systems (1) and the target systems (15), the closed loop has the same decay rate as the target systems. Thus the theorem has been proven.
By Theorem 2.1, the port-Hamiltonian system (1) can be exponentially stabilized if the con-ditions in the theorem are met. Notice that if the port-Hamiltonian system (1) hasncontrols, then there are only controlled boundary conditions and the homogeneous boundary condi-tions (3) are absent. It follows that in this case the condicondi-tions in the theorem are satisfied, thus such a system is exponentially stabilizable.
Unfortunately, many linear port-Hamiltonian systems are not covered by Theorem 2.1. Consider for example the linear port-Hamiltonian system for the uniform Timoshenko beam where the constant coefficients have been chosen equal to 1. This example is described by the following PDE [3]:
∂x(t,ζ) ∂t =P1
∂
∂ζ(x(t,ζ))+P0x(t,ζ) (44)
=
0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0
∂
∂ζx(t,ζ)+
0 0 0 −1 0 0 0 0 0 0 0 0 1 0 0 0
x(t,ζ). (45)
SinceP1andP0do not commute, that isP0P16=P1P0, they are not simultaneously diagonaliz-able [2]. Therefore Theorem 2.1 cannot be applied. Notice though thatP0is skew symmetric. The general class of linear port-Hamiltonian systems whereP0is skew symmetric will be stud-ied in Chapter 3.
3 Second class of port-Hamiltonian systems
For the second class of port-Hamiltonian systems we will focus on the two dimensional case. The port-Hamiltonian systems under study are described by the following partial differential equation (PDE):
∂xe(t,ζ)
∂t =P1 ∂
∂ζ(xe(t,ζ))+P0xe(t,ζ), (46)
ζ∈(0, 1), whereP1∈M2(R) is non-singular and symmetric,P0∈M2(R) is skew symmetric and
e
3.1 The absence of a multiplication mapping
By trying to map the port-Hamiltonian system (46) to a generalized form of exponentially stable target systems via a multiplicative operator, it will be shown that a multiplication mapping is not possible. Thus constructing a stabilizing controller of the form found in Chapter 2 is not possible.
Lemma 3.1. The linear port-Hamiltonian system (46) can be rewritten into the form
∂x(t,ζ) ∂t =Λ
∂
∂ζ(x(t,ζ))+M x(t,ζ), (47)
ζ∈(0, 1), whereΛ=
·
λ1 0 0 λ2
¸
∈M2(R)with non-zero elements on the diagonal and
M=
·
0 µ
−µ 0
¸
∈M2(R). The boundary conditions are
u(t)=WB,1
·x(t, 1)
x(t, 0)
¸
(48)
and
0=WB,2
·
x(t, 1) x(t, 0)
¸
. (49)
Proof. SinceP1is is a real, symmetric, nonsingular matrix, (8) holds for a matrixJ∈M2(R) [8]. SinceP0is skew symmetric, it follows that
(JTP0J)T=JTP0TJ= −JTP0J. (50)
ThusJTP0J∈M2(R) also skew symmetric. Since it is a real skew symmetric matrix the entries
on the diagonal equal zero. LetM=JTP0J=J−1P0J=
·
0 µ
−µ 0
¸
.
Definingx=J−1xe=J
T
e
xand using (8) andM as above we can rewrite (46) as (47). For the
boundary conditionsWB,1=WfB,1
·J 0
0 J
¸
andWB,2=WfB,2
·J 0
0 J
¸
.
We are not interested in the case whenP1andP0of (46) are simultaneously diagonalizable, since this coincides with the case discussed in Chapter 2. SinceP1is symmetric andP0is skew symmetric, they are both diagonalizable per definition. HereP0 is only diagonalizable over
C, not overR. They are simultaneously diagonalizable if and only if they commute, that is if P1P0=P0P1[2]. Thus we assume from now on thatP1andP0do not commute. IfP1andP0do not commute, then this is equivalent with thatΛ=J−1P1J andM=J−1P0J do not commute. To ensure thatΛandM do not commute, we have to assume from now on thatλ16=λ2and µ6=0.
