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GLOBAL STABILIZATION OF LINEAR SYSTEMS
WITH BOUNDED FEEDBACK
BY YUDI YANG
A dissertation submitted to the Graduate School--New Brunswick Rutgers, The State University of New Jersey
in conjunction with The Graduate School of Biomedical Sciences the University of Medicine and Dentistry of New Jersey
in partial fulfillment of the requirements for the
Joint Degree of Doctor of Philosophy
Graduate Program in Mathematics
Written under the direction of
H'ector J. Sussmann
and approved by
New Brunswick, New Jersey
October, 1993 Pass1: Page 2
c fl 1993
Yudi Yang ALL RIGHTS RESERVED Pass1: Page 2
ABSTRACT OF THE DISSERTATION Global Stabilization of Linear Systems
with Bounded Feedback
by Yudi Yang, Ph.D. Dissertation Director: H'ector J. Sussmann
There are two parts in the thesis. The first part deals with the problem of global sta- bilization of linear systems with bounded feedback. A linear system can be globally stabilized by a bounded feedback if and only if all eigenvalues of the system matrix have nonpositive real parts and all eigenvalues of the uncontrollable part have strictly negative real parts. If a linear system satisfies this condition, then we can provide an algorithm to find a bounded stabilizing feedback. The design employs linear combina- tions and compositions of linear functions and saturations. We also show that if we use only a saturated linear feedback, then a linear system cannot be globally stabilized in general. Some applications, such as output feedback stabilization, the stabilization of cascade systems, and the stabilization of flight control, are also presented here.
The second part deals with questions of global stabilizability of nonlinear systems. Based on the use of control-Lyapunov functions, we obtain a class of stabilizing feedback laws. A sufficient condition for such feedbacks to be continuously differentiable is presented. We then apply this condition to a wide class of two- and three-dimensional systems, extending some recent results on stabilization.
ii Pass1: Page 3
Acknowledgements During my research, I received much help from my advisor, Professor H'ector J. Suss- mann. Without his help, this thesis could not have been finished. I cannot sufficiently thank him for his invaluable help during the past years.
I also had many discussions with Professor Eduardo D. Sontag, and received constant suggestions from him during the research. I would like to thank him, especially for his introductory courses on systems and control theory.
I would like to thank the US Air Force and NSF for partially supporting my gradu- ate study under grants AFOSR-91-0343, AFOSR-91-0346, DMS-8902994, and DMS92- 02554.
A special thanks goes to my wife for her encouragement and understanding in the past three years.
iii Pass1: Page 4
Dedication This thesis is dedicated to my parents in China.
iv Pass1: Page 5
Table of Contents Abstract : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : ii Acknowledgements : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : iii Dedication : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : iv
1. Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1
1.1. Part One : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1 1.2. Part Two : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5 1.3. The Organization : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 7
2. Existence Results : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9
2.1. Preliminary Definitions and Results : : : : : : : : : : : : : : : : : : : : 9 2.2. Some Properties of Analytic Functions : : : : : : : : : : : : : : : : : : : 12 2.3. The Proof of Theorem 2.1 : : : : : : : : : : : : : : : : : : : : : : : : : : 17
3. The Naive Design : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 31
3.1. Some Special Cases : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 31 3.2. A Negative Result : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 33 3.3. The proof of Theorem 3.1 : : : : : : : : : : : : : : : : : : : : : : : : : : 34
4. A Design Using Hidden Layers : : : : : : : : : : : : : : : : : : : : : : : : 48
4.1. Constructions of Feedback : : : : : : : : : : : : : : : : : : : : : : : : : : 48 4.2. Technical lemmas : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 51 4.3. The Proof of Theorem 4.1 : : : : : : : : : : : : : : : : : : : : : : : : : : 71 4.4. An Algorithm : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 79 4.5. Multiple Integrators : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 87
v Pass1: Page 6
5. Applications : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 91
5.1. Output Feedback : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 91 5.2. Cascaded Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 93 5.3. F-8 Longitudinal Flight Control : : : : : : : : : : : : : : : : : : : : : : : 95
5.3.1. The Model : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 96 5.3.2. Trim Conditions : : : : : : : : : : : : : : : : : : : : : : : : : : : 98 5.3.3. Controller Design : : : : : : : : : : : : : : : : : : : : : : : : : : : 99 5.3.4. Stability of the Linear System : : : : : : : : : : : : : : : : : : : : 101 5.3.5. Simulation Results : : : : : : : : : : : : : : : : : : : : : : : : : : 106
6. Discrete-Time Linear Systems : : : : : : : : : : : : : : : : : : : : : : : : 112
6.1. Statement of the Main Results : : : : : : : : : : : : : : : : : : : : : : : 112 6.2. Preliminaries : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114 6.3. The Proof of Theorem 6.1 : : : : : : : : : : : : : : : : : : : : : : : : : : 121 6.4. The Proof of Theorem 6.2 : : : : : : : : : : : : : : : : : : : : : : : : : : 125
7. Nonlinear Affine Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : 126
7.1. General Theory : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 126 7.2. Desingularizing Functions : : : : : : : : : : : : : : : : : : : : : : : : : : 133 7.3. Applications to Planar Systems : : : : : : : : : : : : : : : : : : : : : : : 136 7.4. Applications to Three-dimensional Systems : : : : : : : : : : : : : : : : 146 References : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 151 Vita : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 154
vi Pass1: Page 1
1 Chapter 1 Introduction
There are two parts in the thesis. In the first part we completely solve the problem on global stabilization of linear systems with bounded feedback. In the second part we present some results on global stabilization of nonlinear affine systems with continuously differentiable feedback.
1.1 Part One We consider linear time-invariant continuous-time systems
\Sigma : .x = Ax + Bu ; (1.1.1) where A 2 IR
n\Theta n
, B 2 IR
n\Theta m
for some integers n (the dimension of the system) and m
(the number of inputs), and for which the control values u(t) are restricted to lie in a bounded set U ` IR
m
. We assume that U contains zero in its interior.
The study of such systems is motivated by the possibility of actuator saturation or constraints on actuators, reflected sometimes also in bounds on available power supplies or rate limits. These systems cannot be naturally dealt with within the context of standard (algebraic) linear control theory but are ubiquitous in control applications. To quote the recent textbook [23] (page 171): "saturation is probably the most commonly encountered nonlinearity in control engineering." Mathematically, control questions become nontrivial, since only control values in U are allowed into the underlying linear system.
We will present results on global stabilization of systems of this form. Precisely, we want to find a bounded vector-valued function k : IR
n
! IR
n
such that the closed-loop
system of (1.1.1) with the feedback u = k(x) has the origin as a global asymptotically Pass1: Page 2
2 stable equilibrium. Of course, this cannot always be achieved for general linear systems, even for controllable ones. For instance, if the feedback is required to be bounded by 1, then no trajectory of the one-dimensional open-loop system .x = x + u starting in the region fx : jxj ? 1g can approach zero. So, we need to determine conditions under which the system (1.1.1) is globally asymptotically stabilizable.
The theory of controllability of linear systems with bounded controls is a well- studied topic; see e.g. the fundamental paper [21], as well as the different, more algebraic approach discussed in [24]. In these two references it was proved that such asymptotic null-controllability of the system (that is, the existence of open-loop controls that steer each state to the origin, in the limit as t ! 1 ) is equivalent to the following pair of algebraic conditions:
(a) all eigenvalues of A have nonpositive real part, and (b) all eigenvalues of the uncontrollable part of \Sigma have strictly negative real parts
(that is, the pair (A; B) is stabilizable in the ordinary sense).
Note that under Conditions (a) and (b) there may very well be nontrivial Jordan blocks corresponding to critical eigenvalues, so that the system .x = Ax need not be asymp- totically stable, or even Lyapunov-stable. This is what makes the problem interesting and allows inclusion of examples of practical importance such as systems involving integrators.
The first work on global stabilization of linear systems with bounded feedback ap- peared in [28]. In that paper it was shown that the above two conditions are equivalent to the global stabilizability of (1.1.1) with bounded smooth feedback. Using the same technique as in [28], we can even show that under Conditions (a) and (b), there exists a bounded analytic feedback that stabilizes (1.1.1). The proof is by induction. The induction starts with presenting an explicit stabilizing feedback for diagonable A of (1.1.1). Then, if A is not diagonable, we obtain the system matrix by adding Jordan blocks one by one to a diagonable system matrix without changing its eigenvalues and use the inductive hypothesis each time we add a Jordan block. Since the resulting feed- back relies on many submanifolds defined by integrals corresponding to some system Pass1: Page 3
3 trajectories, this approach is not constructive.
In very special cases, including all one- and two-dimensional systems, stabilization is possible by simply using a saturated linear feedback law of the type:
u = oe(F x) ; (1.1.2) where F is an m \Theta n matrix and oe is a function that computes a saturation in each coordinate of the vector F x, for instance, u
i
= tanh((F x)
i
) or u
i
= sat ((F x)
i
) where,
sat (s) = sign (s) minfjsj; 1g : (1.1.3) As stated above, this is also possible in the case of systems for which the Jordan form of A has no off-diagonal ones, i.e., neutrally stable systems. (See [15] and [22].) Thus it is natural to ask if such simple control laws as that of Equation (1.1.2) can also be used for more general systems. This was negatively answered in a paper by A.T. Fuller as far back as the late 1960's. He showed in [13] that already for triple integrators such saturated linear feedback is not sufficient, provided that certain assumptions are satisfied by the saturation oe. In [31] we gave an independent proof of such a negative result, which applies to basically arbitrary oe's. Our result shows that if oe : IR ! IR is a locally Lipschitz function for which both limits lim
s!\Sigma 1
oe(s) exist and are nonzero,
then there is no stabilizing feedback of the form (1.1.2) for n-dimensional integrators with n * 3.
The fact that linear feedback laws when saturated can lead to instability has moti- vated a large amount of research. See for instance [17] and [18], and references there, for estimates of the size of the regions of attraction that result when using linear saturated controllers. Rather than working with linear saturated control laws u = oe(F x) and trying to show that they are globally stabilizing, or trying to estimate their domains of attraction, we allow more general bounded (and hence necessarily nonlinear) laws. Since saturated linear feedbacks suffice for up to two dimensions, in higher dimensions it is natural to look for other simple control rules, for instance, employing linear combina- tions and compositions of saturation nonlinearities. In the language of neural networks, one wants control laws that are implementable by feedforward nets using "hidden lay- ers" rather than the "perceptrons" represented by (1.1.2). Motivated in part by [28] Pass1: Page 4
4 and [31], Andrew Teel showed in [32] how, in the particular case of single-input multiple integrators, such combinations of saturations are indeed sufficient to obtain stabilizing feedback controllers. In [30] we obtained a general solution of the same type, for the full case treated in [28]. The approach is explicit and constructive. Our solution is inspired by the techniques introduced in [32] for the particular case treated there, but the details are far more complicated, due to the possibilities of having both multiple inputs and (perhaps multiple) purely imaginary eigenvalues and to the need to deal with arbitrary saturations.
Our result was first announced in [29] and [40], where we considered a very special type of feedback for which the saturations are exactly linear near 0. When a system has a pure imaginary eigenvalue, a saturation with three different slopes may be needed. Later, in [30], we extended the result by allowing essentially arbitrary saturations. The only conditions imposed on the saturation functions oe are that
(i) oe is locally Lipschitz, (ii) soe(s) ? 0 whenever s 6= 0, (iii) oe is differentiable at 0 and oe
0
(0) ? 0, and
(iv) lim inf
jsj!1
joe(s)j ? 0.
So, mathematically, the results in [30] show that one can use analytic functions to im- plement feedback laws (the results in [40] and [32] would not give this) and, from an engineering point of view, they insure that rather general components could be em- ployed, subject only to mild conditions which are robustly satisfied. In the terminology of current "artificial neural networks" technology, our results allow the implementation of feedback controllers using very general types of activation (neuron characteristic) functions.
In addition to the design that employs linear combinations and compositions of sat- uration nonlinearities as constructed in [32], we also exhibit a different design that uses linear combinations of saturated linear functions. (In neural network terms, it involves a "single hidden layer net.") So, from the above point of view, we have completely Pass1: Page 5
5 solved the problem of global stabilization of linear systems with bounded feedback on both the theoretical level and the algorithmic level. As applications of our results, we studied the output stabilization problem and the stabilization of cascaded systems. Those results were already shown in [40], [30]. In particular, following the general phi- losophy outlined in [29, 40, 30], we developed in detail in [39] an explicit design for the linearized equations of longitudinal flight control for an F-8 aircraft and tested the resulting controller --via simulations-- on the original nonlinear model. In the process of working out this example, we were able to obtain, in certain particular cases, bounds tighter than those of [40, 30]. With these improved bounds, better performance can be achieved.
1.2 Part Two The purpose of this part is to seek continuously differentiable feedbacks for nonlinear affine systems of the type
\Sigma : .x = f (x) + g(x)u ;
where x 2 IR
n
, u is a scalar input, and f; g 2 C
1
(IR
n
), such that the origin is a globally
asymptotically stable equilibrium of the resulting closed-loop systems. There are many articles dealing with the stabilization problem for \Sigma . (See the references listed below.) Particularly, in [1] Artstein proved that if there exists a control-Lyapunov function V of a nonlinear affine system, then there is a feedback that stabilizes the system. Sontag, in [26], improved upon this result by giving a simple explicit formula for such a feedback in terms of directional derivatives of V . This feedback has the same regularity as L
f
V
and L
g
V outside the origin but generally fails to be continuously differentiable at the
origin. The interest in Part Two is in studying the existence of a stabilizing feedback which is continuously differentiable at the origin.
We first show that the feedback law given in [26] can be replaced by a class of functions expressed in terms of L
f
V and L
g
V , and that when L
f
V and L
g
V are ho-
mogeneous with respect to a one-parameter group of dilations, some of these feedbacks are continuously differentiable at the origin. We then present some applications to the Pass1: Page 6
6 stabilization of two- and three-dimensional systems of the type
.x = f (x; y) ;
.y = u : A well-known result on the stabilization of this composite system is based on the exis- tence of a smooth or differentiable stabilizing feedback for .x = f (x; u). But, our results show that even if the system .x = f (x; u) cannot be stabilized by a continuously differ- entiable feedback, it may still be possible to stabilize the composite system .x = f (x; y),
.y = u by a continuously differentiable feedback. The approach pursued here relies on a "desingularizing function" as defined in [20] whose use will be illustrated below. Since we use a constructive proof, such a feedback can be exactly computed.
Our general approach is applied to particular classes of two- and three-dimensional systems. For example, we consider two-dimensional polynomial systems of the type
.x =
P
l*0;m
0
\Gamma jl*0
c
l
x
m
0
\Gamma jl
y
n
0
+il
;
.y = u ;
(1.2.1)
where i ? 0; j ? 0, m
0
* 2, n
0
* 0, and c
0
6= 0 (e.g., .x = x
2
\Gamma 3xy
2
+ y
4
; .y = u).
Our result in this case is that (1.2.1) can be globally asymptotically stabilized by a continuously differentiable feedback if and only if there exist (x
1
; y
1
) and (x
2
; y
2
)
with x
1
? 0, x
2
! 0 such that f (x
1
; y
1
) ! 0 and f (x
2
; y
2
) ? 0, where f (x; y) =
P
l*0;m
0
\Gamma jl*0
c
l
x
m
0
\Gamma jl
y
n
0
+il
. This generalizes the result obtained in [12] for homoge-
neous planar systems.
We also analyze the stabilizability with continuously differentiable feedback of the system
.x = (x \Gamma y
5
)(x
2
+ jyj
k
) ;
.y = u :
(1.2.2)
We show that (1.2.2) is stabilizable with continuously differentiable feedback if k ? 4, and (1.2.2) is not stabilizable with continuously differentiable feedback if k ! 4.
As a third example, we consider three-dimensional polynomial systems of the type
.x = f (x; y) ; .y = g(x; y) + z
q
;
.z = u ;
(1.2.3) Pass1: Page 7
7 where both f and g are homogeneous polynomials of degree p, and p; q are odd numbers. We show that if p ? 1, then (1.2.3) can be globally asymptotically stabilized by a continuously differentiable feedback if and only if there exists (x
1
; y
1
) with x
1
? 0 such
that f (x
1
; y
1
) ! 0. When p = 1 there is a similar result on the stabilization with a
continuous feedback. When f (x; y) = x
p
and g(x; y) = 0, the system (1.2.3) reduces
.x = y
p
;
.y = z
q
;
.z = u ;
(1.2.4)
where p; q are odd numbers. So, in particular for (1.2.4), we are able to extend the results of [5] and [11], where only the case of p = q was considered.
1.3 The Organization The thesis is organized as follows. Chapters 2 through 6 discuss the problem of global stabilization of linear systems with bounded feedback. In Chapter 2 we de- velop some conditions that are equivalent to the global stabilizability of linear systems with bounded feedback using a nonconstructive approach.
In Chapter 3 we present a simple design, based on linear saturated feedback, for some special cases such as double integrators and systems whose system matrices are diagonable. We also show that the linear saturated feedback design does not work for general linear systems.
In Chapter 4 we exhibit a constructive design and a precise algorithm to find a bounded stabilizing feedback for a given linear system when such a feedback exists. Our analysis is for general linear systems. To study a special system, we can either apply the algorithm to the system or just follow the general philosophy of the approach and thus obtain a simpler feedback. So, we also exhibit a particular design that applies to scalar-input multiple integrators.
In Chapter 5 we present the applications to output feedback stabilization, the sta- bilization of cascade systems, and F-8 aircraft control. Pass1: Page 8
8 Finally, in Chapter 6 we discuss the global stabilizability of linear discrete-time systems with bounded feedback. We present a constructive approach for general linear discrete-time systems.
The second part of the thesis is included in Chapter 7, which is totally independent from the first six chapters. Section 7.1 presents the main results of the general theory. We describe a class of stabilizing feedback laws that use a control-Lyapunov function design and give a sufficient condition for feedbacks in the class to be continuously differentiable. In Section 7.2 we introduce desingularizing functions and state a simple, easy to apply version of Lemma 1 of [20]. Finally, in Sections 7.3 and 7.4 we apply the theory of Sections 7.1 and 7.2 to some examples of two- and three-dimensional systems. Pass1: Page 9
9 Chapter 2 Existence Results
This chapter develops some conditions that are equivalent to the global stabilizability of linear systems with bounded feedback.
2.1 Preliminary Definitions and Results Definition 2.1.1 Let \Sigma be a finite-dimensional linear system .x = Ax + Bu; x 2 IR
n
; u 2 IR
m
. Let k : IR
n
! IR
m
be a locally Lipschitz function. Then we say that k
stabilizes \Sigma if 0 is a globally asymptotically stable (GAS) equilibrium of the closed-loop system .x = Ax + Bk(x). 2
We use \Sigma
k
to denote the closed-loop system: .x = Ax + Bk(x). If k(x) = Kx for all
x 2 IR
n
, where K is a matrix of an appropriate size, we use \Sigma
K
to denote the closed-loop
system.
