On Finite-Gain Stabilizability of Linear Systems Subject to Input Saturation
32
Transcript of On Finite-Gain Stabilizability of Linear Systems Subject to Input Saturation
On Finite Gain Stabilizability of
Linear Systems Subject to Input Saturation
�
Wensheng Liu Yacine Chitour Eduardo Sontag
Dept. of Mathematics, Rutgers University, New Brunswick, NJ 08903
abstract
This paper deals with (global) �nite-gain input/output stabilization of linear systems with satu-
rated controls. For neutrally stable systems, it is shown that the linear feedback law suggested by
the passivity approach indeed provides stability, with respect to every L
p
-norm. Explicit bounds on
closed-loop gains are obtained, and they are related to the norms for the respective systems without
saturation.
These results do not extend to the class of systems for which the state matrix has eigenvalues
on the imaginary axis with nonsimple (size > 1) Jordan blocks, contradicting what may be expected
from the fact that such systems are globally asymptotically stabilizable in the state-space sense; this
is shown in particular for the double integrator.
Keywords: input saturations, linear systems, �nite gain stability, Lyapunov functions, passivity
1 Introduction
In this work we are interested in those nonlinear systems which are obtained when cascading a linear
system with a memory-free input nonlinearity:
_x = Ax+ B�(u); y = Cx : (�)
The nonlinearity � is of a \saturation" type (de�nitions are given later). Figure 1 shows the type of
system being considered, where the linear part has transfer function W (s) and the function � shown
is the standard semilinear saturation (results will apply to more general �'s).
�
�
---
yu
W (s)
�
Figure 1: Input-Saturated Linear System
Linear systems with actuator saturation constitute one of the most important classes of nonlinear
systems encountered in practice. Surprisingly, until recently few general theoretical results were avail-
able regarding global feedback design problems for them. One such general result was given in [14],
which showed that global state-space stabilization for such systems is possible under the assumptions
�
This research was supported in part by US Air Force Grant AFOSR-91-0346.
E-mail addresses: fwliu,chitour,[email protected]
1
that all the eigenvalues of A are in the closed left-hand plane, plus stabilizability and detectability of
(A;B; C). (These conditions are best possible, since they are also necessary. The controller consists
of an observer followed by a smooth static nonlinearity.) For more recent work, see [20], which showed
|based upon techniques introduced in [16] for a particular case| how to simplify the controller that
had been proposed in [14]. See also [8] for closely related work showing that such systems can be
semiglobally (that is, on compacts) stabilized by means of linear feedback.
In this paper, we are interested in studying not merely closed-loop state-space stability, but also
stability with respect to measurement and actuator noise. This is the notion of stability that is
often found in input/output studies. The problem is to �nd a controller C so that the operator
(u
1
; u
2
) 7! (y
1
; y
2
) de�ned by the standard systems interconnection
y
1
= P (u
1
+ y
2
)
y
2
= C(u
2
+ y
1
)
is well-posed and �nite-gain stable, where P denotes the input/output behavior of the original plant
�. See Figure 1.
m
m�
-
6
�
?
-
C
P
y
2
u
2
y
1
u
1
Figure 2: Standard Closed-Loop
(In our main results, we will take for simplicity the initial state to be zero. However, nonzero initial
states can be studied as well, and some remarks in that regard are presented in a latter section of
the paper.) Once that such i/o stability is achieved, geometric operator-theoretic techniques can be
applied; see for instance [3] and references there. For other work on computing norms for nonlinear
systems in state space form, see for instance [18] and the references given there.
We focus in this paper in a case which would be trivial if one would only be interested in state
stability, namely that in which the original matrix A is stable, that is, there are no nontrivial Jor-
dan blocks for eigenvalues in the imaginary axis. (The whole point of [14] and [20] was of course
to deal with such possible nontrivial blocks, e.g. multiple integrators.) In this case, a standard pas-
sivity approach suggests the appropriate stabilization procedure. For instance, assume that � is the
identity (so the original system is linear), A + A
T
� 0, and C = B
T
. Then the system is passive,
with storage function V (x) = kxk
2
=2, since integrating the inequality dV (x(t))=dt � y(t)
T
u(t) gives
R
t
0
y(s)
T
u(s)ds � V (x(t)) � V (x(0)). Thus the negative feedback interconnection with the identity
(strictly passive system), that is, u = �y, results in �nite gain stability. For this calculation, and more
on passivity, see for instance [7] and the references given there. (For the use of the same formulas for
just state-space stabilization, but applying to linear systems with saturations, see [5] and [9]; see also
the discussion on the \Jurdevic-Quinn" method in [13].)
2
In this paper, we essentially generalize the passivity technique to systems with saturations. We
�rst establish �nite gain stability in the various p-norms, using linear state feedback stabilizers. Then
we show how outputs can be incorporated into the framework. Our work is very much in the spirit
of the well-known \absolute stability" area, but we have not been able to �nd a way to deduce our
results from that classical literature.
These results do not extend to the class of systems for which the state matrix has eigenvalues
on the imaginary axis with nonsimple (size> 1) Jordan blocks, contradicting what may be expected
from the fact that such systems are globally asymptotically stabilizable in the state-space sense; this
is shown in particular for the double integrator.
One remark on terminology. In the operator approach to nonlinear systems, see e.g. [19], a \system"
is typically de�ned as a partially de�ned operator between normed spaces, and \stability" means that
the domain of this operator is the entire space. In that context, �nite gain stability is the requirement
that the operator be everywhere de�ned and bounded; the norm of the operator is by de�nition the
gain of the system. In this paper, we use simply the term L
p
-stability to mean this stronger �nite gain
condition.
The reader is referred to the companion paper [2] for complementary results to those in this paper,
dealing with Lipschitz continuity (\incremental gain stability") and continuity of the operators in
question. The two papers are technically independent.
Organization of Paper
Section 2 provides de�nitions and statements of the main results, as well as some related comments.
Section 3 discusses the meaning of the assumptions on saturations. Proofs of the main results are
given in Section 4. Section 5 estimates gains in terms of the corresponding gains for systems without
saturation, in particular for p=2 (H
1
-norms). Results regarding nonzero initial states and global
asymptotic stability of the origin are collected in Section 6. Section 7 shows how to enlarge the class
of input nonlinearities even more, so as to include in one result non-saturations as well. The paper
closes with Section 8, which contains the double integrator counterexample.
2 Statements of Main Results
We introduce now the class of saturation functions to be considered, and state the main results on
�nite-gain stability. Some remarks are also provided. Proofs are deferred to a later section.
2.1 Saturation Functions
We next formally de�ne what we mean by a \saturation." Essentially, we ask only that this be a
function which has the same sign as its argument, stays away from zero at in�nity, is bounded, is not
horizontal near zero, and has a derivative that goes to zero at in�nity at a rate at least 1=t:
De�nition 1 We call � : IR! IR a saturation function if it satis�es the following two conditions:
(i) � is absolutely continuous, bounded and (1 + jtj)j�
0
(t)j is bounded for almost all t 2 IR;
(ii) t�(t) > 0 if t 6= 0, lim inf
t!0
�(t)
t
> 0 and lim inf
jtj!1
j�(t)j > 0.
3
For convenience we will simply call a saturation function � an S-function. We say that � is an
IR
n
-valued S-function if � = (�
1
; : : : ; �
n
)
T
where each component �
i
is an S-function and
�(x)
def
= (�
1
(x
1
); : : : ; �
n
(x
n
))
T
for x = (x
1
; : : : ; x
n
)
T
2 IR
n
. Here we use (� � �)
T
to denote the transpose of the vector (� � �).
Remark 1 It follows directly from De�nition 1 as well as the characterizations to be given in Section 3
that most reasonable saturation-type functions are indeed S-functions in our sense. Included are
arctan(t), tanh(t), and the standard saturation function �
0
(t) = sign(t)minfjtj; 1g, i.e.,
�
0
(t) =
8
>
<
>
:
1 if t > 1 ;
t if jtj � 1 ;
�1 if t < �1 :
2.2 L
p
-Stability
Consider the initialized control system (�) given by
_x = f(x; u) ;
x(0) = 0 ;
where the state x and the control u take respectively values in IR
n
and IR
m
. As this is not restrictive
for our purposes, and it simpli�es matters, we assume that the function f : IR
n
� IR
m
! IR
n
is globally
Lipschitz with respect to (x; u). (Most results will still be valid under Carath�eodory-type assumptions.)
Terminology for systems will be as in any standard reference, such as [13]. Notice that under the above
Lipschitz condition, the solution x of (�) corresponding to any input u 2 L
p
([0;1); IR
m
) for 1 � p � 1
is well de�ned for all t 2 [0;1), as an absolutely continuous function.
Throughout this paper, if � is a point in IR
n
, we use k�k = (
P
n
i=1
�
2
i
)
1=2
to denote the usual
Euclidean norm. For each matrix S, the norm kSk is the Frobenius norm given by kSk=Tr(SS
T
)
1=2
,
where Tr(�) denotes trace.
For each 1 � p � 1 and each p-th integrable (essentially bounded, for p = 1) vector-valued
function x 2 L
p
([0;1); IR
n
), we let kxk
L
pdenote the usual L
p
-norms:
kxk
L
p=
�
Z
1
0
kx(t)k
p
dt
�
1=p
if p <1 and
kxk
L
1= ess sup
0�t<1
kx(t)k :
De�nition 2 Let 1 � p � 1 and 0 �M � 1. We say that (�) has L
p
-gain less than or equal to M
if for any u 2 L
p
([0;1); IR
m
), the solution x of (�) corresponding to u satis�es
kxk
L
p�Mkuk
L
p:
The in�mum of such numbers M will be called the L
p
-gain of (�). We say that system (�) is L
p
-stable
if its L
p
-gain is �nite.
4
By a neutrally stable n� n matrix A we mean one for which all solutions of _x = Ax are bounded;
equivalently, A has no eigenvalues with positive real part and each Jordan block corresponding to
a purely imaginary eigenvalue has size one. Another well-known characterization of such matrices
is that they are the ones for which there exists a symmetric positive de�nite matrix Q such that
A
T
Q+ QA � 0.
We now state our main result:
Theorem 1 Let A;B be n�n; n�m matrices respectively. Let � be an IR
m
-valued S-function. Assume
that A is neutrally stable. Then there exists an m� n matrix F such that the system
_x = Ax+B�(Fx + u) ;
x(0) = 0 ;
(1)
is L
p
-stable for all 1 � p � 1.
Theorem 1 is an immediate consequence of the more general technical result contained in The-
orem 2. In order to state that theorem in great generality, we recall �rst a standard notion. Let
(�) _x = Ax + Bu be a linear system, where x and u take values in IR
n
and IR
m
respectively. For
each measurable and locally essentially bounded u : [0;1) ! IR
m
and each x
0
2 IR
n
, let x
u
(t; x
0
)
be the solution of (�) corresponding to u with x
u
(0; x
0
) = x
0
. Following the terminology of [6], the
stabilizable subspace S(A;B) of (A;B) is the subspace of IR
n
which consists of all those initial states
x
0
2 IR
n
for which there is some u so that x
u
(t; x
0
) ! 0 as t ! 1. In other words, S(A;B) is the
subspace of IR
n
made up of all the states that can be asymptotically controlled to 0 (so this includes
in particular the reachable subspace). Observe that the pair (A;B) is stabilizable (asymptotically null
controllable) i� S(A;B) = IR
n
.