To show that a multiplication mapping to an exponentially stable target system is not pos-sible, the chosen target system will be of a generalized form instead of the more specific form (15) used in Chapter 2. The target system is
∂ez(t,ζ)
∂t =P1 ∂
whereP2∈M2(R) is such thatxTP2x>0 for anyx∈R2withx6=0 but not necessarily symmetric. It follows thatP2+P2T is positive definite. The target system has the homogeneous boundary conditions
0=
· f
WB,1
f
WB,2
¸ · e
z(t, 1)
e
z(t, 0)
¸
. (52)
Using Assumption 2.1, it can proven that the target system is exponentially stable. This proof can be found in Appendix A.1.
Sincex=J ydefines a 1-1 correspondence betweenx∈R2andy∈R2, we conclude from
xTP2x=yTJTP2J y=yTP y (53)
thatP=J−1P2J=JTP2J∈M2(R) also has the property thatxTP x>0 for anyx∈R2withx6=0.
LetP =
·
a b c d
¸
. Definingz=J−1
e
z=JTzeand using (8) andP as above the target system can equivalently be formulated as the target system
∂z ∂t =Λ
∂
∂ζ(z)−P z, (54)
with homogeneous boundary conditions
0=
·W
B,1 WB,2
¸ ·z(t, 1)
z(t, 0)
¸
(55)
Now the rewritten port-Hamiltonian systems (47) will be mapped to the rewritten target system (54). Consider the coordinate transformation
z(t,ζ)=x(t,ζ)+Q(x(t,ζ)), (56)
whereQ is a bounded linear mapping fromL2((0, 1);R2) toL2((0, 1);R2) independent oft. By taking the partial derivative with respect to time one obtains
∂z ∂t =
∂x ∂t +Q
µ∂x
∂t
¶
=Λ∂x
∂ζ+M x+Q
µ
Λ∂x ∂ζ
¶
+Q(M x). (57)
On the other hand ∂z
∂t =Λ ∂z
∂ζ−P z=Λ ∂x ∂ζ+Λ
∂Q(x)
∂ζ −P x−PQ(x). (58)
Thus the following equation should be satisfied
Λ∂Q(x) ∂ζ =Q
µ
Λ∂x ∂ζ+M x
¶
+PQ(x)+(P+M)x. (59)
In order to solve the equation above, we first try to solve it for specialxof the formx(ζ)=
esζvwiths∈C,v∈C2. Thus one obtains
Λ∂Q(e
sζv)
∂ζ =Q
³
(sΛ+M)esζv´+PQ(esζv)+(P+M)esζv. (60)
Consider the particular solutions given byv1=£1 0¤
T
andv2=£0 1¤
T
and write:
Q(esζv1)=
·
fs,1(ζ)
gs,1(ζ)
¸
Q(esζv2)=
·
fs,2(ζ)
gs,2(ζ)
¸
. (62)
Using this notation it follows that forx(ζ)=esζv1
Λ∂Q(esζv1)
∂ζ =Q(sλ1e
sζv
1−µesζv2)+PQ(esζv1)+aesζv1+(c−µ)esζv2 (63)
and thus
λ1∂fs,1(ζ)
∂ζ =(sλ1+a)fs,1(ζ)−µfs,2(ζ)+bgs,1(ζ)+aesζ (64)
and
λ2∂gs,1(ζ)
∂ζ =(sλ1+d)gs,1(ζ)−µgs,2(ζ)+c fs,1(ζ)+(c−µ)e
sζ. (65)
Similarly forx(ζ)=esζv2
Λ∂Q(e
sζv2)
∂ζ =Q(sλ2e
sζv
2+µesζv1)+PQ(esζv2)+(b+µ)esζv1+d esζv2 (66)
and thus
λ1∂fs,2(ζ)
∂ζ =(sλ2+a)fs,2(ζ)+µfs,1(ζ)+bps,2(ζ)+(b+µ)e
sζ (67)
and
λ2∂gs,2(ζ)
∂ζ =(sλ2+d)gs,2(ζ)+µgs,1(ζ)+c fs,2(ζ)+d e
sζ. (68)
It follows that to find the desired mapping the following inhomogeneous linear system of differential equations of dimension 4 and order 1 has to be solved:
∂ ∂ζ