Definition 2.1.2 A linear system \Sigma is asymptotically controllable (AC) if for every initial condition _x 2 IR
n
there exists a bounded measurable open-loop control u :
[0; 1) ! IR
m
with the property that the trajectory t ! x(t) of \Sigma that corresponds to
u(\Delta ) and satisfies x(0) = _x is such that lim
t!1
x(t) = 0.
ffl If there is a fixed constant C such that for every _x the control u(\Delta ) can be chosen
in such a way that jju(t)jj ^ C for all t * 0, then we say that \Sigma is asymptotically controllable with bounded control (ACBC).
ffl If in addition C can be chosen arbitrarily small, then we say that \Sigma is asymptot-
ically controllable with small controls (ACSC). 2 Pass1: Page 10
10 Definition 2.1.3 Let \Sigma be a linear system. Then \Sigma is:
ffl bounded feedback stabilizable (BFS) if there exists a bounded locally Lipschitz
feedback k that stabilizes \Sigma .
ffl small feedback stabilizable (SFS) if for every " ? 0 there exists a stabilizing feed-
back k for \Sigma such that jjk(x)jj ^ " for all x, i.e. jjkjj
1
^ ". 2
Definition 2.1.4 Let F be any normed space of feedback controls. Let \Sigma be a linear system. Then \Sigma is:
ffl F -stabilizable if there is a stabilizing feedback k 2 F for \Sigma . ffl small norm F -stabilizable if for every " ? 0 there is a stabilizing feedback k 2 F
for \Sigma with F -norm ! ". 2
If F is the space of all linear feedbacks x ! Kx
def
= k
K
(x), with the norm jjk
K
jj =
jjKjj, then we will use the expressions "linearly stabilizable" and "linearly stabilizable with small norm".
Definition 2.1.5 For an n-tuple ffff
ff
= (ff
1
; \Delta \Delta \Delta ; ff
n
) of nonnegative integers, define jffff
ff
j =
ff
1
+ \Delta \Delta \Delta + ff
n
, and let D
ffff
ff
denote D
ff
1
1
\Delta \Delta \Delta D
ff
n
n
, where D
ff
i
i
=
@
ff
i
@x
ff
i
i
. Let F
w
n;m;r
denote
the space of all analytic functions k : IR
n
! IR
m
such that
jjkjj
C
r
= supfjjD
ffff
ff
k(x)jj : x 2 IR
n
; 0 ^ jffff
ff
j ^ rg
is finite. Then a linear system is analytically stabilizable with small C
r
-norm if it is
small-norm F
w
n;m;r
-stabilizable. 2
For an n-dimensional linear system
\Sigma : .x = Ax + Bu; x 2 IR
n
; u 2 IR
m
; (2.1.1)
IR
n
has a direct sum decomposition IR
n
= E
s
\Phi E
c
\Phi E
u
, where AE
s
` E
s
; AE
c
`
E
c
; AE
u
` E
u
, such that the eigenvalues of the restrictions A
s
, A
c
, A
u
of A to E
s
, E
c
, Pass1: Page 11
11 E
u
are contained, respectively, in fs : Re s ! 0g, fs : Re s = 0g, fs : Re s ? 0g. If we
write x 2 IR
n
as (x
s
; x
c
; x
u
), and Bu = (B
s
u; B
c
u; B
u
u), then the system equations for
\Sigma are written in the form
.x
s
= A
s
x
s
+ B
s
u; x
s
2 E
s
;
.x
c
= A
c
x
c
+ B
c
u; x
c
2 E
c
;
.x
u
= A
u
x
u
+ B
u
u; x
u
2 E
u
:
These three systems will be denoted, respectively, by \Sigma
s
, \Sigma
c
and \Sigma
u
, and referred to as
the stable, critical and unstable parts of \Sigma . We will write \Sigma
u
= 0, \Sigma
s
= 0 if E
u
= f0g
or E
s
= f0g.
Theorem 2.1 Let \Sigma be a finite-dimensional linear system. Then the following condi- tions are equivalent:
(1) \Sigma
u
= 0 and \Sigma
c
is controllable,
(2) \Sigma is ACBC, (3) \Sigma is ACSC, (4) \Sigma is BFS, (5) \Sigma is SFS, (6) \Sigma is analytically stabilizable with small C
1
-norm,
(7) \Sigma is linearly stabilizable with small norm.
We remark that (6) implies that there exists an analytic stabilizing feedback which is globally Lipschitz. This property is needed to deal with output feedback stabilization; see [28]. In Chapter 4 we will give another equivalent condition, namely, \Sigma is analytically stabilizable with small C
0
-norm, and the resulting closed-loop system has the small-
input small-state property. As will be shown in chapter 5, the small-input small-state property makes output stabilization considerably simpler. Pass1: Page 12
12 2.2 Some Properties of Analytic Functions The most delicate part of the proof of Theorem 2.1 is the implication (1) ) (6). Before we prove Theorem 2.1, we need some properties of analytic functions.
Lemma 2.2.1 Let f : IR
n
! IR
n
and g : IR
n+1
! IR
n
be two real analytic functions
such that
(i) f (0) = 0, (ii) all eigenvalues of the linearization of f (x) at the origin have negative real parts, (iii) jg(x; t)j ^ Cjxj for x in a neighborhood of 0 in IR
n
, where C is a constant.
Then there exists a neighborhood U of the origin in IR
n
such that the function , : U !
IR
n
given by
,(x) =
Z
1
0
g(fl(x; t); t)dt (2.2.1)
is well-defined and analytic, where t ! fl(x; t) denotes the trajectory of
. OE = f (OE)
starting at x.
Proof. We consider the differential equation .z = f (z) for z 2 C
n
. For z near 0 in
C
n
, let ,(z) be defined by the expression (2.2.1), where t ! fl(z; t) is the trajectory
of
. OE = f (OE) starting at z. Note that the system .z = f (z) is locally asymptotically
stable because its linearization at the origin is stable. Therefore, fl(z; t) exists for all t * 0 provided that z is sufficiently close to 0. We will show that the integral (2.2.1) converges when z is near 0 and , is analytic at the origin. Since the restriction of , to IR
n
is a real-valued function, if follows that , is analytic at the origin.
First, we need to give a precise meaning to the formula for ,(z) for z 2 C
n
. Note
that f : IR
n
! IR
n
is analytic at 0. Using the power series of f at 0, we can extend f to
a neighborhood O
1
of the origin in C
n
. Similarly, since g : IR
n+1
! IR
n
is an analytic
function, given (x; t) 2 IR
n+1
, we can use the Taylor's series of g at (x; t) to extend g
to a neighborhood of (x; t) in C
n+1
. Therefore, we can extend g to an open set O
2
in
C
n+1
, and IR
n+1
2 O
2
. So, if t ! fl(z; t), z 2 C
n
, is a trajectory of
. OE = f (OE) such that Pass1: Page 13
13 fl(z; t) 2 O
1
and (fl(z; t); t) 2 O
2
for all t * 0 and
R
1
0
g(fl(x; t); t)dt converges, then ,(z)
is well-defined.
Let A denote the linearization matrix of f (x) at the origin. From the assumption, we know that all eigenvalues of A have negative real parts. Therefore there exist two positive constants K and oe such that je
At
j ^ Ke
\Gamma oet
for t * 0. Write f (z) = Az + h(z)
for z near 0 in C
n
. Then h is analytic at the origin in C
n
, and h(z) = O(jzj
2
) for z near
the origin. So there is a ffi ? 0, such that jh(z)j ^
oe
2K
jzj for jzj ! ffi.
Let z 2 C
n
be such that jzj ! ffi. Then for T ? 0, if fl(z; t) exists for 0 ^ t ! T , we
have
fl(z; t) = e
At
z +
Z
t
0
e
A(t\Gamma s)
h(fl(z; s))ds :
It follows that if jfl(z; t)j ! ffi for 0 ^ t ! T , then
jfl(z; t)j ^ Ke
\Gamma oet
jzj + K
Z
t
0
e
\Gamma oe(t\Gamma s)
jh(fl(z; s))jds ^ Ke
\Gamma oet
jzj +
oe
2
Z
t
0
e
\Gamma oe(t\Gamma s)
jfl(z; s)jds :
Multiplying e
oet
to the above inequality, we get
e
oet
jfl(z; t)j ^ Kjzj +
oe
2
Z
t
0
e
oes
jfl(z; s)jds :
From Grownwall's inequality, we conclude that
e
oet
jfl(z; t)j ^ Kjzje
oe
2
t
:
Therefore, if fl(z; t) exists for 0 ! t ! T , where T ? 0, and jfl(z; t)j ! ffi, then
jfl(z; t)j ^ Kjzje
\Gamma
oe
2
t
: (2.2.2)
Now let z 2 C
n
be such that jzj ! minfffi;
ffi
K
g. It is obvious that there is a solution
for small t such that jfl(z; t)j ! ffi. Therefore, (2.2.2) holds for such t. Let T be the infimum t such that (2.2.2) fails. Then for t 2 [0; T ), we have jfl(z; t)j ^ Kjzj ! ffi. Therefore lim sup
t!T
jfl(z; t)j ^ Kjzj ! ffi. If T ! 1, then the trajectory can be extended
at (fl(z; T ); T ) and consequently (2.2.2) holds for some t ? T , which contradicts with the assumption that (2.2.2) fails for t ? T . Thus T = 1.
To summarize, we have shown that for z 2 B
ffi
= fz 2 C
n
: jzj ! ffi; jzj ! ffi=Kg,
fl(z; t) exists for all t and (2.2.2) is satisfied. Since f is analytic on fz 2 C
n
: jzj ! ffig, Pass1: Page 14
14 it follows that the solutions of
. OE = f (OE) starting from B
ffi
analytically depend on t and
the initial values (see [6], Theorem 7.2, for instance). So, for z 2 B
ffi
, fl(z; t) is analytic
in z and t (* 0).
Recall that
jg(z; t)j ^ Cjzj (2.2.3)
for z near the origin in C
n
. Let ffi be sufficiently small such that (2.2.3) is satisfied for
z 2 B
Kffi
and (2.2.2) is satisfied for z 2 B
ffi
. Then for z 2 B
ffi
we have
jg(fl(z; t); t)j ^ Cjfl(z; t)j ^ CKjzje
\Gamma
oe
2
t
:
Thus the integral
R
1
0
g(fl(z; t); t)dt converges. Write ,(z) =
1 P
n=0
u
n
(z), where u
n
(z) =
R
n+1
n
g(fl(z; t); t)dt. Note that ju
n
(z)j ^
R
n+1
n
CKjzje
\Gamma
oe
2
t
dt ^
2Cffi
oe
e
\Gamma
oe
2
n
. It follows that
1 P
n=0
u
n
(z) converges uniformly on z 2 B
ffi
by the dominated convergence theorem. Since
every u
n
is analytic on B
ffi
, we conclude that , is analytic on B
ffi
. 2
Remark 2.2.2 It is obvious that if we replace (2.2.1) by
,(x) =
Z
1
0
g(fl(x; t); t + T )dt ;
where T is a positive constant, then the conclusion of Lemma 2.2.1 is still true. In particular, we can find a neighborhood of the origin so that , is analytic there. From the proof of the lemma, we also see that the neighborhood can be chosen independently of T , because B
ffi
depends only on the value of K and the region where (2.2.3) is true
in the proof. We will use this fact in the proof of the next lemma. 2
Lemma 2.2.3 Let f : IR
n
! IR
n
and g : IR
n+1
! IR
n
be two real analytic functions
and satisfy the conditions (i)-(iii) of Lemma 2.2.1. Suppose that the origin is a globally asymptotically stable equilibrium of the system .x = f (x). Let t ! fl(x; t) denote the trajectory of
. OE = f (OE) with fl(x; 0) = x. Then the function , defined by
,(x) =
Z
1
0
g(fl(x; t); t)dt
is analytic on IR
n
. Pass1: Page 15
15 Proof. From the assumptions of the lemma, we know that for any x 2 IR
n
, fl(x; t)
converges to zero exponentially as t ! 1. Since g(x; t) is bounded by Cjxj when jxj is small, it follows that g
i
fl(x; t); t
j
also converges to zero exponentially as t ! 1. So
,(x) is well-defined for every x 2 IR
n
.
Let T be an arbitrary positive number and use x
T
to denote fl(x; T ). Let ,
T
(x) and
,
T
(x) denote the following two integrals:
,
T
(x) =
Z
T
0
g(fl(x; t); t)dt ;
,
T
(x) =
Z
1
T
g(fl(x; t); t)dt =
Z
1
0
g(fl(x
T
; t); T + t)dt :
Then ,(x) = ,
T
(x) + ,
T
(x). From Remark 2.2.2, we know that there is a neighborhood
B of the origin, which is independent of T , such that
~ ,
T
(x) =
R
1
0
g
i
fl(x; t); T + t
j
dt is
analytic in B.
Let x
0
be an arbitrary point in IR
n
. Since fl(x
0
; t) approaches zero as t ! 1,
it follows that, for sufficiently large T , fl(x
0
; T ) 2 B. We conclude that there is a
neighborhood B
x
0
of x
0
such that fl(x; T ) 2 B for x 2 B
x
0
. (See Theorem 4.1 in [7].)
It is clear that fl(\Delta ; T ) : B
x
0
! B is analytic at x
0
. So ,
T
(\Delta ) =
~ ,
T
(fl(\Delta ; T )) is analytic
at x
0
for sufficiently large T . Also, since g and fl: IR
n+1
! IR
n
are analytic functions,
it is obvious that ,
T
: IR
n
! IR
n
is analytic for any T ? 0. Therefore, by choosing
sufficiently large T , we see that , = ,
T
+ ,
T
is analytic at x
0
. 2
Lemma 2.2.4 Suppose h is an increasing function from IR
+
to IR
+
. Then there exists
an analytic function ' from IR to IR
+
such that '(x) ^ h(jxj)
\Gamma 1
and j'
0
(x)j ^ h(jxj)
\Gamma 2
.
Proof. We first prove that there exists a positive entire function h
1
: IR ! IR
+
such
that h(jxj) ^ h
1
(x).
Let h
1
(x) be a power series in the form
P
1
i=1
c
i
x
2k
i
, where c
i
; k
i
; i = 1; 2; \Delta \Delta \Delta , are
given inductively below:
ffl c
1
= h(2); k
1
= 0;
ffl if c
i
; k
i
are defined for i = 1; 2; \Delta \Delta \Delta ; n \Gamma 1, then let k
n
be an integer greater than
k
n\Gamma 1
satisfying n
k
n
? h(n + 1), and set c
n
= n
\Gamma k
n
. Pass1: Page 16
16 To see that
P
1
i=1
c
i
x
2k
i
converges for all x 2 C, we write the series as
P
1
k=0
a
n
x
n
. Then
for n = 2k
i
, we have a
1 n n
= c
1 2k
i
i
= i
\Gamma
1
2
; otherwise, a
1 n n
= 0. So, a
1 n n
! 0 as n ! 1.
From Hadamard's theorem about the convergence radius of series, we see that
P
c
i
x
2k
i
converges for all x 2 C
Now given any x, suppose n ^ jxj ! n + 1. Then
h
1
(x) * h
1
(n) * c
n
n
2k
n
= n
k
n
? h(n + 1) * h(jxj) :
So h
1
is an entire function as required.
Let h
2
(x) = h
1
(x) + e
x
. Then h
2
(x) ! 1 as x ! +1, and h
0
2
(0) = 1. Write
h
2
(0) = a (? 0). Then h
\Gamma 1
2
(the inverse of h
2
) is an increasing analytic function from
(a; +1) to IR
+
. Notice that h
0
2
is also an increasing analytic function from (0; +1) to
(1; +1), so x ! [h
0
2
(h
\Gamma 1
2
(1=x))]
\Gamma 1
defines an increasing analytic function from (0; a
\Gamma 1
)
to (0; 1). Let
g(x) =
Z
x
0
[h
0
2
(h
\Gamma 1
2
(t
\Gamma 1
))]
\Gamma 1
dt :
Then g is also an analytic function on (0; a
\Gamma 1
), and 0 ! g(x) ^ x. (To see that g is
analytic on (0; a
\Gamma 1
), we take an arbitrary point x
0
2 (0; a
\Gamma 1
). Then g(x) is written as a
sum of a constant and
R
x
x
0
[h
0
2
(h
\Gamma 1
2
(1=t))]
\Gamma 1
dt. Since the integrand function is analytic,
it can be expanded as a power series in a neighborhood of x
0
, and therefore the integral
is a power series in the same neighborhood.) Now, we claim that the function ' given by
'(x) = g(
1
2
h
2
(x)
\Gamma 1
)
satisfies all the requirements described in this lemma.
In fact, x !
1
2
h
2
(x)
\Gamma 1
is an analytic function from (\Gamma 1; +1) to (0;
1
2a
). Thus the
composition x ! g(
1
2
h
2
(x)
\Gamma 1
) defines an analytic function from IR to (0; 1). To verify
the inequalities, since g(x) ^ x, we have '(x) ^
1
2
h
2
(x)
\Gamma 1
^ h(jxj)
\Gamma 1
. Also,
'
0
(x) = \Gamma g
0
(
1
2
h
2
(x)
\Gamma 1
) \Delta
h
0
2
(x)
2h
2
(x)
= \Gamma
1
h
0
2
(h
\Gamma 1
2
(2h
2
(x)))
\Delta
h
0
2
(x)
2h
2
2
(x)
:
Since both h
\Gamma 1
2
and h
0
2
are increasing, we have h
0
2
(h
\Gamma 1
2
(2h
2
(x))) * h
0
2
(h
\Gamma 1
2
(h
2
(x))) =
h
0
2
(x). So j'
0
(x)j ^ h
2
(x)
\Gamma 2
^ h(jxj)
\Gamma 2
. The proof is complete. 2 Pass1: Page 17
17 2.3 The Proof of Theorem 2.1 The implications 6 ) 5 ) 4 ) 2 and 5 ) 3 ) 2 are clear. To show that 2 ) 1 we first observe that, if \Sigma is ACBC, then, to begin with, \Sigma is AC, so \Sigma
c
\Phi \Sigma
u
must be
controllable. To show that E
u
must be f0g we use the fact that, since \Gamma A
u
is Hurwitz,
there is a * \Theta * matrix P ? 0 such that P A
u
+ A
y
u
P
def
= Q ? 0, where * = dim E
u
. If
C ? 0 and we restrict ourselves to controls bounded in norm by C, then it is clear that the derivative of the function V (x
u
) = hP x
u
; x
u
i along trajectories of \Sigma is given by
. V (x
u
) = hQx
u
; x
u
i + 2hP x
u
; B
u
ui * hQx
u
; x
u
i \Gamma 2CjjP jj jjB
u
jj jjx
u
jj :
So there exists R ? 0 such that
. V (x
u
) ? 0 whenever jjx
u
jj * R. Therefore, if we choose
the initial value of x
u
sufficiently far from 0, then the trajectory that starts at such an
x
u
cannot possibly go to zero.