Theorem 2 Let A and B be n � n and n �m matrices respectively. Let S(A;B) be the stabilizable
subspace of (A;B). Let � be an IR
m
-valued S-function and let � : IR
k
! S(A;B) � IR
n
be a globally
Lipschitz and bounded function with �(0) = 0, where k > 0 is some integer. Assume that A is neutrally
stable. Then there exist an m� n matrix F and an " > 0 such that the system
_x = Ax+B�(Fx + u) + "�(v) ;
x(0) = 0 ;
(2)
is L
p
-stable for each 1 � p � 1, i.e. there exists a �nite constant M
p
> 0 such that for any
u 2 L
p
([0;1); IR
m
); v 2 L
p
([0;1); IR
k
),
kxk
L
p�M
p
(kuk
L
p+ kvk
L
p) :
The proof is given later.
Theorem 2 implies Theorem 1 (just take � � 0) as well as a result dealing with small \nonmatching"
state perturbations. It is possible to make the result even more general, by weakening the Lipschitz
assumption on �, but there does not appear to be a need for that generality.
2.3 Output Stabilization
Consider the initialized linear input=output system
(�
ao
) _x = Ax +B�(u) ;
x(0) = 0 ;
y = Cx ;
5
where A;B; C are respectively n�n, n�m, r�n matrices. Assume that system (�
ao
) is asymptotically
observable (that is, it is detectable). Our main result for input/output systems is as follows:
Theorem 3 Assume that system (�
ao
) is asymptotically observable and A is neutrally stable. Then
there exist an m � n matrix F and an n � r matrix L such that the following property holds. Let
1 � p � 1. Pick any u
1
2 L
p
([0;1); IR
m
) and u
2
2 L
p
([0;1); IR
r
), and consider the solution
x = (x
1
; x
2
) of
_x
1
= Ax
1
+B�(y
2
+ u
1
) ; y
1
= Cx
1
;
_x
2
= (A+ LC)x
2
+ B�(Fx
2
)� L(y
1
+ u
2
) ; y
2
= Fx
2
;
with x(0) = 0. Consider the total output function y = (y
1
; y
2
) = (Cx
1
; Fx
2
). Then y is in
L
p
([0;1); IR
r+m
) and
kyk
L
p�M
p
(ku
1
k
L
p+ ku
2
k
L
p)
for some constant M
p
> 0.
2.4 Not Every Feedback Stabilizes
One may ask if any F which would stabilize when the saturation is not present also provides �nite
gain for (1). Not surprisingly, the answer is negative. In order to give an example, we need �rst a
simple technical remark.
Lemma 1 Consider the system _x = Ax + B�(Fx + u), where the matrix A is assumed to have all
eigenvalues in the imaginary axis and where each component of � is a continuous function whose range
contains a neighborhood of the origin (this holds, for instance, if it is an S-function). Furthermore,
assume that the pair (A;B) is controllable. Then, given any state x
0
2 IR
n
, there is some measurable
essentially bounded control u steering the origin to x
0
in �nite time.
PROOF: Since all eigenvalues of A have zero real part and the pair (A;B) is controllable, for each
" > 0 there is some control v
0
for the system _x = Ax + Bu so that jv
0
(t)j < " for all t and which
drives in �nite time the origin to x
0
(see e.g. [12]). Since the range of � contains a neighborhood of the
origin, and using a measurable selection (Fillipov's Theorem), it is also true that there is a measurable
control v which achieves the same transfer, for the system _x = Ax + B�(u). Now let, along the
corresponding trajectory, u(t) = v(t) � Fx(t). It follows that this achieves the desired transfer for
_x = Ax+B�(Fx + u),
The next two examples show that even if A is neutrally stable, Theorem 1 may not be true if F
only satis�es the condition that A +BF is Hurwitz.
Example 1 Let
A =
�
0 �1
1 0
�
; B =
�
0
1
�
; F = �
�
1=2 ; 1
�
;
and any � so that �(1=2) = 1. Then both the origin and (�1; 0) are equilibrium points of the system
_x = Ax +B�(Fx):
By Lemma 1, there is some input u
0
which steers the origin to (�1; 0) in some �nite time T
0
. Consider
the input u
1
equal to u
0
for 0 � t � T
0
and to 0 for t > T
0
. Then if x is the trajectory of (2)
corresponding to u
1
, we have that x(t) = (�1; 0) for all t � T
0
. Clearly, for any 1 � p < 1,
u
1
2 L
p
([0;1); IR) and x 62 L
p
([0;1); IR
2
). Therefore, system (2) is not L
p
-stable for any 1 � p <1.
6
If we use multiple inputs, a di�erent example that includes p =1 is as follows.
Example 2 Assume that m = n = 2. Let
A =
0 0
0 0
!
; B =
1 0
0 1
!
; F =
�3 7
�1 2
!
:
Then A+BF = F is Hurwitz. Let � = (�
0
; �
0
)
T
, where �
0
is the standard saturation function. Then
the system
_x = �(Fx+ u) ;
x(0) = (0; 0)
T
(3)
is not L
p
-stable for any 1 � p � 1. To see this, take a control v on some interval [0; T ] that steers
(0; 0)
T
to (1; 1)
T
. Let u = v on [0; T ] and u = (0; 0)
T
on (T;1). Let x = (x
1
; x
2
)
T
be the solution of
(3) corresponding to u. Then on [T;1), we have x
1
(t) = x
2
(t) = t� T + 1. Thus (3) is not L
p
-stable
for any 1 � p � 1. (In fact, the trajectory is not even bounded for a bounded input.)
3 Some Characterizations of S-functions
Lemma 2 A function � : IR ! IR satis�es Condition (i) in De�nition 1 if and only if it satis�es the
following two conditions:
(a) � is globally Lipschitz;
(b) there exists C
1
� 0 such that for all t; s 2 IR,
jtj j�(t+ s)� �(t)j � C
1
jsj : (4)
PROOF: First Condition (a) implies that � is absolutely continuous and j�
0
(t)j is bounded for almost
all t 2 IR. Assume that � satis�es (b). Letting s = �t in (4) we get that j�(t)j � j�(0)j+ C
1
. On the
other hand, since for s 6= 0
jtj
�
�
�
�
�(t+ s)� �(t)
s
�
�
�
�
� C
1
;
letting s ! 0, we see that jt�
0
(t)j � C
1
for almost all t 2 IR, and then conclude that Condition (i)
holds.
Conversely assume that Condition (i) holds. Since � is absolutely continuous and �
0
(t) is bounded
for almost all t 2 IR, we see that � is globally Lipschitz. To see that (b) holds, we consider two cases.
(1): If jtj � 2jsj, since � is bounded, (b) is clearly true.
(2): If jtj > 2jsj, then t(t + s) > 0. Condition (i) implies that there exists a constant
~
C
1
> 0 such
that j�
0
(t)j �
~
C
1
=jtj for almost all t 2 IR. We have (assume that s 6= 0),
jtj j�(t+ s)� �(t)j = jtj
�
�
�
Z
t+s
t
�
0
(�) d�
�
�
�
�
~
C
1
jtj
�
�
�
Z
t+s
t
1
�
d�
�
�
�=
~
C
1
jtj
�
�
�log(1 +
s
t
)
�
�
�=
~
C
1
jsj
�
�
�
�
�
log(1 +
s
t
)
s
t
�
�
�
�
�
:
7
Now since the function
�
�
�
log(1+�)
�
�
�
�is bounded for j�j � 1=2, we conclude that (b) holds.
Remark 2 It is easy to see that if � satis�es a bound j�(t)j � M jtj in a neighborhood of 0 (in
particular if (i) in De�nition 1 holds), then Condition (ii) in De�nition 1 is equivalent to the following
condition:
(c) There exist positive numbers a; b;K and a measurable function � : IR ! [a; b] such that for all
t 2 IR we have j�(t)� �(t)tj � Kt�(t).
It is clear that (c) implies (ii). To see the converse, let � > 0 be such that j�(t)j � M jtj for jtj � �.
Then just let
�(t) =
8
>
>
>
<
>
>
>
:
1 if t = 0 ;
�(t)
t
if t 2 [��; �]=f0g ;
�(�)
�
if t > � ;
�
�(��)
�
if t < �� ;
It is easily veri�ed that there exist positive constants a; b;K such that (c) holds for this � .
De�nition 3 We say that a constantK > 0 is an S-bound for � if there exist a; b > 0 and a measurable
function � : IR! [a; b] such that
(i) j�(t)� �(s)j � Kjt� sj,
(ii) jtjj�(t+ s)� �(t)j � Kjsj,
(iii) j�(t)� �(t)tj � Kt�(t).
The above discussion shows that such (�nite) S-bounds always exist. Note that if K is an S-bound for
�, then j�(t)j � maxfK;Kjtjg. A constant K > 0 is called an S-bound for a vector valued S-function
� if K is an S-bound for each component of �.
4 Proofs of the Main Results
For notational convenience (to avoid having too many negative signs in the formulas) we will prove
the main theorem for systems written in the form
_x = Ax�B�(Fx + u) + "�(v) ;
x(0) = 0 :
(5)
A trivial remark is needed before we start.
Remark 3 Assume that �
1
: IR
k
1
! IR
m
and �
2
: IR
k
2
! IR
n
each satis�es a growth estimate of the
type k�
1
(u)k � Ckuk, k�
2
(v)k � Ckvk for u 2 IR
k
1
; v 2 IR
k
2
. It follows from classical linear systems
theory that if the system _x = Ax is globally asymptotically stable {that is, A is a Hurwitz matrix{
then the controlled system _x = f(x; u; v) = Ax + B�
1
(u) + �
2
(v) is automatically also L
p
-stable for
all 1 � p � 1. We will be interested in the case in which A is merely stable, but this remark will be
used at various points.
We now prove Theorem 2. First we note that we can assume that (A;B) is controllable.
8
4.1 Proof of Theorem 1 Assuming Controllability
Suppose Theorem 2 is already known to be true for controllable (A;B); we show how the general case
follows. It is an elementary linear systems exercise to show that the stabilizable subspace S(A;B),
for any two A;B, is invariant under A; this follows for instance from its characterization as a sum
of the reachable subspace and the space of stable modes. Thus the restriction of A to S(A;B) is
well-de�ned, and it is again neutrally stable. Now since � takes values in S(A;B), the trajectories of
(5) lie in S(A;B). So we may assume that (A;B) is stabilizable, i.e. S(A;B) = IR
n
, since otherwise
we can restrict ourselves to S(A;B). Then, up to a change of coordinates, we may assume that
A =
A
1
A
2
0 A
3
!
; B =
B
1
0
!