fs,1(ζ)
gs,1(ζ)
fs,2(ζ)
gs,2(ζ)
=
sλ1+a
λ1
b
λ1
−µ
λ1 0
c
λ2
sλ1+d
λ2 0
−µ
λ2
µ
λ1 0
sλ2+a
λ1
b
λ1 0 λµ
2
c
λ2
sλ2+d
λ2
fs,1(ζ)
gs,1(ζ)
fs,2(ζ)
gs,2(ζ)
+ a λ1 c−µ λ2
b+µ
λ1 d λ2
esζ (69)
This linear system can be rewritten as follows:
∂qs(ζ)
∂ζ = µ
1 0 0 0 0 λ1
λ2 0 0 0 0 λ2
λ1 0 0 0 0 1
s+ a λ1 b λ1 −µ λ1 0
c
λ2
d
λ2 0
−µ
λ2
µ λ1 0
a
λ1
b
λ1 0 λµ
2 c λ2 d λ2 ¶
qs(ζ)+
a λ1 c−µ λ2
b+µ
λ1 d λ2
esζ (70)
=(K s+A)qs(ζ)+B esζ (71)
whereqs=£fs,1 gs,1 fs,2 gs,2¤
T
of the formqs(ζ)=Y(ζ)esζ, with the functionY(ζ)=£y1(ζ) y2(ζ) y3(ζ) y4(ζ)¤∈C4not
de-pending ons. Imposingqs(ζ)=Y(ζ)esζyields
d Y(ζ) dζ e
sζ+sY(ζ)esζ=(K s+A)Y(ζ)esζ+B esζ (72)
thus
d Y(ζ)
dζ +sY(ζ)=((K s+A)Y(ζ)+B. (73)
SinceY(ζ) is not allowed to depend ons, it follows that we require that
Y(ζ)=K Y(ζ) (74)
and this statement is only true ifλ1=λ2 or ify2=y3=0. Ifλ1=λ2, thenP1 andP0 from the port-Hamiltonian system (46) are simultaneously diagonalizable and this case has already been solved in Chapter 2. This is why we have assumed thatλ16=λ2. If y2=y3=0, then (73) reduces to d dζ y1 0 0 y4 = a λ1 b λ1 −µ λ1 0
c
λ2
d
λ2 0
−µ
λ2
µ λ1 0
a
λ1
b
λ1 0 λµ
2 c λ2 d λ2 y1 0 0 y4 + a λ1 c−µ λ2
b+µ
λ1 d λ2 . (75)
Thus there are two differential equations and two regular equations:
d y1 dζ =
a λ1y1+
a
λ1 (76)
d y4 dζ =
d λ2y4+
d
λ2 (77)
0= c
λ2y1− µ λ2y4+
c−µ
λ2 (78)
0= µ
λ1y1+ b λ1y4+
b+µ
λ1 (79)
Remember that the valuesa,b,candd from the matrixP are not defined other than the property thatxTP x>0 for anyx∈R2withx6=0 must hold. Thus it is possible to takeb=c=0, since ifa>0 andd>0 that property still holds. Since we have assumed thatµ6=0 to ensure that P1andP0are not simultaneously diagonalizable, it follows from (78) and (79) thaty1=y4= −1, which also satisfies (76) and (77). Thus
Y(ζ)=
−1 0 0 −1 . (80)
Returning to the bounded linear mapping Q, it follows that for this particular solution
Q(esζv1)=
·
fs,1(ζ)
gs,1(ζ)
¸ = · −1 0 ¸
Q(esζv2)=
·
fs,2(ζ)
gs,2(ζ)
¸ = · 0 −1 ¸
esζ. (82)
Using the linearity of Q we obtain
Q(esζv)=
·
−1 0 0 −1
¸
esζv. (83)
SinceQis linear and the subspace spanned by exponential functions lies dense inL2((0, 1);R2), Qmust be a multiplicative operator (see Appendix A.2), thus
Q(x(t,ζ))=
·
−1 0 0 −1
¸
x(t,ζ)= −I x(t,ζ). (84)
The coordinate transformation becomes
z(t,ζ)=x(t,ζ)+Q(x(t,ζ))=x(t,ζ)−I x(t,ζ)=0. (85)
Clearly this is not the desired result, since this does not give a useful mapping to determine the stabilizing input. Thus when mapping the port-Hamiltonian system (46) to the exponen-tially stable target system (51), it is impossible to find a multiplication mapping.