The implication 7 ) 1 is also easy to prove. Indeed, if (7) holds then in particular \Sigma is stabilizable. Moreover, we may pick, for each " ? 0, a matrix K
"
such that
jjK
"
jj ! " and A + BK
"
is stable. Letting " ! 0, and using the continuous dependence
of the eigenvalues of a matrix, we conclude that the spectrum of A is contained in fs : Re s ^ 0g.
So, to prove the theorem it suffices to show that 1 ) 6 and 1 ) 7. We will actually show the following fact:
(A) Suppose (1) holds. Then for every " ? 0, there exists a stabilizing feedback
k 2 F
w
n;m;1
such that
(i) jjkjj
C
1
^ ",
(ii) there exists an analytic Lyapunov function V of \Sigma
k
satisfying
. V (x) ^
\Gamma jjk(x)jj
2
for all x, where
. V denotes the derivative of V along the trajectories
of \Sigma
k
,
(iii) the linearization Kx of k(x) at the origin also stabilizes \Sigma and the quadratic
approximation V
q
of V at the origin is a Lyapunov function of \Sigma
K
. Pass1: Page 18
18 To be precise, by "analytic Lyapunov function" for a differential equation E : .x = f (x) on IR
n
we mean here an analytic function V : IR
n
! IR such that V (x) ? 0 if
x 6= 0, V (0) = 0, lim
jjxjj!1
V (x) = 1, and h5V (x); f (x)i ^ 0 for all x. The LaSalle
Invariance Principle says that, if V is a Lyapunov function for E, then 0 is GAS equilibrium of E if and only if the following Invariance Principle Condition (IPC) holds: whenever t ! x(t); \Gamma 1 ! t ! +1 is a complete orbit of E on which V is a constant, then it follows that x(t) j 0.
It is clear that the above fact implies (6). In addition, from (iii) of Fact A, we know that the linearization Kx of k(x) at 0 also stabilizes \Sigma . Since the norm of the matrix K is bounded by the C
1
-norm of k, we see that the above fact also implies (7). So, we
only need to prove Fact A.
Before we prove the above fact, we first observe that we can always assume that \Sigma
s
= 0. Indeed, a general system satisfying the condition \Sigma
u
= 0 can be written in the
form
.x
s
= A
s
x
s
+ B
s
u ;
.x
c
= A
c
x
c
+ B
c
u :
Assume that our theorem is known to be true when \Sigma
s
= 0. Then we can find, for every
" ? 0, an analytic stabilizing feedback u = k(x
c
) such that (i) jjkjj
C
1 ^ ", (ii) there
is an analytic Lyapunov function V for (\Sigma
c
)
k
which satisfies
. V (x
c
) ^ \Gamma jjk(x
c
)jj
2
, and
(iii) the linearization Kx
c
of k(x
c
) also stabilizes \Sigma
c
, and the quadratic approximation
V
q
(x
c
) of V (x
c
) is a Lyapunov function for (\Sigma
c
)
K
. It is then easy to see that u = k(x
c
)
and u = Kx
c
also stabilize \Sigma , and that there is a Lyapunov function W with the desired
properties.
In fact, let P and Q be positive definite matrices such that P A
s
+ A
y
s
P = \Gamma 2Q.
Define
W (x
s
; x
c
) = hP x
s
; x
s
i + ffV (x
c
) ;
where ff is chosen so that ff ? 1 + jjB
y
P Q
\Gamma 1
P Bjj. It is clear that W (0) = 0, W (x) ? 0
if x 6= 0, and W (x) ! 1 as jjxjj ! 1. The derivative
. W of W along trajectories of
\Sigma
k
is
. W = \Gamma h2Qx
s
; x
s
i + 2 hx
s
; P Bk(x
c
)i + ff
. V (x
c
) : Pass1: Page 19
19 Note that
2 hx
s
; P Bk(x
c
)i = 2
D
Q
1 2
x
s
; Q
\Gamma
1
2
P Bk(x
c
)
E
^ jjQ
1 2
x
s
jj
2
+ jjQ
\Gamma
1
2
P Bk(x
c
)jj
2
= hQx
s
; x
s
i +
D
k(x
c
); B
y
P Q
\Gamma 1
P Bk(x
c
)
E
;
and
. V (x
c
) ^ \Gamma jjk(x
c
)jj
2
. Since ff ? 1 + jjB
y
P Q
\Gamma 1
P Bjj, we have
. W (x) ^ \Gamma hQx
s
; x
s
i \Gamma jjk(x
c
)jj
2
:
In particular,
. W ^ 0, so W is a Lyapunov function for \Sigma
k
which clearly satisfies the
bound
. W ^ \Gamma jjk(x
c
)jj
2
. To prove the asymptotic stability, we establish the IPC. Let
t ! x(t) = (x
s
(t); x
c
(t)) be a complete orbit of \Sigma
k
on which W is constant. Then
. W (x(t)) j 0, so hQx
s
(t); x
s
(t)i = 0. Since Q ? 0, it follows that x
s
(t) j 0 and we have
. W (x(t)) j ff
. V (x
c
(t)). So x
c
(\Delta ) is an orbit of (\Sigma
c
)
k
on which V is constant. Therefore,
x
c
(t) j 0 by the asymptotic stability of (\Sigma
c
)
k
. To see that the linearization Kx
c
of
k(x
c
) stabilizes \Sigma , we notice that the closed-loop system \Sigma
K
with the feedback u = Kx
c
is a linear system whose spectrum is the union of that of A
s
and A
c
+B
c
K. Since (\Sigma
c
)
K
is assumed to be asymptotically stable, it follows that the spectrum of A
c
+ B
c
K is
contained in the set f* : Re * ! 0g. The spectrum of A
s
is also contained in the set
f* : Re * ! 0g, so all eigenvalues of the linear system \Sigma
K
have negative real parts. It is
also obvious that the quadratic approximation of W (x) is a Lyapunov function of \Sigma
K
.
Now we give the general proof for our fact with the assumption \Sigma
s
= 0. First we
consider the simple case when A is diagonable over the complex numbers. In that case there is a real non-singular matrix M such that M AM
\Gamma 1
is skew-symmetric. This
means that we can make a linear change of coordinates x ! M x and assume that A itself is skew-symmetric.
If A is skew-symmetric, then we can stabilize \Sigma as follows. Let
V (x) =
1
2
jjxjj
2
:
We are trying to find a feedback control u = k(x) such that V is a Lyapunov function of the closed-loop system \Sigma
k
. For any feedback k, the derivative
. V of V along the
trajectories of \Sigma
k
is given by
. V (x) = hAx; xi +
m X
i=1
k
i
(x) hb
i
; xi ; Pass1: Page 20
20 where the b
i
are the columns of the matrix B, and the k
i
are the components of k. Since
A is skew-symmetric, the inner product hAx; xi vanishes. So we will have the inequality
. V ^ 0 provided that we choose the function k
i
in such a way that k
i
(x) hb
i
; xi ^ 0 for
all i's. An obvious choice is to take
k
i
(x) = \Gamma '(hb
i
; xi) ;
where ' : IR ! IR is any analytic function such that '(0) = 0 and s'(s) ? 0 whenever s 6= 0. We claim that the resulting closed-loop system has 0 as a GAS equilibrium.
Indeed, our choice of the k
i
suffices to imply the inequality
. V ^ 0. To prove the
asymptotic stability we need to verify the IPC. Let t ! x(t) be an orbit of \Sigma
k
on which
V is constant. Since
. V (x(t)) = \Gamma
m P
i=1
hb
i
; x(t)i '(hb
i
; x(t)i), the equality
. V (x(t)) =
0 implies hb
i
; x(t)i = 0 for all t, and i = 1; : : : ; m. The equation .x(t) = Ax(t) +
m P
i=1
k
i
(x(t))b
i
then implies .x(t) = Ax(t). Hence, repeated differentiation of hb
i
; x(t)i = 0
yields
D
b
i
; A
k
x(t)
E
= 0 for all integers k ? 0. The skew-symmetry of A then implies D
A
k
b
i
; x(t)
E
= 0 for all k. But the vectors A
k
b
i
; i = 1; \Delta \Delta \Delta ; m; k = 0; 1; \Delta \Delta \Delta ; n \Gamma 1,
span the whole space, because (A; B) is controllable. Therefore x(t) j 0, and the IPC follows.
In the above paragraph we choose ' arbitrarily with the properties that '(0) = 0 and s'(s) ? 0 if s 6= 0. Particularly, if we choose '(s) = cs, where c is a positive constant, then the linear feedback k
i
= c hb
i
; xi ; i = 1; \Delta \Delta \Delta ; m, will stabilize \Sigma . More
generally, if we choose ' such that
d'
ds
(0) = c, then the linearization of k
i
is c hb
i
; xi and
therefore stabilizes \Sigma . The inequality
. V (x) ^ \Gamma jjk(x)jj
2
follows as long as ' satisfies
j'(s)j ^ jsj for all s, which follows automatically if j
d'(s)
ds
j ^ 1.
To conclude the proof for the diagonable case, we have to show that for any " ? 0 we can choose an analytic function ' such that '(0) = 0,
d'
ds
(0) ? 0, j
d'
ds
j ^ 1, and
jjk
i
jj
C
1 ! ". In fact, for any bounded analytic function '
1
with the properties '
1
(0) = 0,
d'
1
ds
(0) ? 0 and j
d'
1
ds
j ^ 1, namely '
1
(s) = tanh(s), it follows that jj'
1
(hb
i
; \Delta i)jj
C
1 is
bounded. (Here, jj'
1
(hb
i
; \Delta i)jj
C
0 is bounded because '
1
is bounded, and jj'
1
(hb
i
; \Delta i)jj
C
1
is bounded because jj'
1
(hb
i
; \Delta i)jj
C
0 is bounded and the derivatives of '
1
(hb
i
; xi) are
bounded by j
d'
1
ds
j multiplied by a constant.) Therefore, by taking ' to be a sufficiently Pass1: Page 21
21 small multiple of '
1
, we have jjk
i
jj
C
1
! ". The proof for the diagonable case is now
complete.
We now turn to the general case. The proof will be by induction on n. To be precise, assume our desired conclusion is not always true. Pick the smallest possible n for which the conclusion fails for some m. So there is a system \Sigma with n-dimensional state space for which \Sigma
u
= 0 and \Sigma
c
is controllable but our desired conclusion does not hold. From
our assumption at the beginning of the proof of our conclusion, we have \Sigma = \Sigma
c
. Also,
the above discussion for diagonable systems shows that the system matrix of \Sigma cannot be diagonable. This implies that there exists an eigenvalue * of A for which there is a nonzero eigenvector v and a vector ~v such that A~v = *~v + v. We pick *, v, ~v, and keep them fixed in the following discussion.
The eigenvalue * is either 0 or of the form i!, with ! 2 IR, ! 6= 0. Let us consider first the case when * = 0. In this case, the vectors v and ~v can be chosen to be real. Now let S,
~ S
0
denote, respectively, the real linear spans of v and ~v, and let oe = 1,
J = 0. Let V,
~ V
0
be the sets fvg, f~vg, respectively. Then
(i) the spaces S,
~ S
0
are oe-dimensional,
(ii) S "
~ S
0
= f0g, AS ` S, A
~ S
0
`
~ S
0
+ S,
(iii) V,
~ V
0
are bases of S,
~ S
0
,
(iv) J is a oe \Theta oe skew-symmetric matrix, (v) if we write Ay =
~ A
0
y +
~ G
0
y for y 2
~ S
0
, with
~ A
0
y 2
~ S
0
,
~ G
0
y 2 S, and let A
S
denote the restriction of A to S, then J is the matrix of A
S
with respect to the
basis V and also of
~ A
0
with respect to the basis
~ V
0
, and
~ G
0
maps each vector of
~ V
0
to the corresponding vector of V.
Now, if * = i!, with ! 6= 0, then we can assume ! ? 0. Write v = v
1
+ iv
2
,
~v = ~v
1
+ i~v
2
. Then Av
1
= \Gamma !v
2
, Av
2
= !v
1
, A~v
1
= \Gamma !~v
2
+ v
1
, and A~v
2
= !~v
1
+ v
2
.
So, if we let oe = 2, V = (v
1
; v
2
),
~ V
0
= (~v
1
; ~v
2
), S = span V,
~ S
0
= span
~ V
0
, we see that Pass1: Page 22
22 (i); : : : ; (v) also hold in this case, with
J =
0 B @
0 !
\Gamma ! 0
1 C A
:
(Recall that if V = (v
1
; : : : ; v
n
), W = (w
1
; : : : ; w
N
) are bases of linear spaces V , W , and
A : V ! W is a linear map, then the matrix of A with respect to V, W is the matrix (a
ij
) such that Av
j
=
P
i
a
ij
w
i
. If W = V , then the matrix of A with respect to V is
the matrix of A with respect to V, V.)
So, whether * is real or complex, we have singled out oe, S,
~ S
0
, V,
~ V
0
, J, with
oe = 1 or 2, such that (i); : : : ; (v) hold.
The space
~ S
0
can be extended to a subspace S
0
0
of IR
n
such that
~ S
0
` S
0
0
and
IR
n
= S
0
0
\Phi S. For y 2 S
0
0
, we can write Ay = A
0
0
y + G
0
y, with A
0
0
y 2 S
0
0
, G
0
y 2 S.
Clearly, the restrictions of A
0
0
, G
0
to
~ S
0
are
~ A
0
,
~ G
0
.
Consider an arbitrary complementary subspace S
0
of S, i.e., an arbitrary subspace
S
0
such that IR
n
= S \Phi S
0
. Let ss : IR
n
! S
0
0
be the projection along S, i.e., ss(y) = z
if y = z + x, z 2 S
0
0
, x 2 S. Let _ss be the restriction of ss to S
0
. Then _ss : S
0
! S
0
0
is a
linear one-to-one and onto map. For y 2 S
0
we can write Ay = A
0
y + Gy with A
0
y 2 S
0
,
Gy 2 S.
An easy calculation shows that _ss ffi A
0
= A
0
0
ffi _ss, i.e., the map _ss : S
0
! S
0
0
intertwines
the operators A
0
: S
0
! S
0
and A
0
0
: S
0
0
! S
0
0
. (Let y 2 S
0
. Write y = z + x,
z 2 S
0
0
, x 2 S. Then Ay = Az + Ax = A
0
0
z + G
0
z + Ax. But Ay = A
0
y + Gy. So
A
0
y = A
0
0
z + G
0
z + Ax \Gamma Gy. Since G
0
z + Ax \Gamma Gy 2 S, we have ss(G
0
z + Ax \Gamma Gy) = 0.
Since A
0
0
z 2 S
0
0
, it follows that ss(A
0
0
z) = A
0
0
z. But then _ss(A
0
y) = ss(A
0
y) = A
0
0
z =
A
0
0
ss(y) = A
0
0
_ss(y).)
Let
~ S = _ss
\Gamma 1
(
~ S
0
), and let
~ V be the basis obtained by mapping
~ V
0
via _ss
\Gamma 1
. Let
~ A be
the restriction of A
0
to
~ S. Then
~ A
~ S `
~ S, and _ss gives rise to an isomorphism between
~ S and
~ S
0
that intertwines the operators
~ A and
~ A
0
and maps the basis
~ V to
~ V
0
. From
this it follows in particular that the matrix of
~ A with respect to the basis
~ V is J.
Now let
~ G denote the restriction of G to
~ S, so
~ G :
~ S ! S. We are interested in
computing the matrix
~ \Gamma of
~ G with respect to the bases
~ V, V. For this purpose it is Pass1: Page 23
23 convenient to introduce the linear map L : S
0
0
! S given by Lz = _ss
\Gamma 1
z \Gamma z. (If z 2 S
0
0
then _ssz = z, and so _ssLz = z \Gamma _ssz = 0, so Lz 2 S.)
If z 2 S
0
0
, let y = _ss
\Gamma 1
z = z + Lz. Then Ay = A
0
y + Gy. But
A
0
y = A
0
_ss
\Gamma 1
z = _ss
\Gamma 1
A
0
0
z = A
0
0
z + LA
0
0
z :
So
Ay = _ss
\Gamma 1
A
0
0
z + Gy = A
0
0
z + LA
0
0
z + Gy : (2.3.1)
On the other hand,
Ay = A(z + Lz) = Az + ALz = A
0
0
z + G
0
z + ALz : (2.3.2)
Equating these two expressions (2.3.1) and (2.3.2) for Ay, and canceling the common term A
0
0
z, we get LA
0
0
z + Gy = G
0
z + ALz. Therefore G_ss
\Gamma 1
z = G
0
z + ALz \Gamma LA
0
0
z.
Now, if we let
~ L denote the restriction of L to
~ S
0
, we get in particular
~ G_ss
\Gamma 1
z =
~ G
0
z + A
S
~ Lz \Gamma
~ L
~ A
0
z : (2.3.3)
Next recall that G
0
maps each vector of the basis
~ V
0
to the corresponding vector
of V, so that the matrix of G
0
with respect to the bases
~ V
0
, V is just the identity. Let
~ \Lambda denote the matrix with respect to
~ V
0
, V. Recall that the matrix of A
S
with respect
to V is J, which is also the matrix of
~ A
0
with respect to
~ V
0
. So, using 1 to denote the
oe \Theta oe identity matrix, from (2.3.3) we find that the matrix of G_ss
\Gamma 1
with respect to
~ V
0
,
V is 1 + J
~ \Lambda \Gamma
~ \Lambda J. Since _ss
\Gamma 1
maps
~ V
0
to
~ V, we conclude that
~ \Gamma = 1 + J
~ \Lambda \Gamma
~ \Lambda J :
Summarizing, we have shown: (\Lambda ) if S
0
is an arbitrary linear subspace of IR
n
such that IR
n
= S \Phi S
0
, and we write
Ay = A
0
y + Gy for y 2 S
0
, with A
0
y 2 S
0
, Gy 2 S, then there exists a oe-
dimensional subspace
~ S of S
0
and a basis
~ V of
~ S, such that if
~ A
0
,
~ G denote the
restrictions of A
0
, G to
~ S, then the matrix of
~ A
0
with respect to
~ V is J, and the
matrix of
~ G with respect to
~ V, V is 1 + J
~ \Lambda \Gamma
~ \Lambda J, for some oe \Theta oe matrix
~ \Lambda . Pass1: Page 24
24 So far the choice of S
0
was arbitrary. We now choose S
0
in a special way, taking into
account the map B. We let * be the dimension of BIR
m
" S (so that 0 ^ * ^ oe). Let
m
0
= m \Gamma *. After a linear change of coordinates in IR
m
, we may assume that the last
* columns of B are in S, and the first m
0
columns b
1
; : : : ; b
m
0
are not. We then pick S
0
such that IR
n
= S \Phi S
0
and b
i
2 S
0
for i = 1; : : : ; m
0
. With this choice of S
0
, if we let B
0
,
B
00
denote the matrices whose columns are, respectively, b
1
; : : : ; b
m
0
and b
m
0
+1
; : : : ; b
m
,
and write x = y + z for a typical x 2 IR
n
, with y 2 S
0
, z 2 S, then our system \Sigma has
the form
.y = A
0
y + B
0
u
0
;
.z = Jz + Gy + B
00
u
00
;
(2.3.4)
where u
0
= (u
1
; : : : ; u
m
0
), u
00
= (u
m
0
+1
; : : : ; u
m
). (If m
0
= m, then u
00
disappears.) We
can then make a linear change of coordinates in IR
n
so that S
0
, S become IR
n\Gamma oe
\Theta f0g
and f0g \Theta IR
oe
, respectively. Then we can simply regard y, z as column vectors of length
n\Gamma oe, oe, and think of A
0
, B
0
, B
00
, G as, respectively, an (n\Gamma oe)\Theta (n\Gamma oe), an (n\Gamma oe)\Theta m
0
,
a oe \Theta *, and an (n \Gamma oe) \Theta oe matrix.