;
where (A
1
; B
1
) is controllable and A
1
is neutrally stable. (Since we are assuming that (A;B) is
stabilizable, we also know that A
3
is Hurwitz, but we do not need that additional fact here.) Assume
that A
1
is an r � r matrix and B
1
is an r �m matrix.
Let
~
� : IR
r
! IR
r
be given by
~
�(�) = (
~
�
0
(�
1
); : : : ;
~
�
0
(�
r
))
T
for � 2 IR
r
where
~
�
0
is the standard
saturation function, i.e.
~
�
0
(t) = sign (t) minf1; jtjg.
By our assumption that the result is known in the controllable case, there exists an m� r matrix
F
1
and "
1
> 0 so that the system
_x
1
= A
1
x
1
�B
1
�(F
1
x
1
+ u) + "
1
~
�(w) ;
x
1
(0) = 0
(6)
is L
p
-stable for all 1 � p � 1. Let �
p
be the L
p
-gain of this system, so kx
1
k
L
p� �
p
(kuk
L
p+ kwk
L
p)
for all u 2 L
p
([0;1); IR
m
) and w 2 L
p
([0;1); IR
r
).
Since (A;B) is stabilizable, we can �nd an m � n matrix E such that A + BE is Hurwitz. Then
the system
_y = (A+ BE)y + v ;
y(0) = 0 :
(7)
is L
p
-stable for any 1 � p � 1. Let
p
be the L
p
-gain of (7), so kyk
L
p�
p
kvk
L
p.
From the properties of �, we know that there is some L > 0 so that k�(�)k � minfL; Lk�kg for
all � 2 IR
k
. (In fact, this is the only property of � that is really needed.) Take an " > 0 such that
"L
1
kBEk � "
1
. Let F = (F
1
; 0). We show that for this choice of F and ", system (5) is L
p
-stable for
any 1 � p � 1. For this purpose, let u 2 L
p
([0;1); IR
m
); v 2 L
p
([0;1); IR
k
). Let x be the solution of
(5) corresponding to u; v. Let y be the solution of
_y = (A+BE)y + "�(v) ;
y(0) = 0 :
(8)
Then we have kyk
L
1� "L
1
and kyk
L
p� "L
p
kvk
L
p. Let z = x� y. Then z satis�es
_z = Az �B�(Fz + Fy + u)� BEy ;
z(0) = 0 :
Write z = (z
1
; z
2
)
T
. Then we have z
2
� 0 and z
1
satis�es
_z
1
= A
1
z
1
�B
1
�(F
1
z
1
+ Fy + u)�B
1
Ey ;
z
1
(0) = 0 :
9
Since kB
1
Eyk
L
1� kB
1
Ek kyk
L
1� "L
1
kB
1
Ek � "
1
, we have
�
B
1
Ey
"
1
=
~
�
�
�
B
1
Ey
"
1
�
;
So z
1
satis�es
_z
1
= A
1
z
1
� B
1
�(F
1
z
1
+ Fy + u) + "
1
~
�(�B
1
Ey="
1
) ;
z
1
(0) = 0 :
By the L
p
-stability of (6) we get that
kzk
L
p= kz
1
k
L
p� �
p
�
kFy + uk
L
p+
B
1
Ey
"
1
L
p
�
� �
p
�
kFk kyk
L
p+ kuk
L
p+
kB
1
Ekkyk
L
p
"
1
�
� �
p
�
kuk
L
p+
�
kB
1
Ek
"
1
+ kFk
�
"L
p
kvk
L
p
�
:
This shows that (5) is L
p
-stable, which concludes the proof that we may assume that (A;B) is con-
trollable.
4.2 Proof in the Controllable Case
From elementary linear algebra, we know that any neutrally stable matrix A is similar to a matrix
A
1
0
0 A
2
!
; (9)
where A
1
is an r� r Hurwitz matrix and A
2
is an (n� r)� (n� r) skew-symmetric matrix. So, up to
a change of coordinates, we may assume that A is already in the form (9). In these coordinates, we
write
B =
B
1
B
2
!
where B
2
is an (n � r) � m matrix, and we write vectors as x = (x
1
; x
2
)
T
and also � = (�
1
; �
2
)
T
.
Consider the feedback law F = (0; B
T
2
). Then system (5), with this choice of F , can be written as
_x
1
= A
1
x
1
�B
1
�(B
T
2
x
2
+ u) + "�
1
(v) ;
_x
2
= A
2
x
2
�B
2
�(B
T
2
x
2
+ u) + "�
2
(v) ;
x
1
(0) = 0; x
2
(0) = 0 :
(10)
Since A
1
is Hurwitz, it will be su�cient to show that there exists an " > 0 such that the x
2
-subsystem
is L
p
-stable (we may think of x
2
as an additional input to the �rst subsystem, and apply Remark 3).
The controllability assumption on (A;B) implies that the pair (A
2
; B
2
) is also controllable. Since
A
2
is skew-symmetric, the matrix
~
A := A
2
� B
2
B
T
2
is Hurwitz. (Just observe that the Lyapunov
equation
~
A
T
I
n�r
+ I
n�r
~
A = �2B
2
B
T
2
holds, and the pair (
~
A;B
2
) is controllable; see [13], Exercise
4.6.7.) Therefore, the theorem is a consequence of the following lemma. This is where the main parts
of our argument lie.
10
Lemma 3 Let �; � be as in Theorem 2. Let A be a skew-symmetric matrix. Assume that
~
A :=
A� BB
T
is Hurwitz. Then there exists an " > 0 such that the system
_x = Ax� B�(B
T
x+ u) + "�(v) ;
x(0) = 0 ;
(11)
is L
p
-stable for all 1 � p � 1.
PROOF: Assume that � = (�
1
; : : : ; �
m
)
T
. Let 0 < a � b < 1; K > 0 be constants and �
i
:
IR ! [a; b]; i = 1; : : : ; m; be measurable functions so that the components �
i
of � satisfy (i), (ii)
and (iii) in De�nition 3 with the respective �
i
's. We may assume that K is large enough such that
k�(�)k � minfK;Kk�kg for all � 2 IR
k
. Let
�
def
= min
i=1;:::;m
lim inf
j�j!1
j�
i
(�)j:
Then � > 0. Let " > 0 satisfy
" <
�
K
1
p
mkBk
; (12)
where
1
is the L
1
-gain of the initialized linear control system
_y = (A�BB
T
)y + u ;
y(0) = 0:
(13)
By (12) there exists a � 2 (0; 1=2] such that
" �
(1� 2�)�
K
1
p
mkBk
:
Let u 2 L
p
([0;1); IR
m
); v 2 L
p
([0;1); IR
k
). Let y be the solution of
_y = (A� BB
T
)y + "�(v) ;
y(0) = 0:
(14)
Let x be the solution of (11) corresponding to u; v and let z = x� y. Then z satis�es
_z = Az � B�(B
T
z + u+ B
T
y) + BB
T
y ;
z(0) = 0:
(15)
Let ~u = u+ B
T
y and ~v = B
T
y. Then we get
k~vk
L
1� kBk kyk
L
1� "kBk
1
k�k
L
1�
(1� 2�)�
p
m
: (16)
Now (15) can be written as
_z = Az � B
�
�(B
T
z + ~u)� ~v
�
;
z(0) = 0:
(17)
(We have brought the problem to one of a \matched uncertainty" type, in robust control terms, if we
think of ~v as representing a source of uncertainty.)
11
For each 1 � p <1, consider the following function:
V
0;p
(z) =
kzk
p+1
p+ 1
:
Along the trajectories of (17), we have
_
V
0;p
(z(t)) = �kz(t)k
p�1
z
T
(t)B
�
�
�
B
T
z(t) + ~u(t)
�
� ~v(t)
�
= �kz(t)k
p�1
�
z
T
(t)B
h
�
�
B
T
z(t) + ~u(t)
�
� �(B
T
z(t))
i
+ z
T
(t)B
�
�(B
T
z(t))� ~v(t)
�
�
:
Since K is an S-bound for �, we know that
�
�
�z
T
(t)B
�
�
�
B
T
z(t) + ~u(t)
�
� �
�
B
T
z(t)
��
�
�
�� K
m
X
i=1
j~u
i
(t)j � K
p
mk~u(t)k (18)
for all t 2 [0;1). Therefore we have the following decay estimate:
_
V
0;p
(z(t)) � �kz(t)k
p�1
z
T
(t)B
�
�(B
T
z(t))� ~v(t)
�
+K
p
mkz(t)k
p�1
k~u(t)k: (19)
We next need to bound the �rst term in this estimate. For that purpose, we will partition [0;1) into
two subsets. By the de�nition of �, there is some M
1
� 1 so that
min
i=1;���;m
inf
j�j�M
1
j�
i
(�)j � (1� �)�:
The �rst subset consists of those t for which kz(t)
T
Bk �M
1
p
m. For such t, trivially:
z(t)
T
B
�
�(B
T
z(t))� ~v(t)
�
� z(t)
T
B�
�
B
T
z(t)
�
�M
1
p
mk~v(t)k : (20)
Next we consider those t for which kz(t)
T
Bk > M
1
p
m. First we note some general facts for any
vector � 2 IR
m
for which
k�
T
Bk > M
1
p
m : (21)
If we pick i
0
so that j(�
T
B)
i
0
j = max
i=1;:::;m
fj(�
T
B)
i
jg, then j(�
T
B)
i
0
j > M
1
, and therefore, by the
choice of M
1
,
�
�
��
i
0
�
(B
T
�)
i
0
�
�
�
�� (1 � �)�. (Here (�
T
B)
i
denotes the i-th component of �
T
B.) We
conclude that if � satis�es (21) then
�
T
B�(B
T
�) � (�
T
B)
i
0
�
i
0
�
(B
T
�)
i
0
�
�
k�
T
Bk
p
m
(1� �)�
or equivalently
k�
T
Bk �
p
m�
T
B�(B
T
�)
(1� �)�
:
From this and (16), if kz
T
(t)Bk > M
1
p
m, we have
z
T
(t)B
�
�(B
T
z(t))� ~v(t)
�
� z
T
(t)B�(B
T
z(t))� kz
T
(t)Bk k~v(t)k
� z(t)
T
B�(B
T
z(t))�
p
mk~vk
L
1
(1� �)�
z
T
(t)B�(B
T
z(t))
�
�
1�
1� 2�
1� �
�
z
T
(t)B�(B
T
z(t))
=
�
1� �
z
T
(t)B�(B
T
z(t)): (22)
12
Note also that
�
1��
� 1 for 0 < � � 1=2. Combining (20) and (22) we have a common estimate valid
for all t � 0:
z(t)
T
B
�
�(B
T
z(t))� ~v(t)
�
�
�
1� �
z(t)
T
B�
�
B
T
z(t)
�
�M
1
p
mk~v(t)k :
Using this and (19) we get
_
V
0;p
(z(t)) � �
�
1� �
kz(t)k
p�1
z
T
(t)B�(B
T
z(t)) +
p
mkz(t)k
p�1
(Kk~u(t)k+M
1
k~v(t)k): (23)
Let � = diag(�
1
; : : : ; �
m
) with �(�) = diag(�
1
(�
1
); : : : ; �
m
(�
m
)) for � 2 IR
m
. Then aI � �(�) � bI
for all � 2 IR
m
. Thus for any � 2 IR
m
,
k�(�)� � �(�)k =
�
m
X
i=1
j�
i
(�
i
)�
i
� �
i
(�
i
)j
2
�
1=2
� K
�
m
X
i=1
�
2
i
(�
i
(�
i
))
2
�
1=2
� K�
T
�(�): (24)
For each 1 < p <1, introduce now the following new function:
V
1;p
(x) =
kxk
p
p
:
Along the trajectory of (17), we have, at each t so that z(t) 6= 0:
dV
1;p
�
z(t)
�
dt
= kz(t)k
p�2
z
T
(t)B
n
� �
�
B
T
z(t) + ~u(t)
�
+ ~v(t)
o
= kz(t)k
p�2
z
T
(t)B
n
� �
�
B
T
z(t)
�
B
T
z(t) +
�
�
�
B
T
z(t)
�
B
T
z(t)� �
�
B
T
z(t)
��
+
�
�
�
B
T
z(t)
�
� �
�
B
T
z(t) + ~u(t)
��
+ ~v(t)
o
� �akz(t)k
p�2
kB
T
z(t)k
2
+KkBk kz(t)k
p�1
z
T
(t)B�
�
B
T
z(t)
�
+KkBk kz(t)k
p�1
k~u(t)k+ kBkkz(t)k
p�1
k~v(t)k : (25)
This calculation can be justi�ed also for z(t) = 0, since V
1;p
is di�erentiable and vanishes at zero; in
that case one should interpret kz(t)k
p�2
z
T
(t) as zero.