3.2 The form of the mapping
As we have seen in the previous section, a multiplication mapping from the port-Hamiltonian system (46) to the exponentially stable target system (51) is not possible ifP1andP0 do not commute and thus are not simultaneously diagonalizable. This is equivalent withλ16=λ2and µ6=0. In this section the mapping will be further investigated. To solve the linear system (69), the Laplace transformation is used. Due to the complexity of this solution, we will focus on the target system whereP2=P =aI witha>0, thusa=d>0 andb=c=0. For this target system the inhomogeneous linear system (69) reduces to the two following inhomogeneous linear systems of dimension 2 and order 1
∂ ∂ζ
·
fs,1(ζ) fs,2(ζ)
¸
=
"sλ1+a
λ1 −
µ λ1
µ λ1
sλ2+a
λ1
#·
fs,1(ζ) fs,2(ζ)
¸ + "a λ1 µ λ1 #
esζ (86)
and
∂ ∂ζ
·g
s,1(ζ) gs,2(ζ)
¸
=
"sλ1+a
λ2 −
µ λ2
µ λ2
sλ2+a
λ2
# ·g
s,1(ζ) gs,2(ζ)
¸
+
"
−λµ 2
a
λ2
#
esζ. (87)
To solve the two linear systems, the Laplace transform will be used, transforming the vari-ableζto the variable². Thus the two linear systems (86) and (87) are transformed into
²
·F
s,1(²) Fs,2(²)
¸
=
"sλ1+a
λ1 −
µ λ1
µ λ1
sλ2+a
λ1
# ·F
s,1(²) Fs,2(²
¸
+
"a
λµ1
λ1
#
1 ²−s+
·f
s,1(0) fs,2(0)
¸
(88)
and
²
·G
s,1(²) Gs,2(²)
¸
=
"sλ
1+a
λ2 −
µ λ2
µ λ2
sλ2+a
λ2
# ·G
s,1(²) Gs,2(²)
¸
+
"
−λµ2
a
λ2
#
1 ²−s+
·g
s,1(0) gs,2(0)
¸
The above can be rewritten into:
·F
s,1(²) Fs,2(²)
¸
= 1
(²−sλ1+a
λ1 )(²−
sλ2+a
λ1 )+
µ2
λ2 1
"
²−sλ2+a
λ1 −
µ λ1
µ
λ1 ²−
sλ1+a
λ1
# " a
λ1(µ²−s)+fs,1(0)
λ1(²−s)+fs,2(0)
#
(90)
and
·
Gs,1(²)
Gs,2(²)
¸
= 1
(²−sλ1+a
λ2 )(²−
sλ2+a
λ2 )+
µ2
λ2 2
"
²−sλ2+a
λ2 −
µ λ2
µ
λ2 ²−
sλ1+a
λ2
# "
−λ2(µ²−s)+gs,1(0)
a
λ2(²−s)+gs,2(0)
#
(91)
Solving (²−sλ1+a
λ1 )(²−
sλ2+a
λ1 )+
µ2
λ2 1 =
0 gives:
α+/α−=
1 2λ1
³
s(λ1+λ2)+2a±
q
s2(λ1−λ2)2−4µ2´. (92)
Similarly, solving (²−sλ1+a
λ2 )(²−
sλ2+a
λ2 )+
µ2
λ2 2=
0 gives:
β+/β−=
1 2λ2
³
s(λ1+λ2)+2a±
q
s2(λ1−λ2)2−4µ2´. (93)
Notice thatα+,α−,β+andβ−are the eigenvalues of the two square matrices in (86) and (87). Also notice that ifλ1=λ2orµ=0, these eigenvalues could be further simplified, allowing the existence of a multiplication mapping.