The controllability hypothesis implies that (A
0
; B
0
) is controllable. Using this, plus
our inductive assumption, we will stabilize (2.3.4) with u
00
j 0. In view of this, from
now on we will ignore the term B
00
u
00
, and simply relabel u
0
as u and m
0
= m. So our
system now has the form
.y = A
0
y + B
0
u; y 2 IR
n\Gamma oe
; u 2 IR
m
;
.z = Jz + Gy; z 2 IR
oe
:
(2.3.5)
It is clear that the spectrum of A
0
is a subset of that of A, and that (A
0
; B
0
) is
controllable, so \Sigma
0
: .y = A
0
y + B
0
u satisfies Condition (1) of Theorem 2.1. We now
use the inductive hypothesis for \Sigma
0
. We conclude that, for every " ? 0, we can get
an analytic feedback k
0
: IR
n\Gamma oe
! IR
m
such that (i) jjk
0
jj
C
1
^
"
2
, (ii) there is an
analytic Lyapunov function V for \Sigma
0
k
0
which satisfies
. V (x) ^ \Gamma jjk
0
(x)jj
2
for all x, (iii)
the linearization K
0
x of k
0
(x) stabilizes \Sigma
0
and the quadratic approximation V
q
of V
at the origin is a Lyapunov function for \Sigma
0
K
0
. Here, and in what follows, we will use
. h, if h is any scalar- or vector-valued function of y, to denote the derivative of h along Pass1: Page 25
25 trajectories of \Sigma
0
k
0
, i.e.,
. h(y) = Dh(y)(A
0
y + B
0
k
0
(y)) ;
where Dh is the Jacobian matrix of h. We will define a new feedback k(y; z) by letting
k(y; z)
def
= k
0
(y) + _(y; z) ;
where _ is a function to be chosen below. We will then use
. h
new
to denote the derivative
of a function h(y; z) along trajectories of the closed-loop system \Sigma
k
corresponding to
the new feedback, so that
. h
new
(x) = Dh(x)(A
0
x + B
0
k(x)). Recall that x = (y; z), so
equivalently, if we use D
y
h, D
z
h to denote the Jacobian matrices of h with respect to
y, z, we have
. h
new
(y; z) = D
y
h(y; z)(A
0
y + B
0
k
0
(y) + B
0
_(y; z)) + D
z
h(y; z)(Jz + Gy) :
In the special case when h only depends on y, we have
. h
new
(y; z) =
. h(y) + Dh(y)B
0
_(y; z) :
We now explain how to choose _. First, we have to study what happens if we only use the "old" feedback k
0
. Given any initial condition _x = (_y; _z), the asymptotic
behavior as t ! +1 of the corresponding trajectory t ! x(t) = (y(t); z(t)) can be described explicitly. Indeed, y(\Delta ) is a trajectory of \Sigma
0
k
0
, so its limit is zero. Moreover,
since \Sigma
0
K
0
is asymptotically stable, the vector-valued function y decays exponentially as
t ! 1. Then z(t) satisfies .z(t) = Jz(t)+Gy(t), so that z(t) = e
tJ
(_z +
R
t
0
e
\Gamma sJ
Gy(s) ds).
Since y(s) decays exponentially as s ! 1, the infinite integral
,(_y)
def
= \Gamma
Z
1
0
e
\Gamma sJ
Gy(s) ds
exists. Then e
\Gamma tJ
z(t) has the limit _z \Gamma ,(_y) as t ! 1. In particular, since J is skew-
symmetric and therefore e
tJ
is norm-preserving, we conclude that lim z(t) = 0 if and
only if _z = ,(_y). In other words, the set M = f(y; z) : z = ,(y)g is exactly the set of points (y; z) that are asymptotically driven to (0; 0) by the "old" feedback k
0
. In
particular, it is clear that M is invariant under the flow of \Sigma
k
0
.
We remark that M is an analytic manifold, since it is the graph of the analytic map ,. To see that , is analytic, observe that ,(y) = \Gamma
R
1
0
e
\Gamma sJ
G\Phi (s; y) ds, where Pass1: Page 26
26 \Phi is the flow of \Sigma
0
k
0
, i.e., s ! \Phi (s; y) is the solution OE(s) of
. OE = A
0
OE + B
0
k
0
(OE) such
that OE(0) = y. Note that \Sigma
0
k
0
and the linearization of \Sigma
0
k
0
are globally asymptotically
stable. Also, since J is skew-symmetric, it follows that e
\Gamma sJ
is norm preserving and
therefore jje
\Gamma sJ
Gxjj ^ jjGjj jjxjj. So applying Lemma 2.2.3 to f (y) = A
0
y + B
0
k
0
(y),
g(y; t) = e
\Gamma tJ
Gy, we conclude that , is an analytic function. Let
W (y; z)
def
=
1
2
jjz \Gamma ,(y)jj
2
+ V (y) : (2.3.6)
From our inductive hypothesis that V is analytic and the fact that , is analytic, it follows that W is analytic. We will show below that W is a Lyapunov function of the closed-loop system for some choice of _.
Note that W (0) = 0 and W (x) ? 0 if x 6= 0. Also, lim
(y;z)!1
W (y; z) = +1. Let
. W
new
be the derivative of W along the trajectories corresponding to the "new" feedback
k(y; z) = k
0
(y) + _(y; z), where the function _ still remains to be chosen. Then in order
to insure that W is a Lyapunov function, we will need the inequality
. W
new
^ 0. It is
clear that
. W
new
=
D
.z
new
\Gamma
. ,
new
(y); z \Gamma ,(y)
E
+
. V
new
(y) :
Now
.z
new
= Jz + Gy ;
. ,
new
(y) =
. ,(y) + D,(y)B
0
_(y; z) ;
and
. V
new
(y) =
. V (y) +
\Omega
5V (y); B
0
_(y; z)
ff
:
Since M is the graph of , and is invariant under the flow of \Sigma
k
0, the function
. , is equal
to J, + Gy. So
.z
new
\Gamma
. ,
new
(y) = J(z \Gamma ,(y)) \Gamma D,(y)B
0
_(y; z) :
Because of the skew-symmetry of J it follows that hJ(z \Gamma ,(y)); z \Gamma ,(y)i = 0 and
D
.z
new
\Gamma
. ,
new
(y); z \Gamma ,(y)
E
= \Gamma
\Omega
D,(y)B
0
_(y; z); z \Gamma ,(y)
ff
:
Therefore,
. W
new
= \Gamma
\Omega
D,(y)B
0
_(y; z); z \Gamma ,(y)
ff
+
. V (y) +
\Omega
5V (y); B
0
_(y; z)
ff
: (2.3.7) Pass1: Page 27
27 If we let b
1
; \Delta \Delta \Delta ; b
m
denote the columns of B
0
, and write
`
i
(y; z)
def
= \Gamma hD,(y)b
i
; z \Gamma ,(y)i + h5V (y); b
i
i ; (2.3.8)
we see that
. W
new
=
. V +
P
m
i=1
`
i
_
i
. So, if we choose
_
i
(y; z) = \Gamma '(y; z)`
i
(y; z) ; (2.3.9)
where '(y; z) is a strict positive function, we will have
. W
new
=
. V \Gamma '
P
m
i=1
`
2
^ 0.
Moreover, if '(y; z) ^ 1 for all (y; z), using
. V (y) ^ \Gamma jjk
0
(y)jj
2
, we will have
. W
new
^ \Gamma jjk
0
(y)jj
2
\Gamma jj_(y; z)jj
2
: (2.3.10)
So we get the bound
. W
new
^ \Gamma jjk(y; z)jj
2
. If we manage to select ' such that jj'`jj
C
1
!
" 2
, then the new feedback k will satisfy jjkjj
C
1
! ". Finally, if ' is chosen to be analytic,
since ,(y) and 5V (y) are analytic, the new feedback will be analytic. The missing conditions are the stabilities of the new system \Sigma
k
and its linearization, and that the
quadratic approximation of W is a Lyapunov function for the linearized system.
So we have to show, to conclude our proof, that (i) ' can be chosen so that ' is analytic, 0 ! '(x) ^ 1 for all x, and jj'`jj
C
1 ^
"
2
, (ii) the resulting feedback k is such
that the Lyapunov function W satisfies the IPC, and (iii) the quadratic approximation of W is a Lyapunov function for \Sigma
K
and satisfies the IPC, where K is the linearization
matrix of k(x) at 0.
First, we define a function h : IR
+
! IR
+
by the formula
h(t)
def
= supfM
i;j
(x) : jxj ^
p
t; j = 1; 2; \Delta \Delta \Delta ; m; i = 1; 2; \Delta \Delta \Delta ; ng ; (2.3.11)
where
M
i;j
(x) = maxf1 + 2jxj; j`
j
(x)j; jD
i
`
j
(x)jg (2.3.12)
and D
i
is an operator that takes derivative with respect to the ith component of x.
It is clear that h(t) is a positive increasing function. From Lemma 2.2.4 we can find an analytic function : IR ! IR
+
such that (t) ^ h(jtj)
\Gamma 1
and j
d
dt
j ^ h(jtj)
\Gamma 2
.
Let '
1
(x) = (x
y
x). Then '
1
is analytic. From the constructions of and h, it
follows that 0 ! '
1
(x) ^ h(0)
\Gamma 1
^ 1 for all x. Also, jj'
1
`jj
C
1 is finite. Indeed, since Pass1: Page 28
28 '
1
(x)`
j
(x) = (x
y
x)`
j
(x) ^
`
j
(x)
h(x
y
x)
, from (2.3.11) and (2.3.12), we see that jj'
1
`jj
C
0 is
finite. Write
D
i
('
1
`
j
) =
d
dt
(x
y
x) \Delta D
i
(x
y
x) `
j
(x) + (x
y
x)D
i
`
j
(x) : (2.3.13)
Then from (2.3.11) and (2.3.12) we see that the second term on the right of (2.3.13) is bounded by 1. The first term on the right of (2.3.13) is bounded by j
D
i
(x
y
x)`
j
(x)
h(x
y
x)
2
j
because j
d
dt
(t)j ^ h(t)
\Gamma 2
. Since j
D
i
(x
y
x)
h(x
y
x)
j ^
2jxj
h(x
y
x)
j ^ 1 and j
`
j
(x)
h(x
y
x)
j ^ 1, it follows that
the first term on the right of (2.3.13) is also bounded by 1. Therefore, jj'
1
`jj
C
1 is finite.
Let ' be a sufficiently small multiple of '
1
. We then get jj'`jj
C
1 ^
"
2
and Part (i)
follows.
We now turn to the proof of Part (ii), namely, the verification of the IPC. Let us pick a complete orbit t ! x(t) = (y(t); z(t)) of \Sigma
k
, on which W is constant. From
(2.3.10) we have _(y(t); z(t)) j 0, which implies that y(\Delta ) is actually a solution of .y = A
0
y(t) + B
0
k
0
(y), i.e., a trajectory of \Sigma
0
k
0
. Then since
. W
new
=
. V \Gamma '
P
m
i=1
`
2
i
^
. V ^ 0,
we see that
. V (y(t)) j 0. So y(\Delta ) is an orbit of \Sigma
0
k
0
on which V is constant. Since 0 is
a GAS equilibrium of \Sigma
0
k
0
, it follows that y(t) j 0. But then t ! z(t) is a solution of
.z = Jz. Moreover, from (2.3.8) we have
`
i
(y(t); z(t)) = \Gamma hD,(y(t))b
i
; z(t) \Gamma ,(y(t))i + h5V (y(t)); b
i
i ;
and from (2.3.9) we have `
i
(y(t); z(t)) j 0. Since 5V (y(t)) j 0; y(t) j 0, it follows
that hD,(0)b
i
; z(t)i j 0 for all i = 1; 2; \Delta \Delta \Delta ; m. Let the linearization of ,(y) at y = 0
be Hy. Then D,(0) = H, so hHb
i
; z(t)i = 0 for all t. Since .z(t) = Jz(t), we also have
hJHb
i
; z(t)i = \Gamma hHb
i
; Jz(t)i = 0. Assume HB
0
6= 0 for the present. Then there exists
an i such that Hb
i
6= 0. In the case oe = 1, it follows immediately that z(t) j 0 since
hHb
i
; z(t)i = 0. In the case oe = 2, since Hb
i
and JHb
i
form a basis of IR
2
, it follows
that the equalities hHb
i
; z(t)i = 0 and hJHb
i
; z(t)i = 0 imply z(t) j 0. So all that
remains of the IPC is to show that HB
0
6= 0.
Consider an arbitrary orbit t ! y(t) of the system \Sigma
0
k
0
. Using
. , = J, + Gy,
. , = (D,) .y, and .y = A
0
y + B
0
k
0
(y), we conclude that
(JH + G)y = H(A
0
y + B
0
K
0
y) : (2.3.14) Pass1: Page 29
29 If HB
0
= 0, then (2.3.14) would imply (HA
0
\Gamma JH)y = Gy. So HA
0
\Gamma JH = G. Let
~ H
be the restriction of H to the oe-dimensional subspace
~ S. Recall that
~ V and V are the
bases of
~ S, S described above and
~ \Gamma is the matrix of G with respect to V. Let H be
the matrix of
~ H with respect to
~ V. Since A
0
~
S `
~ S, and the matrix of
~ A
0
with respect
to
~ V is J, we conclude that
~ \Gamma = HJ \Gamma JH. But we know that
~ \Gamma = 1 + J
~ \Lambda \Gamma
~ \Lambda J. These
two equalities together yield a contradiction. Indeed, it is well known that the trace of any commutator is 0. So the first equality implies that trace
~ \Gamma = 0, whereas the second
one implies trace
~ \Gamma = trace 1 = oe 6= 0. This establishes the inequality HB
0
6= 0. So
Part (ii) is proved.
Finally, to prove Part (iii), we write W (x) = W
q
(x) + W
h
(x), where W
q
is the
quadratic approximation of W at x = 0 and W
h
is its error. Since V
q
is positive
definite, from (2.3.6) it follows that W
q
is also positive definite. In addition, let Kx
be the linearization of k(x) at x = 0. If we use
. W
q;L
to denote the derivative of W
q
along the trajectories of \Sigma
K
and use
. W
q;H
to denote the derivative of W
q
along the
trajectories of the closed-loop system corresponding to the feedback u = k(x) \Gamma Kx, then
. W
new
(x) =
. W
q;L
(x) +
. W
q;H
(x) +
. W
h new
(x) ; (2.3.15)
which is less than \Gamma jjk(x)jj
2
from earlier results. Notice that
. W
q;L
consists of some
quadratic monomials of x, and if x is near zero, then
. W
q;H
and
. W
h new
consist of some
monomials of x with degrees higher than two. If we write
jjk(x)jj
2
= jjKxjj
2
+ high order terms ;
then from (2.3.15) we conclude that
. W
q;L
(x) ^ \Gamma jjKxjj
2
. Therefore W
q
is a Lyapunov
function for \Sigma
K
.
To verify that W
q
satisfies the IPC for \Sigma
K
, note that
. W
new
=
. V +
m X
i=1
`
i
_
i
(2.3.16)
(see (2.3.7) and (2.3.8)), where `
i
(y; z) = \Gamma hD,(y)b
i
; z \Gamma ,(y)i + h5V (y); b
i
i and _
i
=
\Gamma '(x)`
i
(x). Since
. W
q;L
consists of all quadratic terms in (2.3.16), it follows that
. W
q;L
=
. V
q;L
old
\Gamma
d'
dt
(0)
m X
i=1
(\Gamma hHb
i
; z \Gamma Hyi + h5V
q
(y); b
i
i)
2
; (2.3.17) Pass1: Page 30
30 where
. V
q;L
old
denotes the derivative of V
q
along the trajectories of \Sigma
0
K
0
. From the
inductive hypothesis, we know that
. V
q;L
old
^ 0. If t ! (y(t); z(t)) is a complete orbit
of \Sigma
K
along which
. W
q;L
j 0, then from (2.3.17) we have
. V
q;L
old
j 0 (2.3.18)
and
hHb
i
; z(t) \Gamma Hy(t)i + h5V
q
(y); b
i
i j 0 (2.3.19)
for all i = 1; 2; \Delta \Delta \Delta ; m. The inductive hypothesis and (2.3.18) imply that y(t) j 0. So from (2.3.19) we have hHb
i
; z(t)i j 0. From the proof of Part (ii) we conclude that
z(t) j 0. The IPC of W
q
then follows. The proof of Theorem 2.1 is complete. 2 Pass1: Page 31
31 Chapter 3 The Naive Design
In Chapter 2 we have proved that a linear system .x = Ax + Bu is bounded stabilizable if and only if the matrix A has no eigenvalues z such that Re z ? 0, and all the uncontrollable eigenvalues have a strictly negative real part. The proof of Theorem 2.1 involves a complicated recursive construction. One might think that global stabilization might be achievable by a much simpler approach, e.g., by taking a stabilizer of the form u(x) = oe(h(x)) where oe is --assuming the input is scalar-- a "saturation function" such as oe(s) = \Gamma tanh(s) or oe(s) = \Gamma sign(s) min(jsj; 1), and h(x) is a linear feedback. It turns out that this is true in some cases, e.g., when the eigenvalues of A with zero real part are simple (see the proof of Theorem 2.1 for diagonable A), or when the system under consideration is a double integrator .x = y, .y = u. On the other hand, it was proved by Fuller in [13] that the simple approach described above cannot possibly work for the n-th order integrator, if n * 3 and oe is a saturation function of a special kind. In this chapter we extend Fuller's result to more general saturation functions. First we look at some simple examples.