Since
~
A is Hurwitz, for each 1 < p < 1 there exist a di�erentiable function V
2;p
and positive real
numbers a
p
, b
p
and c
p
such that for all x 2 IR
n
,
(P
1
) a
p
kxk
p
� V
2;p
(x) � b
p
kxk
p
;
(P
2
) kDV
2;p
(x)k � c
p
kxk
p�1
;
(P
3
) DV
2;p
(x)
~
Ax � �2kxk
p
;
Moreover, we may choose V
2;p
so that
(P
4
) lim sup
p!1+
c
p
= c
1
<1, and the limit V
2;1
(x) := lim
p!1+
V
2;p
(x) exists for all x 2 IR
n
.
(For instance, one may take V
2;p
(x) = �
p
(x
T
Px)
p=2
, where �
p
> 0 is a proper constant and P > 0
satis�es P
~
A +
~
A
T
P = �I .)
13
Rewriting (17) in the form
_z =
~
Az +B
�
B
T
z � �(B
T
z + ~u) + ~v
�
;
z(0) = 0 ;
(26)
we can compute the derivative of V
2;p
along its trajectories as:
dV
2;p
�
z(t)
�
dt
� �2kz(t)k
p
+ c
p
kBkkz(t)k
p�1
n
B
T
z(t)� �(B
T
z(t))B
T
z(t)
+
�(B
T
z(t))B
T
z(t)� �
�
B
T
z(t)
�
+
�
�
B
T
z(t)
�
� �
�
B
T
z(t) + ~u(t)
�
+ k~v(t)k
o
� �2kz(t)k
p
+ c
p
kBkkz(t)k
p�1
n
(b+ 1)kB
T
z(t)k+Kz
T
(t)B�
�
B
T
z(t)
�
+K k~u(t)k+ k~v(t)k
o
: (27)
For 1 � p <1, let
�
p
=
(b+ 1)
2
c
2
p
kBk
2
4a
; (28)
e
p
=
KkBk(�
p
+ c
p
)(1� �)
�
: (29)
(Observe that these constants do not depend on the particular u and v being considered, but only on
the system and on p.) Finally, consider, for each 1 � p <1, the following function
V
p
= e
p
V
0;p
+ �
p
V
1;p
+ V
2;p
; (30)
where e
p
; �
p
are given in (28), (29). Using (23), (25) and (27), for 1 < p <1, we have along trajectories
of (17),
dV
p
�
z(t)
�
dt
� �
�e
p
1� �
kz(t)k
p�1
z
T
(t)B�
�
B
T
z(t)
�
+ e
p
p
mkz(t)k
p�1
(Kk~u(t)k+M
1
k~v(t)k)
�a�
p
kz(t)k
p�2
kB
T
z(t)k
2
+K�
p
kBk kz(t)k
p�1
z
T
(t)B�
�
B
T
z(t)
�
+K�
p
kBk kz(t)k
p�1
k~u(t)k+ �
p
kBk kz(t)k
p�1
k~v(t)k � 2kz(t)k
p
+c
p
kBkkz(t)k
p�1
n
(b+ 1)kB
T
z(t)k+Kz(t)B
T
�
�
B
T
z(t)
�o
+Kc
p
kBk kz(t)k
p�1
k~u(t)k+ c
p
kBk kz(t)k
p�1
k~v(t)k
� �kz(t)k
p
+ kz(t)k
p�2
n
� kz(t)k
2
+ (b+ 1)c
p
kBkkz(t)k kB
T
z(t)k � a�
p
kB
T
z(t)k
2
o
+K(
p
me
p
+ kBk(c
p
+ �
p
))kz(t)k
p�1
k~u(t)k+ (
p
mM
1
e
p
+ kBk(c
p
+ �
p
))kz(t)k
p�1
k~v(t)k :
For each 1 � p <1 denote:
�
p
= maxfK(
p
me
p
+ kBk(c
p
+ �
p
));
p
mM
1
e
p
+ kBk(c
p
+ �
p
)g :
From all the above, we conclude for each 1 < p <1:
dV
p
�
z(t)
�
dt
� �kz(t)k
p
+ �
p
kz(t)k
p�1
(k~u(t)k+ k~v(t)k) : (31)
14
For any t � 0, integrating (31) from 0 to t, we have:
V
p
�
z(t)
�
+
Z
t
0
kz(s)k
p
ds � �
p
Z
t
0
kz(s)k
p�1
(k~u(s)k+ k~v(s)k)ds :
When p = 1, this inequality is also true as an easy consequence of the Lebesgue Dominated Convergence
Theorem (applied to a sequence fp
j
g
1
j=1
decreasing to 1). Thus the inequality is true for all 1 � p <1.
Applying �rst H�older's inequality to
R
t
0
kz(s)k
p�1
(k~u(s)k + k~v(s)k)ds, we conclude that for all
1 � p <1 and t � 0,
V
p
�
z(t)
�
+ kzk
p
L
p
[0;t]
� �
p
kzk
p�1
L
p
[0;t]
(k~uk
L
p+ k~vk
L
p) : (32)
Since V
p
� 0, we get that z 2 L
p
([0;1); IR
n
) and
kzk
L
p� �
p
(k~uk
L
p+ k~vk
L
p) : (33)
Now since z = x� y, ~u = u +B
T
y; ~v = B
T
y, we have
k~vk
L
p� kBk kyk
L
p� "K
p
kBkkvk
L
p
k~uk
L
p� kuk
L
p+ "K
p
kBkkvk
L
p;
kzk
L
p� kxk
L
p� kyk
L
p� kxk
L
p� "K
p
kvk
L
p;
where
p
is the L
p
-gain of (13). Combining this with (33) we have
kxk
L
p� �
p
kuk
L
p+ "K
p
(1 + 2�
p
kBk)kvk
L
p:
This �nishes the proof of the Lemma, and hence our main theorem, for the case when 1 � p <1.
We now prove the case p = 1. This means that we need to show that system (11) has the
uniform bounded input bounded state property, i.e., there exists a �nite constantM such that kxk
L
1�
M(kuk
L
1+ kvk
L
1) for all u 2 L
1
([0;1); IR
m
) and v 2 L
1
([0;1); IR
k
). It will be useful to use the
function V
2
which was also used for p = 2. Letting p = 2, from (31) we have
dV
2
�
z(t)
�
dt
� �kz(t)k
�
kz(t)k � �
2
(k~uk
L
1+ k~vk
L
1)
�
: (34)
Let � = k~uk
L
1+ k~vk
L
1. Thus,
_
V
2
is negative outside the ball of radius �
2
� centered at the origin. It
follows that
V
2
�
z(t)
�
� sup
k�k��
2
�
V
2
(�) �
e
2
�
3
2
3
�
3
+ (b
2
�
2
2
+
�
2
�
2
2
2
)�
2
:
First assume that � � 1. Then we have
a
2
kz(t)k
2
� V
2
�
z(t)
�
�
�
e
2
�
3
2
3
+
�
2
�
2
2
2
+ b
2
�
2
2
�
�
2
;
which implies that
kzk
L
1�
n
2e
2
�
3
2
+ 6b
2
�
2
2
+ 3�
2
�
2
2
6a
2
o
1
2
� :
If � > 1, we have
e
2
kz(t)k
3
3
� V
2
�
z(t)
�
�
�
e
2
�
3
2
3
+
�
2
�
2
2
2
+ b
2
�
2
2
�
�
3
;
15
We then get that
kzk
L
1�
n
2e
2
�
3
2
+ 6b
2
�
2
2
+ 3�
2
�
2
2
2e
2
o
1
3
� :
Let
�
G
1
= max
�
n
2e
2
�
3
2
+ 6b
2
�
2
2
+ 3�
2
�
2
2
6a
2
o
1
2
;
n
2e
2
�
3
2
+ 6b
2
�
2
2
+ 3�
2
�
2
2
2e
2
o
1
3
�
:
We have kzk
L
1�
�
G
1
�. Now
� = k~uk
L
1+ k~vk
L
1� kuk
L
1+ 2"K
1
kBk kvk
L
1:
and
kzk
L
1� kxk
L
1� "K
1
kvk
L
1:
We conclude that
kxk
1
�
�
G
1
kuk
L
1+ "K
1
(1 + 2
�
G
1
kBk)kvk
L
1:
Now the proof of Lemma 3 is complete.
4.3 Proof of the Output Feedback Theorem
We now provide a proof of Theorem 3. We will show a somewhat stronger statement, namely, that
the state trajectory x also satis�es an estimate as required. The proof will be the usual Luenberger-
observer construction, but a bit of care has to be taken because of the nonlinearities.
Asymptotic observability means that there is some n� r matrix L such that A+ LC is Hurwitz.
Pick F be as in Theorem 2. Let e = x
1
� x
2
. Then (x
1
; e) satis�es
_x
1
= Ax
1
+B�(Fx
1
� Fe + u
1
) ;
_e = (A+ LC)e + B
�
�(Fx
1
� Fe + u
1
)� �(Fx
1
� Fe)
�
+ Lu
2
:
Let ~v = �(Fx
1
� Fe+ u
1
)� �(Fx
1
� Fe). Since k~v(t)k � Kku
1
(t)k and A+LC is Hurwitz, we know
that e is in L
p
([0;1); IR
n
) and kek
L
p� M
0
(ku
1
k
L
p+ ku
2
k
L
p) for some constant M
0
> 0. Then the
conclusion follows from Theorem 2 applied to the x
1
-subsystem.