Now, using partial fraction decomposition, after tedious but straightforward calculation the following is obtained:
Fs,1(²)= −
1 ²−s+
hf,1 ²−α++
hf,2 ²−α−
, (94)
Fs,2(²)=
hf,3 ²−α++
hf,4 ²−α−
, (95)
Gs,1(²)=
hg,1 ²−β++
hg,2 ²−β−
, (96)
Gs,2(²)= −
1 ²−s+
hg,3 ²−β++
hg,4 ²−β−
, (97)
wherehf,i depends on s,fs,1(0) and fs,2(0) fori =1, 2, 3, 4 andhg,i depends on s,gs,1(0) and
gs,2(0) fori=1, 2, 3, 4.
Using the inverse Laplace transformation, it follows that all possible solutions are of the following form:
fs,1(ζ)= −1esζ+hf,1eα+ζ+hf,2eα−ζ, (98)
fs,2(ζ)=hf,3eα+ζ+hf,4eα−ζ, (99)
gs,2(ζ)= −1esζ+hg,3eβ+ζ+hg,4eβ−ζ. (101)
In this form the parts−1esζinfs,1(ζ) andgs,2(ζ) correspond to the solution if the mapping has to be independent ofs, see (81) and (82).
Thus one has to find a mapping which gives results of the form
Q(esζv1)=
·
fs,1(ζ)
gs,1(ζ)
¸
=
·
−1 0
¸
esζ+
·
hf,1eα+ζ+hf,2eα−ζ hg,1eβ+ζ+hg,2eβ−ζ
¸
(102)
and
Q(esζv2)=
·f
s,2(ζ) gs,2(ζ)
¸
=
· 0
−1
¸
esζ+
·h
f,3eα+ζ+hf,4eα−ζ hg,3eβ+ζ+hg,4eβ−ζ
¸
. (103)
Such a mapping will probably contain integrals, since we have assumed thatλ16=λ2and µ6=0, and a Volterra integral coordinate transformation is a good candidate. If a mapping is found, invertibility of the mapping and whether the boundary conditions of the port-Hamiltonian system (46) and the target system (51) can be simultaneously satisfied has to be verified.
4 Conclusion
A backstepping approach has been used to construct a boundary controller for a class of lin-ear port-Hamiltonian systems (1), which has constant parameters and the property that the matricesP1andP0are simultaneously diagonalizable. To construct this boundary controller, an exponentially stable target system (15) has been used. From this a multiplication mapping has been deduced after diagonalizing both the port-Hamiltonian system and the target sys-tem. The condition for the existence of this mapping, and thus of the constructed boundary controller, is an algebraic equation on the boundaries.
A similar approach has been used on a second class of linear port-Hamiltonian systems (46) in order to investigate whether a multiplication mapping could again yield the desired result. This class also has constant parameters, but nowP0is skew symmetric and it is assumed thatP1 and P0 are not simultaneously diagonalizable. A more general exponentially stable target system (51) has been used and again the port-Hamiltonian system and the target system have been diagonalized. It has been shown that a multiplication mapping does not yield a coordinate transformation to an exponentially stable target system.
References
[1] A. Diagne, M. Diagne, S. Tang, and M. Krstic. Backstepping stabilization of the linearized saint-venant–exner model.Automatica, 76:345–354, 2017. cited By 11.
[2] S. Friedberg, A. Insel, and L. Spence.Linear Algebra. Pearson, 2014. 4th ed.
[3] B. Jacob and H.J. Zwart. Linear port-Hamiltonian systems on infinite-dimensional spaces. Birkhäuser Basel, 2012. cited By 61.
[4] M. Krstic, A.A. Siranosian, and A. Smyshlyaev. Backstepping boundary controllers and ob-servers for the slender timoshenko beam: Part i - design.Proceedings of the American Con-trol Conference, 2006:2412–2417, 2006. cited By 58.
[5] M. Krstic, A.A. Siranosian, A. Smyshlyaev, and M. Bernent. Backstepping boundary con-trollers and observers for the slender timoshenko beam: Part ii-stability and simulations. Proceedings of the IEEE Conference on Decision and Control, pages 3938–3943, 2006. cited By 35.