3.1 Some Special Cases Example 3.1.1 Consider the system .x = Ax + Bu. Without loss of generality, let us suppose that all the eigenvalues of A have zero real part and (A; B) is controllable. Assume that all the eigenvalues of A are simple. Then there exists an invertible matrix T such that T
\Gamma 1
AT is skew-symmetric. Therefore the feedback u = oe((T
\Gamma 1
B)
y
T
\Gamma 1
x)
will stabilize the system .x = Ax + Bu, if oe(s) is any saturation function with the property that soe(s) ! 0 whenever s 6= 0. Pass1: Page 32
32 Indeed, if we choose V = jjT
\Gamma 1
xjj
2
, then along the trajectories of the closed-loop
system we have
. V = 2hT
\Gamma 1
Ax; T
\Gamma 1
xi + 2hT
\Gamma 1
Bu; T
\Gamma 1
xi :
The skew-symmetry of T
\Gamma 1
AT implies that the first term is zero. So
. V ^ 0. Using
the controllability of (A; B), we can see that if
. V j 0 along a trajectory t ! x(t), then
x(t) j 0. Global asymptotic stability then follows from LaSalle's Invariance Principle. 2
Example 3.1.2 Consider the double integrator
.x = y ;
.y = u :
Let oe : IR ! IR be any bounded strictly increasing continuous function such that oe(0) = 0. Define \Phi (s) =
R
s
0
oe(t) dt. Then the feedback law u = \Gamma oe(x + y) stabilizes
the system and V (x; y) = \Phi (x) + \Phi (x + y) + y
2
is a strict Lyapunov function of the
resulting closed-loop system.
In fact, a simple calculation shows that
. V = (oe(x) \Gamma oe(x + y))y \Gamma [oe(x + y)]
2
. The
monotonicity of oe then implies that (oe(x) \Gamma oe(x + y))y ! 0 if y 6= 0. So
. V ! 0 unless
(x; y) = (0; 0). 2
Example 3.1.3 Consider the system
.x = ffy ;
.y = \Gamma ffx + u ; .z = fiw ; .w = \Gamma fiz + u ; where ff; fi are two distinct positive numbers.
Here the matrix A has four distinct eigenvalues, so we are in a particular case of the situation described in Example 3.1.1. Let u = \Gamma oe(y+w) and V = x
2
+y
2
+z
2
+w
2
, where
oe(s) is any bounded continuous function with the property that soe(s) ? 0 if s 6= 0. From the analysis of Example 3.1.1, we know that the feedback law u = \Gamma oe(y + w) stabilizes the system and V is a Lyapunov function. 2 Pass1: Page 33
33 Example 3.1.4 Consider the system
.x = y ;
.y = u ;
.z = ffw ; .w = \Gamma ffz + u ; where ff ? 0 is a constant. Notice that now zero is a double eigenvalue and there is in addition a pair of imaginary eigenvalues.
In this case define oe and \Phi as in Example 3.1.2. Let u = \Gamma oe(x + y + z), V = \Phi (x + y +z)+
1
2
y
2
+
ff
2
z
2
+
ff
2
w
2
. Then a simple calculation shows that
. V = \Gamma oe
2
(x+y +z) ^ 0.
Clearly, equality holds if and only if x + y + z j 0. So if
. V j 0 along a trajectory,
then the second equation of the system implies that .y j 0. Differentiating the equation x + y + z = 0 twice, we get y + ffw j 0, \Gamma ff
2
z j 0. Thus z j 0. Differentiating the
equation x + y j 0, we then get y j 0. So x j 0 and w j 0 as well. Once again global asymptotic stability follows from LaSalle's Invariance Principle. 2
3.2 A Negative Result As we said earlier, the simple controller used in the examples in Section 3.1 does not work in general. In this section we show that such a control law fails to stabilize multiple integrators with dimension * 3.
Theorem 3.1 For the n-th order integrator with n * 3, there does not exist a feedback law of the form u = k(x) = oe(h(x)), where h : IR
n
! IR is a linear function and
oe : IR ! IR is a locally Lipschitz function for which both limits lim
s!\Sigma 1
oe(s) exist
and are nonzero, such that for the resulting closed-loop system the origin is a globally attracting point.
Recall that a point p is globally attracting for a system \Sigma : .x = f (x) if every trajectory t ! x(t) of \Sigma satisfies x(t) ! p as t ! +1. In particular, a globally asymptotically stable equilibrium is globally attracting, so our result implies that there does not exist a feedback of the form u = oe(h(x)) which is globally stabilizing in the usual sense. Pass1: Page 34
34 The restriction n * 3 is essential because, as shown in Example 3.1.2, the double integrator can be stabilized with a saturation of a linear feedback.
The result proved by Fuller in [13] says that if oe is a function such that oe(s) = sign(s) for jsj * a, a ? 0, and oe(s) is zero or has the same sign as s and does not exceed the saturation level for jsj ! a, then there is no feedback law of the form u = oe(h(x)) for the multiple integrators with order n * 3, with h : IR
n
! IR a linear function, such
that the resulting closed-loop system is globally asymptotically stable. Theorem 3.1 strengthens the result of [13] by allowing a much larger class of functions oe, such as oe(s) = tanh(s), etc. Also, this result can be used to solve the stabilizability problem for certain systems for which the information provided by the result of [13] does not suffice. For example, consider the system
.x = y ;
.y = z ; .z = sin u : Theorem 3.1 implies that no feedback of the form u = `(ax + by + cz) --where ` is a saturation function given by `(s) = sign(s)minfjsj; Ag-- can possibly be stabilizing. (Indeed, the theorem applies directly, using oe(s) = sin `(s), if A is not an integer multiple of 2ss. And in the remaining case, when A = 2kss for some integer k, the conclusion is trivial.) However, the result of [13] only yields the conclusion for A ^ ss=2.
3.3 The proof of Theorem 3.1 We consider an n-th order integrator
.x
1
= x
2
;
.x
2
= x
3
;
. . .
.x
n\Gamma 1
= x
n
;
.x
n
= u ;
(3.3.1)
where n * 3, and a feedback control of the form k(x) = oe(h(x)), where
h(x) = a
1
x
1
+ a
2
x
2
+ : : : + a
n
x
n
(3.3.2) Pass1: Page 35
35 is a linear function, and oe : IR ! IR is a locally Lipschitz function such that the limits of oe(s) as s ! \Sigma 1 exist and are not equal to zero. We use \Sigma to denote the closed-loop system obtained from (3.3.1) by plugging in u = k(x). Our goal is to prove that the origin cannot be globally attracting for \Sigma . The proof will be by contradiction, so from now we assume that 0 is globally attracting.
We first observe that, if a stabilizing feedback of the above form exists, then neces- sarily a
1
6= 0. (Otherwise, if we take two trajectories with initial conditions all whose
components coincide except for the first one, then the corresponding functions t ! x
1
(t)
will differ by a nonzero constant, so at least one of the trajectories will not go to 0 as t ! +1.) It then follows that oe(s) 6= 0 for s 6= 0. (Indeed, assume that oe(_s) =0, _s 6= 0. Then we can find _x
1
such that a
1
_x
1
= _s. The point (_x
1
; 0; : : : ; 0) is then an equilibrium
of \Sigma , contradicting the fact that 0 is globally attracting.)
Moreover, oe(s) must change sign. Indeed, if oe(s) was * 0 for all s, then it would follow that x
n
is nondecreasing along every trajectory, which is obviously impossible if
the origin is globally attracting. The possibility that oe(s) ^ 0 for all s is ruled out in a similar way. Since we know that oe(s) 6= 0 for s 6= 0, it follows that the change of sign must occur at s = 0.
So there are only two possibilities, namely, that soe(s) ! 0 for all s 6= 0, or that soe(s) ? 0 for all s 6= 0. In the latter case, the feedback k can also be expressed as k(x) = ~oe(
~ h(x)), where ~oe(s) = oe(\Gamma s) and
~ h(x) = \Gamma h(x), and it is clear that s~oe(s) ! 0
for all s 6= 0. So we may assume without loss of generality that we are in the first case.
In view of the above considerations, we will assume from now on that
(I) soe(s) ! 0 for s 6= 0, (II) a
1
6= 0.
(III) the limits lim
s!\Gamma 1
oe(s) = L, lim
s!+1
oe(s) = \Gamma M exist and satisfy L ? 0, M ? 0.
It follows in particular that oe is bounded, so we can pick an m such that
(IV) m * 1 and joe(s)j ^ m for all s. Pass1: Page 36
36 We first prove that the coefficient a
1
in (3.3.2) is positive. Assume that a
1
! 0. Define
a polynomial p(t) by
p(t) = \Gamma
a
1
M
2n!
t
n
\Gamma m
`
ja
2
j
t
n\Gamma 1
(n \Gamma 1)!
+ \Delta \Delta \Delta + ja
n
jt
'
+ a
1
_x
1
;
where _x
1
is a number to be chosen. Since M ? 0 and a
1
! 0, the leading coefficient of
p(t) is positive. Let A ? 0 be such that oe(s) ! \Gamma
M
2
for s ? A. Then it is possible to
choose _x
1
such that _x
1
! 0 and p(t) ? 2A for all t * 0. With such a choice of _x
1
, we
let _x = (_x
1
; 0; \Delta \Delta \Delta ; 0).
Now consider the trajectory x(t) of \Sigma starting at x(0) = _x. It is clear that h(x(0)) = a
1
_x
1
? 2A. Therefore there exists a maximal interval [0;
^
t) (with 0 !
^ t ^ +1) such
that h(x(t)) ? A for t 2 [0;
^
t). Then for t 2 [0;
^ t) we have \Gamma m ! oe(h(x(t))) ! \Gamma
M
2
. So
by successive integrations of (3.3.1), we see that, as long as t 2 [0;
^ t) :
\Gamma mt ^ x
n
(t) ^ \Gamma
M
2
t ;
\Gamma m
t
2
2
^ x
n\Gamma 1
(t) ^ \Gamma
M
2
t
2
2
;
. . .
\Gamma m
t
n\Gamma 1
(n\Gamma 1)!
^ x
2
(t) ^ \Gamma
M
2
t
n\Gamma 1
(n\Gamma 1)!
;
_x
1
\Gamma m
t
n
n!
^ x
1
(t) ^ _x
1
\Gamma
M
2
t
n
n!
:
Notice that the right-hand sides of the above inequalities are nonpositive. If we multiply them by ja
n
j; ja
n\Gamma 1
j; \Delta \Delta \Delta ; ja
2
j; a
1
, then all the inequalities are preserved except
for the last one, which gets reversed because a
1
! 0. Adding the resulting inequalities,
we get
h(x(t)) * a
1
x
1
(t) +
n X
i=2
ja
i
jx
i
(t) * a
1
_x
1
\Gamma a
1
M
2
t
n
n!
\Gamma m
n X
i=2
ja
i
j
t
n\Gamma i+1
(n \Gamma i + 1)!
whose right-hand side is greater than 2A by our choice of _x
1
. If
^ t was finite, then we
could let t !
^ t and conclude that h(x(
^ t)) * 2A. But then by continuity we would have
h(x(
^ t)) ? A for t in some larger interval [0;
^ t + ff), ff ? 0, contradicting the choice of
^ t.
So
^ t = +1, and therefore the functions x
i
(t), i = 1; 2; \Delta \Delta \Delta ; n are negative and decreasing
on the whole interval (0; +1). So x(t) cannot tend to zero. This contradiction shows that a
1
must be positive. Pass1: Page 37
37 From now on we shall assume that a
1
? 0. We let ae = maxfja
2
j; ja
3
j; \Delta \Delta \Delta ; ja
n
jg.
Let H denote the hyperplane defined by h(x) = 0, so that H can be identified with IR
n\Gamma 1
via the map x ! (x
2
; x
3
; : : : ; x
n
). For each C ? 0 we consider within H two
regions R
\Gamma
(C), R
+
(C). We define R
\Gamma
(C) to be the set of all those points x 2 H that
satisfy
x
n
! \Gamma C ;
x
n\Gamma 1
!
ae+2
a
1
x
n
;
x
n\Gamma 2
!
ae+2
a
1
(x
n\Gamma 1
+ x
n
) ;
. . .
x
2
!
ae+2
a
1
(x
3
+ x
4
+ \Delta \Delta \Delta + x
n
) :
Similarly, R
+
(C) is the set of those x 2 H such that \Gamma x 2 R
\Gamma
(C), i.e. that
x
n
? C ;
x
n\Gamma 1
?
ae+2
a
1
x
n
;
x
n\Gamma 2
?
ae+2
a
1
(x
n\Gamma 1
+ x
n
) ;
. . .
x
2
?
ae+2
a
1
(x
3
+ x
4
+ \Delta \Delta \Delta + x
n
) :
It is clear that R
\Gamma
(C) and R
+
(C) are not empty. We let R(C) = R
\Gamma
(C) [ R
+
(C).
We will prove our conclusion by studying the first return map \Phi associated with H. Precisely, we will show that, if C is sufficiently large, then the trajectory fl
x
from
any x 2 R(C) returns to H in a finite time o/ (x) and does so by hitting H at a point
~x 2 R(C), and that ~x 2 R
+
(C) if and only if x 2 R
\Gamma
(C). From this it follows easily
that a trajectory that starts at some point of R(C) can never go to zero. (Indeed, since the last coordinate of fl
x
(t) must change by at least 2C and joe(s)j is bounded by m,
it follows that o/ (x) *
2C
m
. So the times o/
j
(x), defined inductively by o/
1
(x) = o/ (x) and
o/
j+1
(x) = o/ (fl
x
(o/
j
(x))), go to infinity as j ! 1 and, clearly, fl
x
(o/
j
(x)) does not go to
zero.)
We will only study the first return of trajectories from points in R
\Gamma
(C). (The other
case is similar, or can be reduced to the first one by the change of variables x ! \Gamma x.)
It will be convenient to work instead with the hyperplane H
A
defined by h(x) + A = Pass1: Page 38
38 0, where A is such that
(1 \Gamma ffi)L ! oe(s) ! (1 + ffi)L ; if s ! \Gamma A ; \Gamma (1 + ffi)M ! oe(s) ! \Gamma (1 \Gamma ffi)M ; if s ? A ;
and ffi ? 0 satisfies two inequalities:
1\Gamma ffi 1+ffi
n
2
n
2
\Gamma 1
? 1 ;
1\Gamma ffi 1+ffi
i
n \Gamma
1
2
j
?
9
4
:
Notice that it is possible to choose ffi ? 0 so as to satisfy the last inequality because n * 3. (This is in fact the only point in the whole proof where the condition n * 3 is used.)
We define R
A;1
\Gamma
(C) to be the set of all those points x 2 H
A
that satisfy
x
n
! \Gamma C ;
x
n\Gamma 1
!
ae+1
a
1
x
n
;
x
n\Gamma 2
!
ae+1
a
1
(x
n\Gamma 1
+ x
n
) ;
. . .
x
2
!
ae+1
a
1
(x
3
+ x
4
+ \Delta \Delta \Delta + x
n
) ;
and R
A;3
+
(C) to be the set of all those points x 2 H
A
that satisfy
x
n
? C ;
x
n\Gamma 1
?
ae+3
a
1
x
n
;
x
n\Gamma 2
?
ae+3
a
1
(x
n\Gamma 1
+ x
n
) ;
. . .
x
2
?
ae+3
a
1
(x
3
+ x
4
+ \Delta \Delta \Delta + x
n
) :
As explained before, the constant C will be required to be "sufficiently large." The Pass1: Page 39
39 precise meaning of this is that C should satisfy the following inequalities:
C ? 5 ; C ? mA ; C ? a
1
+ mae ;
C ? 3m(ae + 2) ; 3C ? A + 4mae ; C \Gamma 1 ? ae(1 + ffi)max(L; M ) ; 3(C \Gamma 1) ? (2ae + 3)(1 + ffi)max(L; M ) ; a
1
(C \Gamma 1) ? 2nae(1 + ffi)max(L; M ) ;
a
1
(C \Gamma 1) ? 3n(1 + ae)(1 + ffi)max(L; M ) :
Our first task is to make sure that
(I) if we start at a point x(0) = _x 2 R
\Gamma
(C), then the resulting trajectory t ! x(t)
reaches R
A;1
\Gamma
(C \Gamma 1) at some positive time,
and (II) a trajectory starting at an x(0) = _x 2 R
A;3
+
(C) gets to R
+
(C) at some later time.
To prove (I) and (II), notice that the derivative
. h
\Lambda
(t) of h
\Lambda
(t) = h(x(t)) is given by
. h
\Lambda
(t) = a
1
x
2
(t) + : : : + a
n\Gamma 1
x
n
(t) + a
n
oe(h
\Lambda
(t)) :
If x(0) 2 R
\Gamma
(C), then h
\Lambda
(0) = 0 and
. h
\Lambda
(0) ^ a
1
_x
2
+ aej_x
3
+ \Delta \Delta \Delta + _x
n
j + mae, which is
! 0 because a
1
_x
2
! \Gamma (ae + 2)j_x
3
+ \Delta \Delta \Delta + _x
n
j and j_x
n
j ? mae (since C ? mae). Then
. h
\Lambda
(t) ! 0 for t in some interval of the form (0;
^
t). Choose
^ t to be as large as possible.
(In principle
^ t could be infinite.) Clearly, h
\Lambda
(t) ! 0 and therefore .x
n
(t) = oe(h
\Lambda
(t)) ? 0
for 0 ! t !
^ t. On the other hand, .x
n
(t) ^ m, so that on an interval of length T it is not
possible for x
n
(t) to increase by more than mT . If we let
~ t = 1=m, we see that on the
interval I = [0;
~
t] the function x
n
(t) satisfies x
n
(t) ^ 1 \Gamma C. In particular (since C ? 1),
x
n
(t) ! 0 on I. Since .x
n\Gamma 1
= x
n
, we see that x
n\Gamma 1
is decreasing on I, so x
n\Gamma 1
(t) ^ 0
on I, because x
n\Gamma 1
(0) ^ 0. Continuing in this way, we find that all the functions x
j
(t), Pass1: Page 40
40 for j = 1; 2; : : : ; n \Gamma 1 are decreasing on I, and all the x
j
(t), for j = 2; 3; : : : ; n, are
negative on I.
Now consider the trajectory x(t) on [0;
~ t]. Since 0 ! oe(h
\Lambda
(t)) ^ m on [0;
~
t], by
repeated integrations we get the following inequalities
0 ^ fl
n
(t) ^ mt ;
0 ^ fl
n\Gamma 1
(t) ^ m
t
2
2
;
. . .
0 ^ fl
2
(t) ^ m
t
n\Gamma 1
(n\Gamma 1)!
;
0 ^ fl
1
(t) ^ m
t
n
n!
;
(3.3.3)
where
fl
n
(t) = x
n
(t) \Gamma _x
n
;
fl
n\Gamma 1
(t) = x
n\Gamma 1
(t) \Gamma (_x
n\Gamma 1
+ _x
n
t) ;
. . .
fl
2
(t) = x
2
(t) \Gamma (_x
2
+ _x
3
t + \Delta \Delta \Delta + _x
n
t
n\Gamma 2
(n\Gamma 2)!