Note that the conclusion of this Theorem can be restated in terms of the �nite gain stability of a
standard systems interconnection
y
1
= P (u
1
+ y
2
)
y
2
= C(u
2
+ y
1
)
where P denotes the input/output behavior of the original system � and C is the input/output
behavior of the controller with state space x
2
and output y
2
.
4.4 Operator Stability Among Di�erent Norms
We can actually prove a stronger result than that stated in Theorem 2, namely that the input to state
operator (u; v)! x from L
p
([0;1; IR
m
)�L
p
([0;1); IR
k
) to L
p
([0;1); IR
n
) is a bounded operator from
L
p
([0;1; IR
m
)� L
p
([0;1); IR
k
) to L
q
([0;1); IR
n
), for any q � p.
16
Remark 4 Let A be skew-symmetric and
~
A = A � BB
T
be Hurwitz. Then from (32), (33) we get
that, for u 2 L
p
([0;1); IR
m
); v 2 L
p
([0;1); IR
k
) and t � 0,
a
p
kz(t)k
p
� V
p
�
z(t)
�
� �
p
kzk
p�1
L
p
k (k~uk
L
p+ k~vk
L
p) � �
p
p
(k~uk
L
p+ k~vk
L
p)
p
;
and then, kzk
L
1� C
1
(k~uk
L
p+ k~vk
L
p) with C
1
= �
p
a
�1=p
p
. Therefore we obtain for q � p,
kzk
q
L
q
� kzk
q�p
L
1
kzk
p
L
p
� C
q�p
1
�
p
p
(k~uk
L
p+ k~vk
L
p)
q
: (35)
From this one can easily deduce that for any q � p the solution x of (11) satis�es
kxk
L
q�M
p;q
(kuk
L
p+ kvk
L
p)
for some constants M
p;q
> 0. The same results then hold for the original system in Theorem 2, as is
clear from the reduction to (11). That is, for any u 2 L
p
([0;1); IR
m
); v 2 L
p
([0;1); IR
k
), the solution
x of (5) satis�es a similar inequality.
4.5 A Remark on Robustness of a Linear Feedback
It is worth pointing out that the same method used to prove Lemma 3 allows establishing a result
regarding time-varying multiplicative uncertainties on a linear feedback law u = �B
T
x:
Lemma 4 Let A be an n�n skew-symmetric matrix and B an n�m matrix. Assume that A�BB
T
is Hurwitz. Let D(t) be an m � m matrix with bounded entries. Assume also that there exists a
constant a > 0 such that D(t) +D
T
(t) � aI for almost all t in [0;1). Then the following initialized
system
(
~
�) _x =
�
A(t)x+ u ;
x(0) = 0;
where u 2 L
p
([0;+1); IR
n
) and
�
A(t) := A�BD(t)B
T
, is �nite gain L
p
-stable, and the L
p
-gain depends
only on p; a; A B and M , where M = supfkD(t)k : t 2 [0;1)g.
PROOF: Let V
p
= �
p
V
1;p
+ V
2;p
with �
p
=
2(M+1)
2
c
2
p
kBk
2
a
, where V
1;p
; V
2;p
are de�ned in the proof of
Lemma 3. Then one can show that along the trajectories of (
~
�)
_
V
p
(x(t)) � �kx(t)k
p
+ ��
p
kx(t)k
p�1
ku(t)k ;
for some constant ��
p
> 0. Then everything else follows exactly the same as in the proof of Lemma 3.
5 Comparison with Linear Gains
From the proof of Lemma 3, we can also obtain explicit bounds for the L
p
-gain for (11). For simplicity,
we deal only with the case � � 0, as in Theorem 1. Speci�cally, we will compare these bounds with
the L
p
-gain of the linear control system
_x =
~
Ax+Bu ;
x(0) = 0 ;
(36)
where
~
A = A� BB
T
. For the cases p = 1; 2 we have the following:
17
Corollary 1 Let A;B; � be as in Lemma 3. Let G
1
and G
2
be respectively the L
1
and L
2
-gains of the
system
_x = Ax� B�(B
T
x+ u) ;
x(0) = 0 :
(37)
Let
1
and
2
be respectively the L
1
and L
2
-gains of (36). Then we have
1. G
1
� K(1 +
p
mK)
1
,
2. G
2
� 2
p
nK(1 +
p
mK)
2
.
(In the literature,
2
is called the \H
1
-norm" of (36) and is usually denoted by kWk
1
, where W (s)
is the transfer matrix for system (36).)
PROOF: For p = 1 one can easily get from (19)
Z
1
0
x
T
(s)B�
�
B
T
x(s)
�
ds � K
p
mkuk
L
1: (38)
(This also follows by integrating (55) below from 0 to 1.) Let v(t) = B
T
x(t) � �
�
B
T
x(t) + u(t)
�
.
Then, we have
Z
1
0
kv(s)kds �
Z
1
0
n
B
T
x(s)� �
�
B
T
x(s)
�
+
�
�
B
T
x(s)
�
� �
�
B
T
x(s) + u(s)
�
o
ds
� K
Z
1
0
n
x
T
(s)B�
�
B
T
x(s)
�
+ ku(s)k
o
ds
� K(
p
mK + 1)kuk
L
1:
Now (37) can be written as
_x =
~
Ax +Bv(t) ;
x(0) = 0 :
By the de�nition of
1
we have kxk
L
1�
1
kvk
L
1� K(
p
mK + 1)
1
kuk
L
1. Therefore
G
1
� K(
p
mK + 1)
1
;
and Conclusion 1 is then proved.
Now we show Conclusion 2. Since
~
A is Hurwitz, we can choose the explicit V
1;2
= x
T
Px where P
is the positive de�nite symmetric matrix, satisfying
~
A
T
P + P
~
A = �I : (39)
Take
V
2
(x) =
ckxk
3
3
+ x
T
Px
with c = 2KkPBk. Similar to the proof of Lemma 3, we can show that
G
2
� 2KkPBk(1 +
p
mK) : (40)
Therefore we need to compare kPBk with kWk
1
. For this purpose we consider the Hankel norm
for system (36). The matrix P is also called the observability Gramian for (36) in the literature (the
18
output is just the state in our case). The controllability Gramian for system (36) is de�ned to be the
symmetric matrix that satis�es
~
AQ+Q
~
A
T
+BB
T
= 0 : (41)
We know that the Hankel norm for (36) is equal to
kWk
hankel
=
�
�
max
(PQ)
�
1
2
; (42)
where �
max
(�) denotes the largest eigenvalue, cf. [1]. We also know that the H
1
-norm for (36) is
related to the Hankel norm by the following inequalities:
kWk
1
� (2n+ 1)kWk
hankel
� (2n+ 1)kWk
1
: (43)
Now in our case, since
~
A = A�BB
T
and
~
A
T
= �A�BB
T
, the controllability Gramian Q is equal to
1
2
I . Therefore the Hankel norm for (36) is just
kWk
hankel
=
�
�
max
(P=2)
�
1
2
:
Since P satis�es
(A
T
�BB
T
)P + P (A�BB
T
) + I = PA � AP � BB
T
P � PBB
T
+ I = 0 ;
multiplying to the right by P to both sides, we get
PAP �APP � BB
T
PP � PBB
T
P + P = 0 : (44)
Now taking trace to both sides of (44), we get that
kPBk
2
= Tr(P=2) :
On the other hand we know that Tr(P=2) is equal to the sum of all the eigenvalues of P=2. Therefore
we have kPBk �
p
n�
1
2
max
(P=2) =
p
nkWk
hankel
�
p
nkWk
1
. We end up with
G
2
� 2
p
nK(1 +
p
mK)kWk
1
= 2
p
nK(1 +
p
mK)
2
:
Remark 5 If we use the classical L
1
-norm, i.e., kxk
1
def
=
P
n
i=1
R
1
0
jx
i
(s)jds for functions x 2
L
1
([0;1); IR
n
), and for controls u 2 L
1
([0;1); IR
m
) too, then the estimate in Conclusion 1 of Corol-
lary 1 can be replaced by K(1 + K)
1
. In this manner, the dimension of the control space does
not appear in the bound. We suspect also that the estimate for G
2
should be independent of the
dimensions of both the state and the control spaces.
6 Nonzero Initial States
We now turn to nonzero initial states. We start with an easy observation.
Remark 6 Consider systems as in Theorem 2, but without controls, that is, any system (S) given by
_x = Ax+B�(Fx), where A;B; � are as in Theorem 2 and F is chosen as in its proof. It is well-known
that the origin is globally asymptotically stable, assuming for instance controllability of the matrix
pair (A;B). It is interesting to see that this fact also can be shown as a consequence of our arguments.
From the proof of Theorem 2, it is enough to show that the system _x = Ax�B�(B
T
x), with A skew-
symmetric and (A;B) controllable, is globally asymptotically stable with respect to the origin. But this
follows trivially from (31), since we have along the trajectories of (S
0
) that dV
2
(x(t))=dt � �kx(t)k
2
.
Thus V
2
is a strict Lyapunov function for this system without controls.
19
The previous remark suggests the study of relationships between L
p
-stability and global asymptotic
stability of the origin. We prove below that, even for nonlinear feedback laws, L
p
-stability for �nite p
implies asymptotic stability.
6.1 Relations Between State-Space Stability and L
p
-Stability
Lemma 5 Let A be an n � n matrix, B an n � m matrix, � an IR
m
-valued S-function and f a
locally Lipschitz continuous function from IR
n
to IR
m
such that f(0) = 0. We assume that (A;B) is
controllable. Consider the system of di�erential equations
_x = Ax+ B�(f(x)) (45)
and the control system
_x = Ax +B�(f(x) + u) ;
x(0) = 0 :
(46)
We have the following conclusions:
(i) if system (46) is L
p
-stable for some p 2 [1;1), then system (45) is locally asymptotically stable
with respect to the origin;
(ii) if the reachable set from 0 of (46) is equal to IR
n
and if system (46) is L
p
-stable for some p 2 [1;1),
then system (45) is globally asymptotically stable with respect to the origin;
(iii) if � and f are di�erentiable at 0, let F = D(� � f)(0). Then, if system (46) is L
p
-stable for some
p 2 [1;1), A +BF is a Hurwitz matrix.
The proof of this Lemma uses the following result, which is of independent interest.
Lemma 6 Let f : IR
n
� IR
m
! IR
n
be a locally Lipschitz continuous function such that kf(x; u)k �
K
1
+ K
2
kxk for all (x; u) 2 IR
n
� IR
m
, where K
1
; K
2
are some positive constants. Assume that the
system
_x = f(x; u); x(0) = 0 (47)
is L
p
-stable for some p 2 [1;1). Then there exists a constant C
p
> 0 so that for any u 2
L
p
([0;1); IR
m
), if x
u
is the corresponding solution of (47),
kx
u
k
L
1� C
p
max
n
kuk
L
p; kuk
p
p+1
L
p
o
: (48)
Moreover, x
u
(t)! 0 as t!1.