[6] H. Ramirez, H. Zwart, Y. Le Gorrec, and A. Macchelli. On backstepping boundary control for a class of linear port-hamiltonian systems. 2017 IEEE 56th Annual Conference on Decision and Control, CDC 2017, 2018-January:658–663, 2018. cited By 0.
[7] A. Smyshlyaev, E. Cerpa, and M. Krstic. Boundary stabilization of a 1-d wave equation with in-domain antidamping.SIAM Journal on Control and Optimization, 48(6):4014–4031, 2010. cited By 95.
[8] F. Ulrich, W. Kern, and G. Still.Mathematical Optimization. Department of Applied Math-ematics, Faculty EWI, University of Twente, 2015.
A Appendix
A.1 Exponential stability of the target systems
Consider the system ∂ez(t,ζ)
∂t =P1 ∂
∂ζ(ez(t,ζ))−c Ize(t,ζ), (104)
ζ∈(0, 1), whereP1∈Mn(R) is non-singular and symmetric,zetakes values inR
nandc
∈R,c>0. It has homogeneous boundary conditions
0=WfB
· e
z(t, 1)
e
z(t, 0)
¸
(105)
where the matrixWfB is ann×2nmatrix of full rank.
Define the state spaceX asX=L2((0, 1);Rn) with the standard inner product〈xe1,ex2〉. The
Sobolev space of orderkis denoted byHk((0, 1),Rn). We define the operatorExe=P1
d
dζ(xe) with
domain
D(E)=
½ e
x∈H1((0, 1),Rn)|
· e
x(1)
e
x(0).
¸
∈kerWfB
¾
. (106)
We will only consider strong or classical solutions of the system (104), thusez∈D(E) for allze.
Theorem A.1. If〈Eez,ez〉 ≤0forze∈D(E), then system (104) is globally exponentially stable.
Proof. Consider the following candidate Lyapunov function, which is continuously differen-tiable and positive definite:
V(t)=1
2
Z 1
0 e
z(t,ζ)2dζ=1
2||ez(t)||
2
. (107)
Forze∈D(E) the time derivative ofV is
dV d t =
1 2
Z 1
0
2ezT∂ez
∂tdζ (108)
=
Z 1
0 e
zT³P1∂e z ∂ζ−c Ize
´
dζ (109)
=
Z 1
0 e zTP1∂e
z ∂ζdζ−c
Z 1
0 e
z2dζ (110)
= 〈Eez,ez〉 −c||ze(t)||
2 (111)
≤ −c||ez(t)||
2
= −2cV <0. ifez6=0 (112)
ThusdVd t is globally negative definite andV is a Lyapunov function. Furthermore we have that
dV
d t ≤ −2cV (113)
As a consequence of Grönwalls lemma
V(t)≤e−2c tV(0) (114)
which is equivalent to
||ze(t)||
2
≤e−2c t||ze(0)||
2 (115)
thus
||ze(t)|| ≤e
−c t
||ez(0)||. (116)
Now consider the system
∂ez(t,ζ)
∂t =P1 ∂
∂ζ(ez(t,ζ))−P2ez(t,ζ), (117)
whereP2∈M2(R) is such thatxTP2x>0 for anyx∈R2withx6=0 and otherwise the same as system (104).
Theorem A.2. If〈Eez,ez〉 ≤0forze∈D(E), then system (117) is globally exponentially stable.
Proof. Consider the following candidate Lyapunov function, which is continuously differen-tiable and is positive definite:
V(t)=1
2
Z 1
0 e
z(t,ζ)2dζ=1
2||ez||
2. (118)
Now the time derivative ofV is
dV d t =
1 2
Z 1
0
2ezT∂ez
∂tdζ (119)
=
Z 1
0 e
zT³P1∂e z ∂ζ−P2ze
´
dζ (120)
=
Z 1
0 ez
TP
1∂e z ∂ζdζ−
Z 1
0 ez
TP
2zde ζ (121)
= 〈Eez,ez〉 −
Z 1
0 e
zTP2ezdζ (122)
≤ −
Z 1
0 e
zTP2ezdζ. (123)
By the given property ofP2, there exists a constantc>0 such thatxTP2x≥xTc I x=c x2for all x∈Rn. It follows that
dV d t =≤ −
Z 1
0 e
zTP2ezdζ (124)
≤ −c
Z 1
0 e
z2dζ= −c||ez(t)||
2
= −2cV <0. ifez6=0 (125)
ThusdVd t is globally negative definite andV is a Lyapunov function. Similarly as in the proof of Theorem A.1, we have that
dV
d t ≤ −2cV (126)
thus
||ze(t)|| ≤e
−c t
||ez(0)||. (127)
and the system is globally exponentially stable.