) ;
fl
1
(t) = x
1
(t) \Gamma (_x
1
+ _x
2
t + \Delta \Delta \Delta + _x
n
t
n\Gamma 1
(n\Gamma 1)!
) :
(3.3.4)
From (3.3.3), we see that x
j
(t) *
P
n
k=j
_x
k
t
k\Gamma j
(k\Gamma j)!
. Therefore, for i = 2; 3; \Delta \Delta \Delta ; n \Gamma 1, we
have
n X
j=i+1
x
j
(t) *
n X
j=i+1
n X
k=j
_x
k
t
k\Gamma j
(k \Gamma j)!
=
n\Gamma 1
X
j=i
(_x
j+1
+ _x
j+2
+ \Delta \Delta \Delta + _x
n
)
t
j\Gamma i
(j \Gamma i)!
: (3.3.5)
Since x
j
(t) ! 0 on [0;
~
t] for j = 2; 3; \Delta \Delta \Delta ; n, we have
. h
\Lambda
(t) ^ a
1
x
2
(t) + aejx
3
(t) + \Delta \Delta \Delta + x
n
(t)j + mae :
From (3.3.5), in particular for i = 2, we have
jx
3
(t) + \Delta \Delta \Delta + x
n
(t)j ^ \Gamma
n\Gamma 1
X
i=2
(_x
i+1
+ \Delta \Delta \Delta + _x
n
)
t
i\Gamma 2
(i \Gamma 2)!
:
Also, a
1
x
2
(t) =
P
n
i=2
a
1
_x
i
t
i\Gamma 2
(i\Gamma 2)!
+ a
1
fl
2
(t). So we have the estimate:
. h
\Lambda
(t) ^
n X
i=2
(a
1
_x
i
\Gamma ae(_x
i+1
+ \Delta \Delta \Delta + _x
n
))
t
i\Gamma 2
(i \Gamma 2)!
+ a
1
m
t
n\Gamma 1
(n \Gamma 1)!
+ mae: Pass1: Page 41
41 Since _x 2 R
\Gamma
(C), it follows that all the coefficients in the summation of the above
inequality are negative. In particular, for t 2 I = [0;
~
t] we have
. h
\Lambda
(t) ^ a
1
_x
2
\Gamma ae(_x
3
+ \Delta \Delta \Delta + _x
n
) +
a
1
m
2\Gamma n
(n \Gamma 1)!
+ mae;
which is bounded above by 2_x
n
+a
1
+mae (since a
1
_x
2
! (ae+2)(_x
3
+\Delta \Delta \Delta + _x
n
) and m ? 1).
Because _x
n
! \Gamma C and C is greater than a
1
+ mae and mA, we see that
. h
\Lambda
(t) ! \Gamma mA
for t 2 I. So
^ t ?
~ t, and h
\Lambda
(t) is decreasing on I. Moreover, the bound for
. h
\Lambda
, together
with h
\Lambda
(0) = 0, imply that h
\Lambda
(
~ t) ^ \Gamma A. So there exists a
^ t 2 I such that h
\Lambda
(
^ t) = \Gamma A.
To complete the proof of this part, we need to show that x(
^ t) 2 R
A;1
\Gamma
(C \Gamma 1). Here, the
inequality x
n
(
^ t) ^ 1 \Gamma C follows directly from (3.3.3) and the fact that
^ t ^
~ t =
1
m
. We
have to prove the remaining inequalities:
x
i
(
^ t) !
ae + 1
a
1
i
x
i+1
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t)
j
for i = 2; 3; \Delta \Delta \Delta ; n \Gamma 1.
From (3.3.3) we have
x
i
(
^ t) ^
n X
j=i
_x
i
^ t
j\Gamma i
(j \Gamma i)!
+ m
^ t
n\Gamma i+1
(n \Gamma i + 1)!
and from (3.3.5) we have
x
i+1
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t) *
n\Gamma 1
X
j=i
(_x
j+1
+ _x
j+2
+ \Delta \Delta \Delta + _x
n
)
^ t
j\Gamma i
(j \Gamma i)!
:
Therefore for i = 2; 3; \Delta \Delta \Delta ; n, we see that a
1
x
i
(
^ t)\Gamma (ae+1)
n X
j=i+1
x
j
(
^ t) ^
n X
j=i
(a
1
_x
j
\Gamma (ae+1)(_x
j+1
+\Delta \Delta \Delta +_x
n
))
^ t
j\Gamma i
(j \Gamma i)!
+a
1
m
^ t
n\Gamma i+1
(n \Gamma i + 1)!
:
Here all coefficients in the summation on the right-hand side are negative because
_x 2 R
\Gamma
(C). So in particular the right-hand side is bounded above by
a
1
_x
i
\Gamma (ae + 1)(_x
i+1
+ : : : + _x
n
) + a
1
m
^ t
n\Gamma i+1
(n \Gamma i + 1)!
:
Since
^ t ^
~ t =
1
m
, and
a
1
_x
i
^ (ae + 2)(_x
i+1
+ : : : + _x
n
) ^ (ae + 1)(_x
i+1
+ : : : + _x
n
) + _x
n
;
we get in fact the upper bound _x
n
+
a
1
(n\Gamma i+1)!
. The inequalities _x
n
! \Gamma C and C ? a
1
imply in particular that this is negative. So x(
^ t) 2 R
A;1
\Gamma
(C \Gamma 1) and (I) is proved. Pass1: Page 42
42 To prove (II) notice that, if we start a trajectory t ! x(t) at a point _x in R
A;3
+
(C),
then at that point h
\Lambda
= \Gamma A, so that .x
n
(0) ? 0. Again, this implies that there is a
maximal interval J = (0;
^
t) (with
^ t finite or infinite) on which h
\Lambda
! 0, and on that
interval x
n
is increasing and therefore positive, and in fact * C. So x
n\Gamma 1
is increasing
and positive, and then the same is true for x
n\Gamma 2
; : : : ; x
2
. Since
. h
\Lambda
(t) = a
1
x
2
(t) + : : : + a
n\Gamma 1
x
n
(t) + a
n
oe(h
\Lambda
(t)) ;
it follows that
. h
\Lambda
(t) * a
1
x
2
(t) \Gamma aejx
3
(t) + \Delta \Delta \Delta + x
n
(t)j \Gamma mae :
Notice that Formulas (3.3.3) are still true in this case. So, using the upper bound for fl
i
, we get
n X
j=i+1
x
j
(t) ^
n\Gamma 1
X
j=i
(_x
j+1
+ _x
j+2
+ \Delta \Delta \Delta + _x
n
)
t
j\Gamma i
(j \Gamma i)!
+ m
n\Gamma i
X
j=1
t
j
j!
; for t 2 J : (3.3.6)
Also
x
i
(t) *
n X
j=i
_x
j
t
j\Gamma i
(j \Gamma i)!
; for t 2 J : (3.3.7)
Applying (3.3.6) and (3.3.7) for i = 2, we then see that
. h
\Lambda
(t) *
n X
i=2
(a
1
_x
i
\Gamma ae(_x
i+1
+ \Delta \Delta \Delta + _x
n
))
t
i\Gamma 2
(i \Gamma 2)!
\Gamma mae
n\Gamma 2
X
i=1
t
i
i!
\Gamma mae:
Since _x 2 R
A;3
+
(C), it follows that a
1
_x
i
\Gamma ae(_x
i+1
+ \Delta \Delta \Delta + _x
n
) * 3_x
n
? 0 for i =
2; 3; \Delta \Delta \Delta ; n \Gamma 1. Therefore, on J " [0; 1] we have
. h
\Lambda
(t) * 3_x
n
\Gamma mae(e + 1) ? A ( because
_x
n
? C ?
A+4mae
3
), which implies that
^ t ! 1 because otherwise h
\Lambda
(1) ? h
\Lambda
(0) + A = 0
which is a contradiction. So we see that J ae [0; 1],
^ t is finite, and then h
\Lambda
(
^ t) = 0.
To see that x(
^ t) 2 R
+
(C), we need to show that
x
i
(
^ t) ?
ae + 2
a
1
i
x
i+1
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t)
j
for i = 2; 3; \Delta \Delta \Delta ; n \Gamma 1.
This can be shown by a calculation similar to the one we used to prove (I). In fact, (3.3.6) and (3.3.7) imply that
a
1
x
i
(
^ t)\Gamma (ae+2)
n X
j=i+1
x
j
(
^ t) *
n X
j=i
(a
1
_x
j
\Gamma (ae+2)(_x
j+1
+\Delta \Delta \Delta + _x
n
))
^ t
j\Gamma i
(j \Gamma i)!
\Gamma m(ae+2)
n\Gamma i X
j=1
^ t
j
j!
: Pass1: Page 43
43 In particular, the right-hand side is bounded below by
a
1
_x
i
\Gamma (ae + 2)(_x
i+1
+ \Delta \Delta \Delta + _x
n
) \Gamma m(ae + 2)e
^ t
:
Since _x 2 R
A;3
+
, it follows that a
1
_x
i
* (ae + 3)(_x
i+1
+ \Delta \Delta \Delta + _x
n
), and thus a
1
_x
i
\Gamma (ae +
2)(_x
i+1
+ \Delta \Delta \Delta + _x
n
) * _x
n
, so the lower bound is in fact * _x
n
\Gamma 3m(ae + 2) ? 0 (because
^ t ! 1 and _x
n
? C ? 3m(ae + 2)). Therefore x(
^ t) 2 R
+
(C), and (II) is proved.
In view of (I) and (II) above, our conclusion will follow if we prove: (III) a trajectory starting at an x 2 R
A;1
\Gamma
(C \Gamma 1) gets to R
A;3
+
(C) at some later time.
From now on, we write K = C \Gamma 1 and assume that t ! x(t) is a trajectory such that x(0) = _x 2 R
A;1
\Gamma
(K). As before, we let h
\Lambda
(t) = h(x(t)). Then h
\Lambda
(0) = \Gamma A, and
. h
\Lambda
(0) = a
1
_x
2
+ \Delta \Delta \Delta + a
n\Gamma 1
_x
n
+ a
n
oe(\Gamma A) ! 0 (because a
1
_x
2
! (ae + 1)(_x
3
+ \Delta \Delta \Delta + _x
n
) and
_x
n
! \Gamma mae, which follows from the facts that _x
n
! 1 \Gamma C, C ? 3m(ae + 2) and m ? 1).
So there is a maximal interval I = (0;
^ t) on which h
\Lambda
! \Gamma A. Our goal is to prove that
^ t is finite and x(
^ t) 2 R
A;3
+
(C). Since h
\Lambda
(
^ t) must be equal to \Gamma A if
^ t is finite, what we
need is to show that
^ t ! 1 and to prove the inequalities
x
n
(
^ t) ? C ;
x
n\Gamma 1
(
^ t) ?
ae+3
a
1
x
n
(
^ t) ;
x
n\Gamma 2
(
^ t) ?
ae+3
a
1
(x
n\Gamma 1
(
^ t) + x
n
(
^ t)) ;
. . .
x
2
(
^ t) ?
ae+3
a
1
(x
3
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t)) :
(3.3.8)
On I we have (1 \Gamma ffi)L ^ .x
n
(t) ^ (1 + ffi)L, so by successive integrations we get the
following bounds:
(1 \Gamma ffi)Lt ^ fl
n
(t) ^ (1 + ffi)Lt ;
(1 \Gamma ffi)L
t
2
2
^ fl
n\Gamma 1
(t) ^ (1 + ffi)L
t
2
2
;
. . .
(1 \Gamma ffi)L
t
n\Gamma 1
(n\Gamma 1)!
^ fl
2
(t) ^ (1 + ffi)L
t
n\Gamma 1
(n\Gamma 1)!
;
(1 \Gamma ffi)L
t
n
n!
^ fl
1
(t) ^ (1 + ffi)L
t
n
n!
;
(3.3.9) Pass1: Page 44
44 where fl
i
(t), i = 1; 2; \Delta \Delta \Delta ; n are defined in (3.3.4). Notice that both sides of the above
inequalities are positive. If we define
f (t) = a
1
fl
1
(t) + a
2
fl
2
(t) + \Delta \Delta \Delta + a
n
fl
n
(t) ;
then
f (t) * (1 \Gamma ffi)La
1
t
n
n!
\Gamma (1 + ffi)Lae
`
t
n\Gamma 1
(n \Gamma 1)!
+
t
n\Gamma 2
(n \Gamma 2)!
+ \Delta \Delta \Delta + t
'
; (3.3.10)
and
f (t) ^ (1 + ffi)La
1
t
n
n!
+ (1 + ffi)Lae
`
t
n\Gamma 1
(n \Gamma 1)!
+
t
n\Gamma 2
(n \Gamma 2)!
+ \Delta \Delta \Delta + t
'
: (3.3.11)
Since h
\Lambda
(0) = a
1
_x
1
+ a
2
_x
2
+ \Delta \Delta \Delta + a
n
_x
n
= \Gamma A, it is clear that
f (t) = h
\Lambda
(t) + A \Gamma
n\Gamma 1
X
k=1
i
n\Gamma k
X
i=1
a
i
_x
i+k
j
t
k
k!
;
i.e,
h
\Lambda
(t) + A = f (t) +
n\Gamma 1
X
k=1
i
n\Gamma k
X
i=1
a
i
_x
i+k
j
t
k
k!
; (3.3.12)
and
a
1
_x
k+1
+ ae(_x
k+2
+ \Delta \Delta \Delta + _x
n
) ^
n\Gamma k
X
i=1
a
i
_x
i+k
^ a
1
_x
k+1
\Gamma ae(_x
k+2
+ \Delta \Delta \Delta + _x
n
): (3.3.13)
From (3.3.10) and (3.3.12), we see that h
\Lambda
(t) + A * (1 \Gamma ffi)La
1
t
n
n!
+ p(t), where p(t) is
a polynomial of degree n \Gamma 1. If I was infinite it would follow that h
\Lambda
(t) + A ? 0 for
large enough t, contradicting the fact that h
\Lambda
! \Gamma A on I. So I is finite, and of course
h
\Lambda
(
^ t) = \Gamma A. Now all we need is to establish the inequalities (3.3.8).
First, from (3.3.11), (3.3.12) and (3.3.13), for t 2 I, we have
h
\Lambda
(t) + A ^ (1 + ffi)La
1
t
n
n!
+
n\Gamma 1
X
k=1
(a
1
_x
k+1
\Gamma ae(_x
k+2
+ \Delta \Delta \Delta + _x
n
) + (1 + ffi)Lae)
t
k
k!
: (3.3.14)
Denote
ff
k
= a
1
_x
k+1
\Gamma ae(_x
k+2
+ \Delta \Delta \Delta + _x
n
) + (1 + ffi)Lae; k = 1; 2; \Delta \Delta \Delta ; n \Gamma 1:
Since _x 2 R
A;1
\Gamma
(C), we have a
1
_x
k+1
! (ae + 1)(_x
k+2
+ \Delta \Delta \Delta + _x
n
) for k = 1; 2; \Delta \Delta \Delta ; n \Gamma 2 and
it then follows that ff
k
! _x
n
+ (1 + ffi)Lae ! 0 (because j_x
n
j ? C \Gamma 1 ? (1 + ffi)Lae). Also, Pass1: Page 45
45 ff
n\Gamma 1
= a
1
_x
n
+ (1 + ffi)Lae ! 0 (since j_x
n
j ? C \Gamma 1 ?
(1+ffi)Lae
a
1
). So all the ff
k
are negative.
Substituting t =
^ t into (3.3.14), we then see that (1 + ffi)La
1
^ t
n
n!
* \Gamma
P
n\Gamma 1
k=1
ff
k
^ t
k
k!
because
h
\Lambda
(
^ t) = \Gamma A. Therefore,
La
1
^ t
n\Gamma 1
n!
* \Gamma
1
1 + ffi
n\Gamma 1
X
k=1
ff
k
^ t
k\Gamma 1
k!
: (3.3.15)
Since all the ff
k
for k = 1; 2; \Delta \Delta \Delta ; n \Gamma 1 are negative, this implies that
La
1
^ t
n\Gamma 1
n!
* \Gamma
1
1 + ffi
ff
n\Gamma 1
^ t
n\Gamma 2
(n \Gamma 1)!
(3.3.16)
which implies that
^ t *
n
(1+ffi)La
1
(\Gamma ff
n\Gamma 1
). Since ff
n\Gamma 1
= a
1
_x
n
+ (1 + ffi)Lae and j_x
n
j ?
C \Gamma 1 ?
2n(1+ffi)Lae
a
1
, we have (1 + ffi)Lae ! \Gamma
a
1
2n
_x
n
, thus ff
n\Gamma 1
!
(2n\Gamma 1)a
1
2n
_x
n
. Therefore
^ t *
2n\Gamma 1
2(1+ffi)L
j_x
n
j and x
n
(
^ t) * _x
n
+ (1 \Gamma ffi)L
^ t * (
1\Gamma ffi
1+ffi
(n \Gamma
1
2
) \Gamma 1)j_x
n
j, which is greater than
5 4
j_x
n
j because
1\Gamma ffi
1+ffi
(n \Gamma
1
2
) ?
9
4
. We know that j_x
n
j ? K = C \Gamma 1 and C ? 5. Therefore,
x
n
(
^ t) ?
5
4
j_x
n
j ? C. The first inequality in (3.3.8) follows.
Now we establish the other inequalities in (3.3.8), i.e,
x
i
(
^ t) *
ae + 3
a
1
i
x
i+1
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t)
j
;
for i = 2; 3; \Delta \Delta \Delta ; n \Gamma 1. We need to find a positive lower bound for a
1
x
i
(
^ t) \Gamma (ae +
3)(x
i+1
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t)) for i = 2; 3; \Delta \Delta \Delta ; n \Gamma 1.
Since x
j
(
^ t) =
P
n
k=j
_x
k
^ t
k\Gamma j
(k\Gamma j)!
+ fl
j
(
^ t), we have
n X
j=i+1
x
j
(
^ t) =
n X
j=i+1
n X
k=j
_x
k
^ t
k\Gamma j
(k \Gamma j)!
+
n X
j=i+1
fl
j
(
^ t) : (3.3.17)
The first summation on the right-hand side is equal (see (3.3.5)) to
P
n\Gamma 1
j=i
(_x
j+1
+ \Delta \Delta \Delta +
_x
n
)
^ t
j\Gamma i
(j\Gamma i)!
, i.e. to
P
n\Gamma i\Gamma 1
j=0
(_x
j+i+1
+\Delta \Delta \Delta + _x
n
)
^ t
j
j!
. From (3.3.9) it follows that
P
n
j=i+1
fl
j
(
^ t) is
bounded above by
P
n
j=i+1
(1+ffi)L
^ t
n\Gamma j+1
(n\Gamma j+1)!