If in addition the following conditions hold:
(A) f(x; 0) is di�erentiable at 0 and
~
A = D
x
f(0; 0) is Hurwitz, and
(B) there exists M > 0 such that kf(x; u)� f(x; 0)k �Mkuk for all (x; u) 2 IR
n
� IR
m
,
then we can conclude
kx
u
k
L
1� C
p
kuk
L
p; (49)
and therefore for any q � p, kx
u
k
L
q�
~
C
p;q
kuk
L
pfor some constants C
p
;
~
C
p;q
> 0.
20
PROOF: For each T > 0, let �
T
= supfkx
u
(t)k j t 2 [0; T ]g. By the continuity of x
u
, there exist
t
1
� t
2
2 [0; T ] such that kx
u
(t
1
)k =
�
T
2
; kx
u
(t
2
)k = �
T
, and
�
T
2
� kx
u
(t)k � �
T
for t 2 [t
1
; t
2
]. Then
from (47) we have
�
T
2
�
Z
t
2
t
1
(K
1
+K
2
kx
u
(s)k)ds � (K
1
+K
2
�
T
)jt
2
� t
1
j :
Therefore jt
2
� t
1
j �
�
T
2(K
1
+K
2
�
T
)
. Then, if G
p
is the L
p
-gain of (46), we have
�
p+1
T
2
p+1
(K
1
+K
2
�
T
)
�
Z
t
2
t
1
kx
u
(s)k
p
ds � G
p
p
kuk
p
L
p
:
From this we can get
�
T
� C
p
maxfkuk
L
p; kuk
p
p+1
L
p
g
for some constant C
p
> 0 independent of u. Letting T !1, we obtain (48).
Pick now any u 2 L
p
([0;1); IR
m
). From the above conclusion, x
u
is bounded. From the systems
equations, it follows that also k _x
u
k is bounded. Thus, if y(t) = x
p
u;i
is a power of the ith coordinate of
x
u
, we know also that _y is bounded, from which it follows that y is uniformly continuous. Since also
by assumption the system is L
p
-stable, y 2 L
1
([0;1)). By Barbalat's Lemma, lim
t!1
y(t) = 0 and
hence also x
u
(t)! 0.
Now assume that (A) and (B) hold. We can rewrite system (47) as follows:
_x =
~
Ax+H(x) + L(x; u) ;
x(0) = 0 ;
(50)
where the continuous functions H and L satisfy
lim
kxk!0
kH(x)k
kxk
= 0 (51)
and kL(x; u)k = kf(x; u)� f(x; 0)k �Mkuk for all x; u.
Let V
2;p
be the function de�ned in the proof of Lemma 3 with this matrix
~
A. Then (51) implies
that there exists a � > 0 such that kDV
2;p
(x)H(x)k � kxk
p
if kxk � �. On the other hand, (48)
implies that there exists an � > 0 such that if kuk
L
p� �, the solution of (50) satis�es kx
u
(t)k � � for
all t � 0. Thus, if kuk
L
p� �, we have
_
V
2;p
(x
u
(t)) = �2kx
u
(t)k
p
+DV
2;p
(x
u
(t))
�
H(x
u
(t)) + L(x
u
(t); u(t))
�
� �kx
u
(t)k
p
+ c
p
Mkx
u
(t)k
p�1
ku(t)k :
Integrating, and using that V
2
(x
u
(0)) = 0 and V
2
is nonnegative, we conclude that kx
u
k
L
p� c
p
Mkuk
L
p
provided that kuk
L
p� �. Now if kuk
L
p� � we have kuk
p
p+1
L
p
� kuk
L
p�
�
1
p+1
. Then from (48) we have
kx
u
k
L
1� C
p
max
n
1; �
�
1
p+1
o
kuk
L
p:
The fact that (49) implies kx
u
k
L
q�
~
Ckuk
L
pnow follows trivially, as in Equation (35) in the proof of
Remark 4. This completes the proof of the Lemma.
We now return to the proof of the claims made in Lemma 5.
PROOF OF (i): Fix a u 2 L
p
([0;1); IR
m
). Let x
u
be the solution of (46) corresponding to u. From
Lemma 6 we know that x
u
(t)! 0 as t!1.
21
To prove stability, we need some elementary reachability results for linear systems. By our as-
sumption we know that the system
_x = Ax+ Bu (52)
is controllable. Any point x
0
2 IR
n
can be reached from 0 by trajectories of (52) at time 1. Moreover
we can choose a u
x
0
on [0; 1] that steers 0 to x
0
and satis�es ku
x
0
k
L
1
[0;1]
� Ckx
0
k, where C > 0 is a
constant depending on A;B (cf. e.g. [13]). By a measurable selection it is also true that there is a
measurable control v that steers 0 to x
0
for the system _x = Ax + B�(v), provided that x
0
is small
enough. Moreover kvk
L
1
[0;1]
can be made small if kx
0
k is small. So if we let u = v(t) � f(x(t)) on
[0; 1], where x is the solution of (52), then u steers 0 to x
0
for (46) at time 1. Let U be an open
neighborhood of 0. For each � > 0, let �(�) be small enough such that, for each x
0
with kx
0
k � �(�),
there exists a u
x
0
that steers 0 to x
0
for (46) with ku
x
0
k
L
p
[0;1]
< �. If x is the solution of (45) starting
at x
0
, and if we let u(t) = u
x
0
(t) on [0; 1] and u(t) = 0 on (1;1), then the solution x
u
of (46) satis�es
x
u
(t) = x(t � 1) on [1;1). By (48) we can take a � > 0 small enough such that for any x
0
with
kx
0
k � �(�), the solution x of (45) starting at x
0
stays in U . So system (45) is locally stable.
PROOF OF (ii): Local stability follows as in (i). To prove global attraction, note that the reach-
ability assumption implies that any trajectory of (45) can be seen as a part of a trajectory of (46)
corresponding to a control in L
p
. Now Lemma 6 provides that x(t)! 0.
PROOF OF (iii): By (i), we know that A+ BF cannot have eigenvalues with positive real part. We
show that A+ BF cannot have purely imaginary eigenvalues either.
Assume that A+BF has an eigenvalue on the imaginary axis. Let G
p
be the L
p
-gain for (46). Fix
any " > 0 so that "G
p
< 1. Then there is some m�n matrix F
"
so that kF
"
k � " and A+B(F +EF
"
)
has an eigenvalue with positive real part, where E = diag(�
0
1
(0); : : : ; �
0
m
(0)). (This follows as in [12],
or by noticing that one may �rst use a small F to make the system reachable from one input {cf. [13],
Remark 4.1.13{ and then using Ackermann's formula for pole-shifting.) By the Small-Gain Theorem,
the system
_x = Ax+ B�(f(x) + F
"
x+ u) ;
x(0) = 0 ;
(53)
obtained by feeding back F
"
x, is again L
p
-stable. Now from (i) we know that _x = Ax+B�(f(x)+F
"
x)
is locally asymptotically stable. But this is impossible since the linearized system is unstable (see e.g.
[13], Corollary 4.8.8). This contradiction shows that A+BF is Hurwitz.
6.2 Dissipation Inequality and Input to State Stability
Next we give a slightly di�erent proof of Theorem 2, which results in a weaker statement (we now allow
" to depend on p) but which is somewhat simpler. Moreover, it results in a simple dissipation-type
inequality, from which conclusions about nonzero initial states will be evident. We will only sketch
the steps, as they parallel those in the previous proofs.
Assume that A is skew-symmetric and A�BB
T
is Hurwitz. Fix a 1 � p <1 �rst. Let �; a; b;K,
V
0;p
; V
1;p
; V
2;p
be as in the proof of Lemma 3. Let
�
p
=
(b+ 1)
2
c
2
p
kBk
2
4a
;
e
p
= KkBk(�
p
+ c
p
) ;
"
p
=
1
2Ke
p
:
22
Consider the system
_x = Ax�B�(B
T
x+ u) + "
p
�(v) (54)
where the initial states are now arbitrary. Along the trajectories of (54),
_
V
0;p
(x(t)) = �kx(t)k
p�1
x
T
(t)B�
�
B
T
x(t) + u(t)
�
+ "
p
kx(t)k
p�1
x
T
(t)�(v(t))
� �kx(t)k
p�1
x
T
(t)B�
�
B
T
x(t)
�
+K
p
mkx(t)k
p�1
ku(t)k+K"
p
kx(t)k
p
: (55)
(Compare this with (23).) Similar to (25) and (27) we can get (for p > 1)
_
V
1;p
(x(t)) � �akx(t)k
p�2
kB
T
x(t)k
2
+KkBk kx(t)k
p�1
x
T
(t)B�
�
B
T
x(t)
�
+KkBk kx(t)k
p�1
ku(t)k+ "
p
Kkx(t)k
p�1
kv(t)k : (56)
and
_
V
2;p
(x(t)) � �2kx(t)k
p
+ c
p
kBk kx(t)k
p�1
n
(b+ 1)kB
T
x(t)k+Kx
T
(t)B�
�
B
T
x(t)
�
g
+ c
p
kx(t)k
p�1
�
KkBk ku(t)k+K"
p
kv(t)k
�
: (57)
Again letting V
p
(x) = e
p
V
0;p
(x) + �
p
V
1;p
(x) + V
2;p
(x), we obtain
_
V
p
(x(t)) � �(1�K"
p
e
p
)kx(t)k
p
+ kx(t)k
p�1
�
(K
p
m+ 1)e
p
ku(t)k+K"
p
(c
p
+ �
p
)kv(t)k
�
= �
1
2
kx(t)k
p
+ kx(t)k
p�1
�
(K
p
m+ 1)e
p
ku(t)k+K"
p
(c
p
+ �
p
)kv(t)k
�
:
Let
�
p
= maxf(K
p
m+ 1)e
p
; K"
p
(c
p
+ �
p
)g :
Thus, for p > 1,
_
V
p
(x(t)) � �
1
2
kx(t)k
p
+ �
p
kx(t)k
p�1
(ku(t)k+ kv(t)k) : (58)
Arguing as in the proof of Lemma 3, this provides L
p
-stability provided that x(0) = 0. But we also
note in this case that it is possible to rewrite (58) in a \dissipation inequality" form, as follows. First,
by Young's inequality, we have for any �; �; � > 0 and p > 1,
�
p�1
� �
p� 1
p
�
p
p�1
�
p
+
�
p
p�
p
:
Let
�
p
=
h
p
4(p� 1)�
p
i
p�1
p
:
Then (58) can be written as
_
V
p
(x(t)) � �
1
4
kx(t)k
p
+
�
p
p�
p
p
(ku(t)k+ kv(t)k)
p
:
So if we let
~
V
p
= 4V
p
, r
p
=
4�
p
p�
p
p
, we �nally conclude, along all solutions of (54):
_
~
V
p
(x(t)) � �kx(t)k
p
+ r
p
(ku(t)k+ kv(t)k)
p
: (59)
This is sometimes called a dissipation inequality ; see [7].