A.2 The multiplication mappingQ
Lemma A.1. The subspace spanned by exponential functions lie dense in L2((0, 1);Rn).
Proof. It can be seen by the Fourier series that the subspace spanned byeiπkζ withk∈Zlie dense inL2(0, 1), since the Fourier series states that for every functionf(ζ)∈L2(0, 1)
f(ζ)= X
k∈Z
fkeiπkζ= lim N→∞
N
X
k=−N
fkeiπkζ (128)
wherefkare the Fourier coefficients.
Now consider a function f(ζ)=£f1(ζ) f2(ζ) ... fn(ζ)¤ T
∈L2((0, 1);Rn). By the Fourier series, there exist Fourier coefficientsfi,kwithi=1, 2, ...,nsuch that for each element off(ζ)
fi(ζ)=
X
k∈Z
fi,keiπkζ= lim
N→∞
N
X
k=−N
fi,keiπkζ. (129)
By takingvk=
£
f1,k f2,k ... f3,k
¤T
, it follows that
f(ζ)= X
k∈Z
eiπkζvk= lim
N→∞
N
X
k=−N
eiπkζvk. (130)
withvk ∈Cn. Thus the subspace spanned byeiπkζv, v∈Cn, lie dense inL2((0, 1);Rn). Since
the subspace spanned byeiπkζvis contained in the subspace spanned byesζv, the subspace spanned byesζvmust also lie dense inL2((0, 1);Rn).
Theorem A.3. If the mapping Q from L2((0, 1);Rn)to L2((0, 1);Rn)is linear and bounded and there exists a function Y(ζ)∈Mn(C)such that Q(esζv)=Y(ζ)esζv for all s∈Cand v∈Cn, then
Q must be a multiplication operator. That is, for any f(ζ)∈L2((0, 1);Rn), Q(f(ζ))=Y(ζ)f(ζ).
Proof. First, we prove that the mappingQmust be continuous, given that it is linear and bounded. Notice that since Q is linear by definition
Q(αx1+βx2)=αQ(x1)+βQ(x2) (131)
for anyx1,x2∈L2((0, 1);Rn) andα,βconstants. Since Q is bounded, it follows that there exists some constantM∈Rsuch that
||Q(x)|| ≤M||x||. (132)
It immediately follows that
||Q(x1)−Q(x2)|| = ||Q(x1−x2)|| ≤M||x1−x2||. (133)
Thus for every²>0 there exists anδ>0, namelyδ=²/M, such that for allx1,x2∈L2((0, 1);Rn)
||x1−x2|| ≤δ⇒ ||Q(x1)−Q(x2)|| ≤². (134)
It follows that the mapping Q is continuous.
Now we will prove that the mapping must be multiplicative. Take anyf(ζ)∈L2((0, 1);Rn). From the lemma above we know that
f(ζ)= X
k∈Z
eiπkζvk= lim N→∞
N
X
k=−N
Thus
Q(f(ζ))=Q( lim
N→∞
N
X
k=−N
eiπkζvk) (136)
= lim
N→∞Q(
N
X
k=−N
eiπkζvk) by continuity (137)
= lim
N→∞ N
X
k=−N
Q(eiπkζvk) by linearity (138)
= lim
N→∞ N
X
k=−N
Y(ζ)eiπkζvk sinceQ(esζv)=Y(ζ)esζv (139)
=Y(ζ) lim
N→∞
N
X
k=−N
eiπkζvk (140)
=Y(ζ)f(ζ). (141)