, which is equal to (1+ffi)L
P
n\Gamma i
j=1
^ t
j
j!
. Therefore,
(3.3.17) implies
n X
j=i+1
x
j
(
^ t) ^ (_x
i+1
+\Delta \Delta \Delta + _x
n
)+
n\Gamma i\Gamma 1
X
j=1
(_x
j+i+1
+\Delta \Delta \Delta + _x
n
+(1+ffi)L)
^ t
j
j!
+(1+ffi)L
^ t
n\Gamma i
(n \Gamma i)!
:
(3.3.18) Also, from (3.3.9) and (3.3.4), we get
x
i
(
^ t) *
n\Gamma i X
j=0
_x
j+i
^ t
j
j!
+ (1 \Gamma ffi)L
^ t
n\Gamma i+1
(n \Gamma i + 1)!
: (3.3.19) Pass1: Page 46
46 Therefore, using (3.3.18) and (3.3.19), it follows that
a
1
x
i
(
^ t) \Gamma (ae + 3)
i
x
i+1
(
^ t) + \Delta \Delta \Delta + x
n
(
^ t)
j
* (1 \Gamma ffi)La
1
^ t
n\Gamma i+1
(n\Gamma i+1)!
+
i
a
1
_x
n
\Gamma (ae + 3)(1 + ffi)L
j
^ t
n\Gamma i
(n\Gamma i)!
+
i
a
1
_x
n\Gamma 1
\Gamma (ae + 3)(_x
n
+ (1 + ffi)L)
j
^ t
n\Gamma i\Gamma 1
(n\Gamma i\Gamma 1)!
+ \Delta \Delta \Delta
+
i
a
1
_x
i+1
\Gamma (ae + 3)(_x
i+2
+ \Delta \Delta \Delta + _x
n
+ (1 + ffi)L)
j
^ t
+a
1
_x
i
\Gamma (ae + 3)(_x
i+1
+ \Delta \Delta \Delta + _x
n
)
= q
i
(
^ t) + r
i
(
^ t) ;
where
q
i
(
^ t) = a
1
(1\Gamma ffi)L
^ t
n\Gamma i+1
(n \Gamma i + 1)!
+ff
n\Gamma 1
^ t
n\Gamma i
(n \Gamma i)!
+\Delta \Delta \Delta +ff
i
^ t+ff
i\Gamma 1
\Gamma (2ae+3)(1+ffi)L
^ t
n\Gamma i
(n \Gamma i)!
;
r
i
(
^ t) = (\Gamma 3_x
n
\Gamma (2ae + 3)(1 + ffi)L)
^ t
n\Gamma i\Gamma 1
(n\Gamma i\Gamma 1)!
+ \Delta \Delta \Delta
+
i
\Gamma 3
P
n
j=i+2
_x
j
\Gamma (2ae + 3)(1 + ffi)L
j
^ t
+
i
\Gamma 3
P
n
j=i+1
_x
j
\Gamma (1 + ffi)Lae
j
* \Gamma (3_x
n
+ (2ae + 3)(1 + ffi)L)
P
n\Gamma i\Gamma 1
k=0
^ t
k
k!
:
It is clear that r
i
(
^ t) * 0 since _x
n
! 1 \Gamma C and 3(C \Gamma 1) ? (2ae + 3)(1 + ffi)L. What we
need is to show q
i
(
^ t) * 0. This time we will require an estimation for
^ t more accurate
than (3.3.16).
Since all ff
k
in the summation of (3.3.15) are negative, if we restrict the summation
to the range k = i \Gamma 1; i; i + 1; \Delta \Delta \Delta ; n \Gamma 1, the inequality still holds. Dividing both sides of the new inequality by
^ t
i\Gamma 2
, we then get
La
1
^ t
n\Gamma i+1
n!
* \Gamma
1
1 + ffi
`
ff
n\Gamma 1
^ t
n\Gamma i
(n \Gamma 1)!
+ ff
n\Gamma 2
^ t
n\Gamma i\Gamma 1
(n \Gamma 2)!
+ \Delta \Delta \Delta + ff
i\Gamma 1
1
(i \Gamma 1)!
'
:
Multiplying both sides of the above inequality by the constant n(n \Gamma 1) \Delta \Delta \Delta (n \Gamma i + 2), we find (since \Pi
i\Gamma 2
j=0
n\Gamma j n\Gamma k\Gamma j
*
n
n\Gamma 1
for k = 1; 2; \Delta \Delta \Delta ; n \Gamma i + 1) that
La
1
^ t
n\Gamma i+1
(n \Gamma i + 1)!
* \Gamma
1
1 + ffi
n n \Gamma 1
`
ff
n\Gamma 1
^ t
n\Gamma i
(n \Gamma i)!
+ ff
n\Gamma 2
^ t
n\Gamma i\Gamma 1
(n \Gamma i \Gamma 1)!
+ \Delta \Delta \Delta + ff
i\Gamma 1
'
:
Therefore,
q
i
(
^ t) *
`
\Gamma
1\Gamma ffi
1+ffi
n n\Gamma 1
+ 1
'`
ff
n\Gamma 1
^ t
n\Gamma i
(n\Gamma i)!
+ ff
n\Gamma 2
^ t
n\Gamma i\Gamma 1
(n\Gamma i\Gamma 1)!
+ \Delta \Delta \Delta + ff
i\Gamma 1
'
\Gamma (2ae + 3)(1 + ffi)L
^ t
n\Gamma i
(n\Gamma i)!
: Pass1: Page 47
47 Since
1\Gamma ffi
1+ffi
n
2
n
2
\Gamma 1
? 1, it follows that
1\Gamma ffi
1+ffi
n n\Gamma 1
\Gamma 1 ?
1
n
. Thus
q
i
(
^ t) * (\Gamma
1
n
ff
n\Gamma 1
\Gamma (2ae + 3)(1 + ffi)L)
^ t
n\Gamma i
(n \Gamma i)!
:
Since ff
n\Gamma 1
= a
1
_x
n
+ (1 + ffi)Lae, we have
q
i
(
^ t) ? (\Gamma
a
1
n
_x
n
\Gamma 3(1 + ffi)(1 + ae)L)
^ t
n\Gamma i
(n \Gamma 1)!
:
The assumption that j_x
n
j * C \Gamma 1 ?
3n(1+ffi)(1+ae)L
a
1
then implies that q
i
(
^ t) ? 0, as
desired. 2 Pass1: Page 48
48 Chapter 4 A Design Using Hidden Layers
In Chapter 2 we have shown that a linear system \Sigma : .x = Ax + Bu is globally asymptotically stabilizable with a bounded feedback if and only if the following pair of algebraic conditions are satisfied:
(a) all eigenvalues of A have nonpositive real part, and (b) all eigenvalues of the uncontrollable part of \Sigma have strictly negative real parts
(that is, the pair (A; B) is stabilizable in the ordinary sense).
In some special cases stabilization is possible by simply using a saturated linear feedback law. But as we see in Chapter 3, such saturated linear feedback does not work for general linear systems. In this chapter we present two designs for the general case. One employs linear combinations and compositions of saturation nonlinearities. The other uses linear combinations of saturated linear functions. Both approaches are explicit and constructive.
4.1 Constructions of Feedback We first define S to be the class of all functions oe from IR to IR such that
ffl oe is locally Lipschitz, ffl soe(s) ? 0 whenever s 6= 0, ffl oe is differentiable at 0 and oe
0
(0) ? 0 ,
ffl lim inf
jsj!1
joe(s)j ? 0. Pass1: Page 49
49 For any finite sequence oeoe
oe
= (oe
1
; \Delta \Delta \Delta ; oe
k
) of functions in S, we define a set F
n
(oeoe
oe
) of
functions f from IR
n
to IR inductively as follows:
ffl if k = 0 (i.e. if oeoe
oe
is the empty sequence), then F
n
(oeoe
oe
) consists of one element,
namely, the zero function from IR
n
to IR;
ffl F
n
(oe
1
) consists of all the functions h : IR
n
! IR of the form h(x) = oe
1
(g(x)),
where g : IR
n
! IR is linear;
ffl for every k ? 1, F
n
(oe
1
; \Delta \Delta \Delta ; oe
k
) is the set of all functions h : IR
n
! IR that are of
the form h(x) = oe
k
(f (x) + cg(x)), with f linear, g 2 F
n
(oe
1
; \Delta \Delta \Delta ; oe
k\Gamma 1
), and c * 0.
We also define G
n
(oeoe
oe
) to be the class of functions h : IR
n
! IR given by
h(x) = a
1
oe
1
(f
1
(x)) + a
2
oe
2
(f
2
(x)) + \Delta \Delta \Delta + a
k
oe
k
(f
k
(x)) ;
where f
1
; \Delta \Delta \Delta ; f
k
are linear functions and a
1
; \Delta \Delta \Delta ; a
k
are nonnegative constants such that
a
1
+ \Delta \Delta \Delta + a
k
^ 1.
Next, for an m-tuple ll
l
= (l
1
; \Delta \Delta \Delta ; l
m
) of nonnegative integers, define jll
l
j = l
1
+\Delta \Delta \Delta +l
m
.
For a finite sequence oeoe
oe
= (oe
1
; \Delta \Delta \Delta ; oe
jll
l
j
) = (oe
1
1
; \Delta \Delta \Delta ; oe
1
l
1
; \Delta \Delta \Delta ; oe
m
1
; \Delta \Delta \Delta ; oe
m
l
m
) of functions in
S, we let F
ll
l
n
(oeoe
oe
) (and respectively G
ll
l
n
(oeoe
oe
) ) denote the set of all functions h : IR
n
! IR
m
that are of the form (h
1
; \Delta \Delta \Delta ; h
m
), where h
i
2 F
n
(oe
i
1
; \Delta \Delta \Delta ; oe
i
l
i
) (and respectively h
i
2
G
n
(oe
i
1
; \Delta \Delta \Delta ; oe
i
l
i
)) for i = 1; 2; \Delta \Delta \Delta ; m. (It is clear that F
ll
l
n
(oeoe
oe
) = F
n
(oeoe
oe
), G
ll
l
n
(oeoe
oe
) = G
n
(oeoe
oe
) if
m = 1.)
Definition 4.1.1 Let ffi ? 0. A function f : [0; +1) ! IR
n
is eventually bounded by ffi
(and write jf j ^
ev
ffi), if there exists T ? 0 such that jf (t)j ^ ffi for all t * T .
Definition 4.1.2 An n-dimensional system E : .x = f (x) is SISS (small-input small- state) if for every " ? 0 there is a ffi ? 0 such that, if e : [0; +1) ! IR
n
is bounded,
measurable, and eventually bounded by ffi, then every solution t ! x(t) of .x = f (x)+e(t) is eventually bounded by ".
Definition 4.1.3 For \Delta ? 0; N ? 0, A system E : .x = f (x) is SISS
L
(\Delta ; N ) if,
whenever 0 ! ffi ^ \Delta , it follows that, if e : [0; +1) ! IR
n
is bounded, measurable, and Pass1: Page 50
50 eventually bounded by ffi, then every solution of .x = f (x) + e(t) is eventually bounded by N ffi. The system is SISS
L
if it is SISS
L
(\Delta ; N ) for some \Delta ? 0; N ? 0.
It is obvious that if a system .x = f (x) is SISS
L
(\Delta ; N), then the system .x = *f (
x
*
) is
SISS
L
(*\Delta ; N ).
Definition 4.1.4 For a system .x = f (x; u), a feedback u = k(x) is stabilizing if 0 is a globally asymptotically stable equilibrium of the closed-loop system .x = f (x; k(x)). If, in addition, this closed-loop system is SISS
L
, then we will say that k is SISS
L
-
stabilizing.
For a square matrix A we let s(A) denote the number of conjugate pairs of purely imaginary eigenvalues of A (including multiplicity) and we let z(A) denote the multi- plicity of zero as an eigenvalue of A. We write _(A) = s(A) + z(A). Our main result is as follows:
Theorem 4.1 Let \Sigma be a linear system .x = Ax + Bu with state space IR
n
and input
space IR
m
. Assume that \Sigma is asymptotically null-controllable and does not have an
unstable part, i.e., that all the eigenvalues of A have nonpositive real parts and all the eigenvalues of the uncontrollable part of A have strictly negative real parts. Let _ = _(A). Let oeoe
oe
= (oe
1
; \Delta \Delta \Delta ; oe
_
) be an arbitrary sequence of bounded functions belonging
to S. Then there exists an m-tuple ll
l
= (l
1
; \Delta \Delta \Delta ; l
m
) of nonnegative integers such that
jll
l
j = _, for which there are SISS
L
-stabilizing feedbacks
u = \Gamma k
F
(x) (4.1.1)
u = \Gamma k
G
(x) (4.1.2)
such that k
F
2 F
ll
l
n
(oeoe
oe
), k
G
2 G
ll
l
n
(oeoe
oe
). Furthermore, the linearizations of the closed-loop
systems are asymptotically stable.
Remark 4.1.5 Even if we were only interested in stabilization, and did not care for the SISS
L
property, our inductive proof of Theorem 4.1 would still require that we prove
SISS
L
at each step in order to carry out the induction. So the SISS
L
property is in Pass1: Page 51
51 any case a byproduct of our proof, and this is one of the reasons why we have chosen to include it in the conclusion of our theorem. In addition, the SISS
L
conclusion is
important for at least one other reason, namely that it makes stability a crucial role in our proof for the partially observed case. It is not hard to see that not all stabilizing feedbacks have the SISS property, even if they are linear near the origin. To illustrate this, consider the double integrator: .x = y; .y = u:
Let oe(s) be an odd continuous function such that soe(s) ? 0 for s 6= 0, oe(s) = s for jsj !
1
2
and oe(s) =
1
s
for s ? 1. Then the feedback u = \Gamma oe(x + y) stabilizes the double
integrator. (This can be established by verifying that
V (x; y) =
Z
x+y
0
oe(s) ds +
1
2
y
2
is a Lyapunov function for the closed-loop system with u = \Gamma oe(x + y), and applying the LaSalle Invariance Principle.) Let
e(t) = oe
i
log(t + 1) +
1
t + 1
j
\Gamma
1
(t + 1)
2
:
Then clearly e(t) ! 0 as t ! 1. But not every trajectory of the system
.x = y ;
.y = \Gamma oe(x + y) + e(t)
(4.1.3)
converges to zero. For instance, x(t) = log(t + 1), y(t) =
1
t+1
is a solution of (4.1.3),
but x(t) ! 1 as t ! 1. 2
We will say that (4.1.1), (4.1.2) are feedbacks of Type F , G, respectively. In Sec- tion 4.4, we will give precise procedures to get these feedbacks. These feedback are also illustrated by diagrams.
4.2 Technical lemmas In this section we present two technical lemmas that will be needed for the proof of Theorem 4.1. Pass1: Page 52
52 Lemma 4.2.1 Consider an n-dimensional linear single-input system
\Sigma : .x = Ax + bu : (4.2.1) Suppose that (A; b) is a controllable pair and that all the eigenvalues of A have zero real part. Fix a * ? 0.
(i) If 0 is an eigenvalue of A, then there is a linear change of coordinates T x =
(y
1
; \Delta \Delta \Delta ; y
n
)
0
= (_y
0
; y
n
)
0
of IR
n
that puts \Sigma in the form
. _y = A
1
_y + b
1
(y
n
+ *u) ;
.y
n
= u ;
(4.2.2)
where the pair (A
1
; b
1
) is controllable and y
n
is a scalar variable.
(ii) If A has an eigenvalue of the form i!, with ! ? 0, then there is a linear change
of coordinates T x = (y
1
; \Delta \Delta \Delta ; y
n
)
0
= (_y
0
; y
n\Gamma 1
; y
n
)
0
of IR
n
that puts \Sigma in the form:
. _y = A
1
_y + b
1
(y
n
+ *u) ;
.y
n\Gamma 1
= !y
n
;
.y
n
= \Gamma !y
n\Gamma 1
+ u ;
(4.2.3)
where the pair (A
1
; b
1
) is controllable and y
n\Gamma 1
; y
n
are scalar variables.
Proof. We first prove (i). If 0 is an eigenvalue of A, then there exists a nonzero n- dimensional row vector v such that vA = 0. Let , : IR
n
! IR be the linear function
x ! vx. Then, along trajectories of \Sigma ,
. , = (vb)u. Controllability of (A; b) implies
that vb 6= 0. So we may assume that vb = 1. Make a linear change of coordinates T x = (_z
0
; z
n
)
0
so that z
n
j ,. Then the system equations are of the form
. _z = A
1
_z + z
n
~
b
1
+ u
~
b
2
;
.z
n
= u :
It is clear that every eigenvalue of A
1
also has zero real part. Now change coordinates
again by letting _y = _z + z
n
~
b
3
, y
n
= z
n
, where the vector
~
b
3
will be chosen below. Then
the system equations become
. _y = A
1
_y + y
n
(
~
b
1
\Gamma A
1
~
b
3
) + u(
~
b
2
+
~
b
3
) ;
.y
n
= u : Pass1: Page 53
53 Choose
~
b
3
to be a solution of
~
b
2
+
~
b
3
= *(
~
b
1
\Gamma A
1
~
b
3
), i.e, *A
1
~
b
3
+
~
b
3
= *
~
b
1
\Gamma
~
b
2
. (This
is possible because *A
1
+ I is nonsingular.) Let b
1
= (
~
b
1
\Gamma A
1
~
b
3
). The equations now
become
. _y = A
1
_y + (y
n
+ *u)b
1
, .y
n
= u, as desired.
We now prove (ii). Let i!, ! ? 0, be an eigenvalue of A. Then \Gamma !
2
is an eigenvalue
of A
2
. So there is a nonzero n-dimensional row vector ~v such that ~vA
2
= \Gamma !
2
~v. Let
~w = !
\Gamma 1
~vA. Then ~wA = \Gamma !~v and ~wA
2
= \Gamma !
2
~w. Moreover, ~w cannot be a multiple
of ~v because, if ~w = *~v, then from the fact that ~w = w
\Gamma 1
~vA we see that *! would be
a nonzero real eigenvalue of A. So the linear span S of ~v and ~w is a two-dimensional subspace all whose members v satisfy vA
2
= \Gamma !
2
v. In particular, we can choose v 2 S
such that vb = 0 but v 6= 0. If we then define w by w = !
\Gamma 1
vA, we have wA = \Gamma !v.
Moreover, wb cannot vanish for, if it did, the subspace fx : vx = wx = 0g would contain b and be invariant under A, contradicting controllability. So, after multiplying both v and w by a constant, if necessary, we may assume that wb = 1. Let ,, j be the linear functionals x ! vx, x ! wx. Then, along trajectories of \Sigma ,
. , = !j and .j = \Gamma !, + u.