23
Take in particular p = 2 and write V =
~
V
p
. The estimate (59) shows that V (x(t)) must decrease
if kx(t)k is larger than d times the input magnitude, d =
p
r
2
. Thus, irrespective of the initial state,
the state trajectory is ultimately bounded, assuming that the inputs u and v are bounded, and this
asymptotic bound depends on an asymptotic bound on u and v. One way to summarize this conclusion
is by means of the estimate
kx(t)k � �(kx(0)k; t) +
�
k(u; v)k
L
1
[0;t]
�
(60)
valid for all x(0), all t � 0, and all essentially bounded u; v, where is a function of class K and �
is a class-KL function (that is, � : IR
�0
� IR
�0
! IR
�0
is so that for each �xed t � 0, �(�; t) is a
class-K function, and for each �xed s � 0, �(s; �) is decreasing to zero as t!1). This is the notion of
ISS-stability discussed in e.g. [11, 10, 17, 15]; Equation (60) is a consequence of (59), which says that
V is a Lyapunov-ISS function. In fact, in our case one can say more about the function , namely: it
can be taken to be linear . Indeed, from the proof (p. 441) in [11] one can take any � �
�1
1
� �
2
� �
4
,
where �
4
(l) = dl and where the �
i
's are class-K functions so that
�
1
(kxk) � V (x) � �
2
(kxk)
for all x 2 IR
n
. Here we can choose �
2
= c�
1
, for some c > 1, where �
1
is of the form �(l) =
a
1
l
2
+ a
2
l
3
, and is thus a convex function. Since for any convex function � and c > 1, and any d > 0,
�
�1
(c�(dl)) � cdl for all l, this gives a linear as claimed.
7 More General Input Nonlinearities
Now we consider a broader class of input nonlinearities, allowing unbounded functions as well. The
main result will be extended to this case.
De�nition 4 We call � : IR ! IR an
e
S-function if it can be written as �(t) = �tg(t) + �(t), where
� � � 0 is a constant;
� g : IR! [a; b] is measurable and a, b are strictly positive real numbers; and
� � : IR! IR is an S-function.
We say that � = (�
1
; : : : ;�
m
)
T
is an IR
m
-valued
e
S-function if each �
i
is an
e
S-function. As before if
� = (�
1
; : : : ; �
m
)
T
2 IR
m
, then �(�) = (�
1
(�
1
); : : : ;�
m
(�
m
)).
With this de�nition we have the following generalization of Theorem 1.
Theorem 4 Let A;B be n�n; n�m matrices respectively and � be an IR
m
-valued
e
S-function. Assume
that A is neutrally stable. Then there exists an m� n matrix F such that the system
_x = Ax+ B�(Fx + u) ;
x(0) = 0 ;
(61)
is L
p
-stable for all 1 � p � 1.
24
PROOF: As in the proof of Theorem 2, we can assume without loss of generality that A is skew-
symmetric and (A;B) is controllable.
Assume that � = (�
1
; : : : ;�
m
)
T
with �
i
= �
i
tg
i
(t) + �
i
(t). Let � = (�
1
; : : : ; �
m
) and G =
diag(�
1
g
1
; : : : ; �
m
g
m
) with G(�) = diag(�
1
g
1
(�
1
); : : : ; �
m
g
m
(�
m
)) for � 2 IR
m
, then �(�) = G(�)� +
�(�).
The �
i
's split into two sets, �
1
= f�
i
; �
i
> 0g and �
2
= f�
i
; �
i
= 0g. We can assume without loss
of generality that
�
1
= f�
1
; : : : ; �
r
g and �
2
= f�
r+1
; : : : ; �
m
g r � m:
Therefore system (61) becomes
_x = Ax +B
0
B
B
B
B
B
B
B
@
�
1
g
1
.
.
.
O O
O
.
.
.
�
r
g
r
O O
1
C
C
C
C
C
C
C
A
(Fx+ u) + B�(Fx+ u):
Consider B = (B
1
; B
2
), F =
�
F
1
F
2
�
, u =
�
u
1
u
2
�
, � =
�
�
1
�
2
�
, and G
1
= diag(�
1
g
1
; : : : ; �
r
g
r
). The
sizes of the matrices B
1
; B
2
; F
1
; F
2
are respectively, n � r; n � (m � r); r � n; (m � r) � n. As for
u
1
; u
2
, they are respectively elements of IR
r
and IR
m�r
. The S-functions �
1
; �
2
are respectively IR
r
and
IR
m�r
-valued. We rewrite (61) as
_x = Ax+ B
1
G
1
(F
1
x+ u)(F
1
x+ u
1
) + B
1
�
1
(F
1
x+ u
1
) +B
2
�
2
(F
2
x+ u
2
);
x(0) = 0:
(62)
Let R(A;B
1
) : IR
rn
! IR
n
be the reachability matrix of (A;B
1
). (Here and below we will identify
matrices with the corresponding linear maps.)
Let D = ImR(A ;B
1
) and H = D
?
. We have D �H = IR
n
. Clearly the subspace D is invariant
under A and Im(B
1
) � D. Since A is skew-symmetric, the subspace H is also invariant under A. So
there exists an orthogonal n � n matrix U such that
UAU
T
=
�
A
1
O
O A
2
�
(63)
where A
1
and A
2
are skew-symmetric and are restrictions of A to G and H respectively. So, up to
an orthonormal change of basis, we can assume that A is already of the form (63). According to this
decomposition, D = Im R(A;B
1
). Let s = dimD = rank R(A;B
1
). Consider now
x =
�
x
1
x
2
�
; B
1
=
�
B
11
B
12
�
; B
2
=
�
B
21
B
22
�
;
F
1
= (F
11
; F
12
); and F
2
= (F
21
; F
22
):
Here, x
1
2 IR
s
; x
2
2 IR
n�s
and the sizes of B
11
; B
12
; B
21
; B
22
and F
11
; F
12
; F
21
; F
22
are respectively
s� r; (n� s)� r; s� (m� r); (n� s)� (m� r) and r� s; r� (n� s); (m� r)� s; (m� r)� (n� s).
Since ImB
1
� D, we have B
12
= 0. Now system (62) becomes
_x
1
= A
1
x
1
+ B
11
G
1
(F
11
x
1
+ F
12
x
2
+ u
1
)(F
11
x
1
+ F
12
x
2
+ u
1
)
+ B
11
�
1
(F
11
x
1
+ F
12
x
2
+ u
1
) +B
21
�
2
(F
21
x
1
+ F
22
x
2
+ u
2
);
_x
2
= A
2
x
2
+ B
22
�
2
(F
21
x
1
+ F
22
x
2
+ u
2
) :
25
Choose now F
12
= F
21
= 0, F
11
= �B
T
11
, and F
22
= �B
T
22
. We obtain
_x
1
=
�
A
1
�B
11
G
1
(�B
T
11
x
1
+ u
1
)B
T
11
�
x
1
+ B
11
G
1
(�B
T
11
x
1
+ u
1
)u
1
+B
11
�
1
(�B
T
11
x
1
+ u
1
) +B
21
�
2
(�B
T
22
x
2
+ u
2
);
_x
2
= A
2
x
2
+B
22
�
2
(�B
T
22
x
2
+ u
2
):
In the previous system, replacing �(�) by ��(��) (still denoted by �), the system becomes
_x
1
=
�
A
1
�B
11
G
1
(�B
T
11
x
1
+ u
1
)B
T
11
�
x
1
+ B
11
G
1
(�B
T
11
x
1
+ u
1
)u
1
� B
11
�
1
(B
T
11
x
1
� u
1
)�B
21
�
2
(B
T
22
x
2
� u
2
);
_x
2
= A
2
x
2
�B
22
�
2
(B
T
22
x
2
� u
2
):
Since (A;B) is controllable, (A
2
; B
22
) is also controllable. It follows from Theorem 2 that the x
2
-
subsystem is L
p
-stable for all 1 � p � 1. So there exists C
1
p
> 0 such that kx
2
k
L
p� C
1
p
ku
2
k
L
p.
Let
v = B
11
G
1
(�B
T
11
x
1
+ u
1
)u
1
�B
11
�
�
1
(B
T
11
x
1
� u
1
)� �
1
(B
T
11
x
1
)
�
� B
21
�
2
(B
T
22
x
2
� u
2
):
We have
kvk � C(ku
1
k+ kx
2
k+ ku
2
k)
for some C > 0. Now for i = 1; : : : ; r, let d
i
(t) = �
1
i
(t)=t if t 6= 0 and d
i
(t) = 0 if t = 0. Let
~
G
1
(�) = diag(d
1
(�
1
); : : : ; d
r
(�
r
)). Then we can rewrite the x
1
-subsystem as
_x
1
=
�
A
1
�B
11
G
1
(�B
T
11
x
1
+ u
1
)B
T
11
�
x
1
�B
11
�
1
(B
T
11
x
1
) + v
=
h
A
1
�B
11
�
G
1
(�B
T
11
x
1
+ u
1
) +
~
G
1
(B
T
11
x
1
)
��
B
T
11
i
x
1
+ v:
If we let
e
D(t) = G
1
(�B
T
11
x
1
(t) + u
1
(t)) +
~
G
1
(B
T
11
x
1
(t)), then the above equation can be written as
_x
1
(t) =
�
A
1
� B
11
e
D(t)B
T
11
�
x
1
(t) + v(t) :
By de�nition of an S-function and an
e
S-function, there exist two real numbers �
1
and �
2
such that
0 < �
1
� �
2
and if
e
D(t) =diag(
~
d
1
(t); � � � ;
~
d
r
(t)), then
�
1
�
~
d
i
(t) � �
2
for i = 1; � � � ; r. Since (A;B) is controllable, (A
1
; B
11
) is controllable too. Then it follows from
Lemma 4 that
kx
1
k
L
p�
�
C
2
p
kvk
L
p;
for some
�
C
2
p
> 0 depending on A
1
; B
11
; �
1
; �
2
and p. But we know that
kvkp� C(ku
1
k
L
p+ ku
2
k
L
p+ kx
2
k
L
p) � Cku
1
k
L
p+ C(1 + C
1
p
)ku
2
k
L
p:
Therefore we have kx
1
k
L
p� C
2
p
kuk
L
pfor some constant C
2
p
> 0.
26
8 Counterexample: The nth Order Scalar Integrator
The next result is a negative one, and it concerns systems such as those in equation (1), except that
the matrix A is not neutrally stable but instead is assumed to have a non-simple Jordan block for
the zero eigenvalue. In that case, we show that for any possible F which stabilizes the corresponding
linear control system
_x = Ax+B(Fx + u) ;
x(0) = 0 ;
the resulting system (�
u
) is not in general L
p
-stable for any 1 � p � 1. We �rst consider the simplest
case, namely the double integrator. The proof is of interest because the origin of the corresponding
system without inputs (but with the saturation) is globally asymptotically stable. Thus the result is
quite surprising. In the end we discuss the n-integrator for n � 3.
Proposition 1 Let 1 � p � 1. Consider the following 2-dimensional initialized control system
(S
a;b
) _x = y ;
_y = ��(ax+ by + u) ;
x(0) = y(0) = 0 ;
where a; b > 0, � is a scalar S-function and inputs u belong to L
p
([0;1); IR). Then (S
a;b
) is not
L
p
-stable.
PROOF: Up to a reparametrization of the time and a linear change of variables, it is enough to
show that the initialized control system
_x = y ;
_y = ���(x+ y + u) ;
x(0) = y(0) = 0 ;
where � > 0, is not L
p
-stable. Now replacing �� by � (note �� is still an S-function) we may assume
that � = 1. Therefore all needed is to show that the system
(S) _x = y ;
_y = ��(x+ y + u) ;
with x(0) = y(0) = 0 is not L
p
-stable. The proof is quite technical, but the idea is not di�cult to
understand. It is based on the fact that the feedback u = �y makes the system (S) have periodic
trajectories, with a control u whose norm is proportional to that of the y coordinate. But the x
coordinate is the integral of y, so the ratio between the p-norms of x and u can be made to be large
for p <1. (For p =1, one modi�es the argument to reach states of large magnitude.)
Let us �rst �x a p in [1;1). Assume that (S) is L
p
-stable. Then the following holds: There exists
C
p
> 0 so that, if u 2 L
p
([0;+1); IR), then
ky
2
u
k
L
p� C
p
kuk
L
p; (64)
where y
u
is the second coordinate of (x
u
; y
u
), the solution of (S) associated to u.
27
To see this, let q = 2(p� 1) � 0 and
V
q
(x; y) = �
xyjyj
q
q + 1
:
Then along the trajectory (x
u
; y
u
) of (S) we have
_
V
q
= �
1
q + 1
jy
u
j
q+2
+ x
u
�(x
u
+ y
u
+ u)jy
u
j
q
:
Therefore,
_
V
q
+
1
q + 1
jy
u
j
q+2
� Kjx
u
jjy
u
j
q
; (65)
where K is an S-bound for �. From Lemma 6 we know that lim
t!1
(x
u
; y
u
) = (0; 0). Integrating (65)
from 0 to t and letting t!1, we end up with
1
q + 1
Z
1
0
jy
u
j
q+2
� K
Z
1
0
jx
u
jjy
u
j
q
:
Therefore if p = 1 we get that ky
2
u
k
L
1� Kkxk
L
1. If p > 1, applying H�older's inequality, we get
1
q + 1
Z
1
0
jy
u
j
2p
� Kkx
u
k
L
p
�
Z
1
0
jy
u
j
qp
p�1
�
p�1
p
:
But q
p
p�1
= 2(p� 1)
p
p�1
= 2p. Therefore
ky
2
u
k
L
p� (2p� 1)Kkx
u
k
L
p: (66)
Since (S) is L
p
-stable, kx
u
k
L
p� G
p
kuk
L
p, where G
p
is the L
p
-gain of (S). So (64) indeed holds.
Now we will construct trajectories of (S) which contradict (64).
We consider the level sets of the following Lyapunov function:
V (x; y) = y
2
+G(x);
where G(x) = 2
R
x
0
�(s)ds.
Let �
1
= 2 inf
jtj�1
j�(t)j > 0 and de�ne H : IR! IR by
H(x) =
(
0 if jxj � 1;
�
1
�
jxj � 1
�
if jxj > 1 :
We have
y
2
+H(x) � V (x; y) � y
2
+ 2Kjxj: (67)
Note that along trajectories of
(S
0
) _x = y ;
_y = ��(x);
V is constant.
Let us �x a constant V
0
� maxf1; 2Kg and let x
�
< 0 and x
+
> 0 be such that G(x
+
) = G(x
�
) =
V
0
. Since (S) is controllable, there exists a u
0
in L
p
([0;1); IR) such that for some time T
1
> 0
(x
u
0
(T
1
); y
u
0
(T
1
)) = (0;
p
V
0
). We can also assume that u
0
(t) = 0 for t > T
1
. For t � 0, consider
28
(�x
0
(t); �y
0
(t)), the solution of (S
0
) starting at the point (0;
p
V
0
). Note that V (�x
0
(t); �y
0
(t)) � V
0
.
Clearly this trajectory is periodic, since it lies in the closed curve V (x) � V
0
and there are no equilibria
there. Assume that the period is T .
Consider the sequence fu
n
g
1
n=1
of inputs de�ned as follows,
u
n
(t) =
8
>
<
>
:
u
0
(t) on [0; T
1
];
��y
0
(t� T
1
) on (T
1
; T
1
+ nT ];
0 on (T
1
+ nT;1) :
Then if (x
n
; y
n
) denotes the solution of (S) associated to u
n
, we have for t 2 [T
1
; T
1
+ nT ],
�
x
n
(t); y
n
(t)
�
=
�
�x
0
(t� T
1
); �y
0
(t� T
1
)
�
:
In this case,
Z
1
0
ju
n
(s)j
p
ds =
Z
T
1
0
ju
0
(s)j
p
ds + n
Z
T
0
j�y
0
(s)j
p
ds;
Z
1
0
jy
2
n
(s)j
p
ds =
Z
1
0
jy
2
u
0
(s)j
p
ds + n
Z
T
0
j�y
2
0
(s)j
p
ds :
We conclude that
lim
n!1
ky
2
n
k
L
p
ku
n
k
L
p
=
�
R
T
0
j�y
2
0
(s)j
p
ds
�
1
p
�
R
T
0
j�y
0
(s)j
p
ds
�
1
p
def
= L
p;V
0
:
According to (64), this quotient should be bounded independently of the choice of V
0
. We next derive
a contradiction by showing that this is not so.
Notice that for any r � 1, since
_
�x(t) = �y(t), we have
Z
T
0
j�y
0
(s)j
r
ds =
Z
T
0
j�y
0
(s)j
r�1
j
_
�x
0
(s)jds:
Since V (x; y) = V (x;�y), we have
Z
T
0
j�y
0
(s)j
r�1
j
_
�x
0
(s)jds = 2
Z
x
+
x
�
j�y(x)j
r�1
dx ; (68)
where j�y(x)j =
p
V
0
� G(x) for x between x
�
and x
+
. (Note that the curve V (x) = V
0
can be written
as the union of the graphs of the functions y(x) = �
p
V
0
� G(x). Thus we can reparameterize the
orbit in each of these two parts in terms of the variable x.
Considering (67), we have jx
�
j � V
0
=(2K) and x
+
� V
0
=�
1
+ 1. Then it follows from (68) that
R
T
0
j�y
0
(s)j
p
ds � 4
�
R
1
0
V
p�1
2
0
dx+
R
V
0
=�
1
+1
1
�
V
0
� �
1
(x� 1)
�
p�1
2
dx
�
� C
1
V
p+1
2
0
;
R
T
0
j�y
2
0
(s)j
p
ds � 4
R
V
0
=(2K)
0
�
V
0
� 2Kx
�
p�
1
2
dx � C
2
V
p+1=2
0
;
where C
1
; C
2
> 0 are some constants. Finally, we get L
p;V
0
� CV
1=2
0
for some C > 0. But according
to (64), L
p;V
0
� C
p
. Therefore, for V
0
large enough we get a contradiction. So (S) cannot be L
p
-stable
for 1 � p <1.
29
There remains to establish the special case p = 1. We use again the level sets of V . Let u
0
on
[0; T
0
] for some T
0
> 0 be an input such that
�
x
u
0(T
0
); y
u
0(T
0
)
�
= (0;
p
V
0
), for some V
0
> 0 that will
be �xed below.
From (0;
p
V
0
), follow the trajectory of
(I) _x = y;
_y = �
2
;
on [T
0
; T
0
+ 1], where �
2
= ��(�1) > 0. The trajectory of (I), hence, reaches (
p
V
0
+ �
2
=2;
p
V
0
+ �
2
).
Let
V
1
= (
p
V
0
+ �
2
)
2
+ G(
p
V
0
+ �
2
=2) � V
0
+ 2�
2
p
V
0
:
Note that also V
1
� V
0
+C(
p
V
0
+1) for some C > 0. Furthermore, the trajectory of (I) can be viewed
as a trajectory of (S) with u
1
(t) = �1� x(t)� y(t) for T
0
< t � T
0
+ 1. Let
u
1
= �u
1
(T
0
+ 1) = 1 +
p
V
0
+ �
2
=2 +
p
V
0
+ �
2
= 2
p
V
0
+ 3=2�
2
+ 1:
Then, for T
0
+1 < t � T
1
, follow the trajectory of (S
0
) from (
p
V
0
+�
2
=2;
p
V
0
+�
2
) until the resulting
trajectory reaches (0;
p
V
1
) at t = T
1
. This trajectory can also be considered as a trajectory of (S)
with u
1
(t) = ��y(t) on (T
0
+ 1; T
1
]. Note that ju
1
(t)j �
p
V
1
for T
0
+ 1 < t � T
1
. Fix V
0
such that
p
V
1
� u
1
� 3
p
V
0
. It is clear that on [T
0
; T
1
] ju
1
(t)j � u
1
.
If we iterate the above construction, we can build three sequences fV
n
g
1
n=0
, fu
n
g
1
n=1
and fT
n
g
1
n=0
such that
(1): V
n+1
= (
p
V
n
+ �
2
)
2
+ G(
p
V
n
+ �
2
=2) � V
n
+ 2�
2
p
V
n
;
(2): u
n
= 2
p
V
n�1
+ 3=2�
2
+ 1 � 3
p
V
n�1
;
(3): on [T
n�1
; T
n
], there exists an input u
n
such that supfju
n
(t)j : t 2 [T
n�1
; T
n
]g = u
n
and the
trajectory of (S) associated to u
n
goes from (0;
p
V
n�1
) to (0;
p
V
n
).
Clearly lim
n!1
V
n
=1 and then lim
n!1
u
n
=1. Furthermore let x
�
n
< 0 be such that G(x
�
n
) = V
n
.
Then jx
�
n
j � 1=(2K)V
n
for n large enough, which implies that lim
n!1
jx
�
n
j =1.
Let f�u
n
g
1
n=0
be the sequence of inputs which equals to the concatenation of u
0
; u
1
; � � � ; u
n
on [0; T
n
]
and 0 for t > T
n
. For n large enough, we have
k(x
�u
n; y
�u
n)k
1
� jx
�
n
j;
k�u
n
k
1
= u
n
:
Since
�
�
�
x
�
n
u
n
�
�
�
� 1=(2K)
p
V
n
for n large enough, (S) is not L
1
-stable.
For n integrators and n > 2, the proof that L
p
-stabilization is not possible is simpler (but the
result is far less interesting). We can argue as follows. Let � be a scalar S-function. It was proved in
[4, 21] that, if n � 3, the n-integrator
_x
1
= x
2
;
.
.
.
_x
n�1
= x
n
;
_x
n
= ��(u)
30
is not globally asymptotically stabilizable by any possible linear feedback. With this, it follows from
Lemma 5 that, if n � 3, the system
_x
1
= x
2
;
.
.
.
_x
n�1
= x
n
;
_x
n
= ��(Fx+ u) ;
x(0) = 0
is not L
p
-stable for any 1 � p <1 and any row vector F .
31
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32