Make a linear change of coordinates T x = (_z
0
; z
n\Gamma 1
; z
n
)
0
so that z
n\Gamma 1
j ,, z
n
j j. Then
the system equations are of the form
. _z = A
1
_z + z
n\Gamma 1
~
b
1
+ z
n
~
b
2
+ u
~
b
3
;
.z
n\Gamma 1
= !z
n
, .z
n
= \Gamma !z
n\Gamma 1
+ u ;
and every eigenvalue of A
1
has zero real part. Now change coordinates again by letting
_y = _z + z
n\Gamma 1
~
b
4
+ z
n
~
b
5
;
y
n\Gamma 1
= z
n\Gamma 1
;
y
n
= z
n
;
where the vectors
~
b
4
,
~
b
5
will be chosen below. Then the system equations become
. _y = A
1
_y + y
n\Gamma 1
(
~
b
1
\Gamma A
1
~
b
4
\Gamma !
~
b
5
) + y
n
(
~
b
2
\Gamma A
1
~
b
5
+ !
~
b
4
) + u(
~
b
3
+
~
b
5
);
.y
n\Gamma 1
= !y
n
;
.y
n
= \Gamma !y
n\Gamma 1
+ u:
(4.2.4)
If we could choose
~
b
4
,
~
b
5
such that
~
b
1
\Gamma A
1
~
b
4
\Gamma !
~
b
5
= 0 (4.2.5) Pass1: Page 54
54 and
~
b
3
+
~
b
5
= *(
~
b
2
\Gamma A
1
~
b
5
+ !
~
b
4
); (4.2.6)
then we could let
b
1
=
~
b
2
\Gamma A
1
~
b
5
+ !
~
b
4
(4.2.7)
and the system equations would become
. _y = A
1
_y + (y
n
+ *u)b
1
;
.y
n\Gamma 1
= !y
n
;
.y
n
= \Gamma !y
n\Gamma 1
+ u ;
as desired. To prove the existence of
~
b
4
and
~
b
5
, we rewrite (4.2.6) as (*A
1
+ I)
~
b
5
=
*
~
b
2
\Gamma
~
b
3
+ *!
~
b
4
, multiply both sides by !, and plug in the value of !
~
b
5
given by (4.2.5),
namely, !
~
b
5
=
~
b
1
\Gamma A
1
~
b
4
. We end up with the equation
(*A
2
1
+ A
1
+ *!
2
I)
~
b
4
= *A
1
~
b
1
+
~
b
1
\Gamma *!
~
b
2
+ !
~
b
3
:
Since the eigenvalues of A
1
have zero real part, the matrix *A
2
1
+A
1
+*!
2
is nonsingular,
so
~
b
4
exists. 2
Lemma 4.2.2 Let ! ? 0. Then for every oe 2 S there exist functions " ! ffi
0
("); ffi
1
("); ffi
2
(") from (0; 1) to (0; 1) such that
(I) whenever e
0
; e
1
; e
2
are bounded measurable real-valued functions on [0; 1) such that
lim sup
t!+1
je
i
(t)j ! ffi
i
(") for i = 0; 1; 2 ; (4.2.8)
then, if fl = (x(\Delta ); y(\Delta )) : [0; 1) ! IR
2
is any solution of the system
.x = !y + e
1
(t) ;
.y = \Gamma !x \Gamma oe(y \Gamma e
0
(t)) + e
2
(t) ;
(4.2.9)
it follows that
lim sup
t!+1
jjfl(t)jj ! " ; (4.2.10)
and Pass1: Page 55
55 (II) there exist an _" ? 0 and *
0
; *
1
; *
2
? 0 such that ffi
i
(") =
"
*
i
for i = 0; 1; 2 and
0 ! " ^ _".
Proof. We first observe that it suffices to find ffi
0
; ffi
1
; ffi
2
such that
(I) Inequality (4.2.10) holds whenever fl : [0; 1) ! IR
2
is a solution of (4.2.9) for some
triple of functions e
0
; e
1
; e
2
such that jje
i
jj
L
1
! ffi
i
for i = 0; 1; 2.
Indeed, if (I) holds, and e
0
; e
1
; e
2
are bounded measurable and satisfy (4.2.8), then there
is a T * 0 such that the restrictions ~e
0
; ~e
1
; ~e
2
of e
0
; e
1
; e
2
to [T; 1) satisfy jj~e
i
jj
L
1
! ffi
i
.
Applying (I) to the restriction ~fl of fl to [T; 1) yields the desired conclusion.
The hypotheses on oe imply that joej is bounded away from 0 on the complement of every interval (\Gamma `; `), ` ? 0. Define
'(`) = inffjoe(s)j : jsj * `g ; (4.2.11) (`) = supfjoe(s)j : jsj ^ 3`g : (4.2.12) Then (`) * '(`) ? 0. If we let
^ = oe
0
(0) ; (4.2.13)
then
lim `!0+
'(`)
`
= ^ (4.2.14)
and
lim `!0+
(`)
`
= 3^ : (4.2.15)
For each triple (ffi
0
; ffi
1
; ffi
2
) with ffi
i
? 0, let \Gamma (ffi
0
; ffi
1
; ffi
2
) denote the set of all fl :
[0; 1) ! IR
2
that are solutions of (4.2.9) for some triple (e
0
; e
1
; e
2
) of functions from
[0; 1) to IR such that jje
i
jj
L
1
! ffi
i
for i = 0; 1; 2.
Now pick ff ? 0, fi ? 0, and define the functions
V
1;ff;fi
(x; y) = (x + ff)
2
+ (y \Gamma fi)
2
; (4.2.16)
V
2;ff;fi
(x; y) = (x + ff)
2
+ (y + fi)
2
: (4.2.17) Pass1: Page 56
56 Let us compute the derivatives
. V
1;ff;fi
,
. V
2;ff;fi
of V
1;ff;fi
, V
2;ff;fi
along a trajectory fl that
is a solution of (4.2.9) for a triple e
0
; e
1
; e
2
. We get, if fl(t) = (x(t); y(t)):
. V
1;ff;fi
= 2(x + ff)(!y + e
1
) + 2(y \Gamma fi)(\Gamma !x \Gamma oe(y \Gamma e
0
) + e
2
) ; (4.2.18)
. V
2;ff;fi
= 2(x + ff)(!y + e
1
) + 2(y + fi)(\Gamma !x \Gamma oe(y \Gamma e
0
) + e
2
) : (4.2.19)
Then
. V
1;ff;fi
= 2x(e
1
+ !fi) \Gamma 2y(oe(y \Gamma e
0
) \Gamma e
2
\Gamma ff!) + 2ffe
1
+ 2fi(oe(y \Gamma e
0
) \Gamma e
2
) (4.2.20)
and
. V
2;ff;fi
= 2x(e
1
\Gamma !fi) \Gamma 2y(oe(y \Gamma e
0
) \Gamma e
2
\Gamma ff!) + 2ffe
1
\Gamma 2fi(oe(y \Gamma e
0
) \Gamma e
2
) (4.2.21)
Now assume that we can pick positive numbers ff; fi; `; ffi
0
; ffi
1
; ffi
2
; A; B; C such that
ffi
0
^ ` ; (4.2.22)
ffi
1
^ 2` ; (4.2.23)
ffi
1
^ !fi ; (4.2.24)
Affi
2
+ ff! ^ '(`) ; (4.2.25)
fi ! 2` ; (4.2.26) ff(ffi
1
+ fi!) ! (2` \Gamma fi)Bffi
2
; (4.2.27)
ff ! Cffi
2
; (4.2.28)
and
A \Gamma 1 * maxfB; Cg : (4.2.29)
Assume that fl 2 \Gamma (ffi
0
; ffi
1
; ffi
2
), i.e., let e
0
; e
1
; e
2
be such that fl is a solution of (4.2.9)
and jje
i
jj
L
1
! ffi
i
for i = 0; 1; 2. Then the coefficient e
1
+ !fi is positive, and e
1
\Gamma !fi is
negative. So we get the a.e. bounds
. V
1;ff;fi
^ 2ffffi
1
+ 2fi(oe(y \Gamma e
0
) \Gamma e
2
) \Gamma 2y(oe(y \Gamma e
0
) \Gamma e
2
\Gamma ff!) ; (4.2.30)
. V
2;ff;fi
^ 2ffffi
1
\Gamma 2fi(oe(y \Gamma e
0
) \Gamma e
2
) \Gamma 2y(oe(y \Gamma e
0
) \Gamma e
2
\Gamma ff!) ; (4.2.31) Pass1: Page 57
57 valid, respectively, when x ^ 0 and x * 0. If, in addition, y * 2`, then y \Gamma e
0
* `, so
oe(y \Gamma e
0
) ? 0. Moreover, since jy \Gamma e
0
j * `, we have oe(y \Gamma e
0
) * '(`). Then
oe(y \Gamma e
0
) \Gamma e
2
\Gamma ff! * '(`) \Gamma ffi
2
\Gamma ff! * (A \Gamma 1)ffi
2
: (4.2.32)
In particular oe(y \Gamma e
0
) \Gamma e
2
? 0, and then the bound (4.2.31) for
. V
2;ff;fi
becomes
. V
2;ff;fi
! 2ffffi
1
\Gamma 4`(A \Gamma 1)ffi
2
; (4.2.33)
provided, of course, that x * 0, y * 2`.
To get a bound for
. V
1;ff;fi
, we rewrite (4.2.30) as
. V
1;ff;fi
^ 2ffffi
1
+ 2fiff! \Gamma 2(y \Gamma fi)(oe(y \Gamma e
0
) \Gamma e
2
\Gamma ff!) : (4.2.34)
Then from (4.2.26) and (4.2.32) we have
. V
1;ff;fi
^ 2ffffi
1
+ 2fiff! \Gamma (4` \Gamma 2fi)(A \Gamma 1)ffi
2
: (4.2.35)
In view of (4.2.28) and (4.2.23), Inequality (4.2.33) implies that
. V
2;ff;fi
! 0 a.e.
provided that x * 0, y * 2`. In view of (4.2.27), it follows from (4.2.35) that
. V
1;ff;fi
! 0
a.e. So we have proved:
(II) if ff; fi; `; ffi
0
; ffi
1
; ffi
2
; A; B; C are positive numbers such that (4.2.22)-(4.2.29) hold,
then V
1;ff;fi
is strictly decreasing along every trajectory fl 2 \Gamma (ffi
0
; ffi
1
; ffi
2
) which is
contained in the region
R
1
(`) = f(x; y) : x ^ 0; y * 2`g ; (4.2.36)
and V
2;ff;fi
is strictly decreasing along every fl 2 \Gamma (ffi
0
; ffi
1
; ffi
2
) which is contained in
R
2
(`) = f(x; y) : x * 0; y * 2`g : (4.2.37)
We now fix ff; fi; `; ffi
0
; ffi
1
; ffi
2
; A; B; C that satisfy (4.2.22)-(4.2.29), and define, for
each L ? 2ff, a closed contour C(L) as follows (see Figure 4.1). We let P
0
(L) be the
point (\Gamma L; 2`). Let
~ C
1
(L) be the circle through P
0
(L) with center (\Gamma ff; fi). Then define
C
1
(L) to be the arc of
~ C
1
(L) obtained by moving clockwise along
~ C
1
(L) from P
0
(L) to Pass1: Page 58
58
(-a,-b)
(a,b) (a,-b)
\Lambda
2q 2q
O x
y
L~ P (L)5
4P (L)
3P (L) 2P (L) 1P (L) 0P (L)
C (L)1
2C (L)
3C (L)
4C (L) 5C (L)
6C (L)
(-a,b) L
Figure 4.1: Contour C(L) the point P
1
(L) where
~ C
1
(L) intersects the half-line f(x; y) : x = 0; y * 2`g. Then let
~ C
2
(L) be the circle through P
1
(L) with center (\Gamma ff; \Gamma fi), and define C
2
(L) to be the
arc of
~ C
2
(L) obtained by moving clockwise along
~ C
2
(L) from P
1
(L) to the point P
2
(L)
where
~ C
2
(L) intersects the half-line f(x; y) : x * 0; y = 2`g. Then let C
3
(L) be the
straight-line segment from P
2
(L) to P
3
(L), where P
3
(L) = (L; \Gamma 2`). The concatenation
C
+
(L) = C
1
(L)[C
2
(L)[C
3
(L) is an arc from P
0
(L) to P
3
(L). By reflecting with respect
to the origin, we get another arc C
\Gamma
(L) = C
4
(L) [ C
5
(L) [ C
6
(L) from P
3
(L) to P
0
(L).
The concatenation C
+
(L)[C
\Gamma
(L) is our closed contour C(L). We let K(L) be the closed
region bounded by C(L), so K(L) is compact.
We now let \Gamma
c
(ffi
0
; ffi
1
; ffi
2
) be the set of those solutions fl of (4.2.9) that correspond
to continuous functions e
i
: [0; 1) ! IR such that jje
i
jj
L
1
! ffi
i
for i = 0; 1; 2. If
fl 2 \Gamma
c
(ffi
0
; ffi
1
; ffi
2
) then .fl(t) exists for all t. If fl(t) 2 C
1
(L) then the derivative at
s = t of the function s ! V
1;ff;fi
(fl(s)) is ! 0, so V
1;ff;fi
(fl(s)) ! V
1;ff;fi
(fl(t)) for all
s ? t sufficiently close to t. This implies that oe(s) 2 Int (K(L)) for s ? t, s near t, except possibly when fl(t) = P
0
(L) or fl(t) = P
1
(L). A similar conclusion follows if Pass1: Page 59
59 fl(t) 2 C
2
(L), using the fact that
d ds
fi fi fi fi
s=t
i
V
2;ff;fi
(fl(s))
j
! 0 ; (4.2.38)
provided that fl(t) 6= P
1
(L) and fl(t) 6= P
2
(L).
If fl(t) = P
1
(L) then the inequalities V
i;ff;fi
(fl(s)) ! V
i;ff;fi
(fl(t)) hold for s ? t, s near
t, both for i = 1 and i = 2. This means that fl(s) belongs to Int(K(L)) for s ? t, s near t, whether it lies in R
1
(`) or in R
2
(`). (Since fl(t) = P
1
(L), then fl(s) must belong to
R
1
(`) [ R
2
(`) for s near t.) Therefore we have shown:
(III) if fl 2 \Gamma
c
(ffi
0
; ffi
1
; ffi
2
) and fl(t) 2
i
C
1
(L) [ C
2
(L)
j
\Gamma fP
0
(L); P
2
(L)g, then fl(s) 2
Int(K(L)) for s ? t, s near t.
It is clear by symmetry that a similar conclusion holds if fl(t) 2
i
C
4
(L) [ C
5
(L)
j
\Gamma
fP
3
(L); P
5
(L)g. We now study the missing case, namely, when fl(t) 2 C
3
(L) [ C
6
(L).
First we need more information about P
2
(L). Write
P
2
(L) = (
~ L; 2`) ; P
1
(L) = (0; \Lambda + 2`) : (4.2.39)
Then
(L \Gamma ff)
2
+ (2` \Gamma fi)
2
= (\Lambda + 2` \Gamma fi)
2
+ ff
2
(4.2.40)
and
(\Lambda + 2` + fi)
2
+ ff
2
= (
~ L + ff)
2
+ (2` + fi)
2
: (4.2.41)
Therefore
L
2
\Gamma 2ffL = \Lambda
2
+ 2(2` \Gamma fi)\Lambda (4.2.42)
and
\Lambda
2
+ 2(2` + fi)\Lambda =
~ L
2
+ 2ff
~ L : (4.2.43)
From (4.2.42) and (4.2.26) we get \Lambda ! L. Then (4.2.42) and (4.2.43) imply:
~ L
2
+ 2ff
~ L = L
2
\Gamma 2ffL + 4fi\Lambda ; (4.2.44)
so that
L
2
\Gamma
~ L
2
= 2ff(
~ L + L) \Gamma 4fi\Lambda (4.2.45) Pass1: Page 60
60 and
L \Gamma
~ L = 2ff \Gamma
4fi\Lambda
L +
~ L
: (4.2.46)
If we now impose the further requirement that
fi ^
ff
4
; (4.2.47)
we can conclude that
4fi\Lambda
L +
~ L
! ff ; (4.2.48)
and then
L \Gamma 2ff ^
~ L ^ L \Gamma ff : (4.2.49)
This implies in particular that the x-coordinate of P
2
(L) is smaller than L, so the
segment C
3
(L) has a negative slope, whose absolute value , satisfies
, =
4`
L \Gamma
~ L
(4.2.50)
so that
2`
ff
^ , ^
4`
ff
: (4.2.51)
The segment C
3
(L) is a level curve of the function V
3;,
, given by
V
3;,
(x; y) = ,x + y : (4.2.52)
The derivative of V
3;,
along trajectories of (4.2.9) is
. V
3;,
= ,!y + ,e
1
\Gamma !x \Gamma oe(y \Gamma e
0
) + e
2
: (4.2.53)
If we let R
3
(`) denote the strip f(x; y) : jyj ^ 2`g, then jy \Gamma e
0
j ^ 3` on R
3
(`), if
jje
0
jj
L
1
! ffi
0
. Therefore
. V
3;,
^ 2,!` + ,ffi
1
+ (`) + ffi
2
\Gamma !x (4.2.54)
along any fl 2 \Gamma
c
(ffi
0
; ffi
1
; ffi
2
), at every time t such that fl(t) 2 R
3
(`). If in addition
fl(t) 2 C
3
(L), then x *
~ L, and we get
. V
3;,
^ 2,!` + ,ffi
1
+ ffi
2
+ (`) \Gamma !
~ L : (4.2.55) Pass1: Page 61
61 Since
~ L * L \Gamma 2ff, and , ^
4`
ff
, we can conclude that
. V
3;,
^
8`
2
!
ff
+
4`ffi
1
ff
+ 2ff! + ffi
2
+ (`) \Gamma !L : (4.2.56)
Let L
crit
(the "critical" L) be defined by
L
crit
=
8`
2
ff
+
4`ffi
1
ff!
+ 2ff +
ffi
2
!
+
(`)
!
: (4.2.57)
Then, if L ? L
crit
, we have
. V
3;,
! 0 : (4.2.58)
So we have shown that, if L ? L
crit
, then
d ds
fi fi fi fi
s=t
i
V
3;,
(fl(s))
j
! 0 (4.2.59)
if fl 2 \Gamma
c
(ffi
0
; ffi
1
; ffi
2
), fl(t) 2 C
3
(L). This implies that fl(s) 2 Int(K(L)) for s ? t, s near
t, except possibly if fl(t) = P
2
(L) or fl(t) = P
3
(L). If fl(t) = P
2
(L) then both functions
s ! V
2;ff;fi
(fl(s)) and s ! V
3;,
(fl(s)) have a negative derivative at s = t, so fl(s) lies
to left of C
3
(L) for s ? t, s near t, if fl(s) 2 R
3
(`), and lies inside the circle
~ C
2
(L) if
fl(s) 2 R
2
(`). In either case, fl(s) 2 Int(K(L)) for s ? t, s near t. A similar argument
works for fl(t) = P
3
(L). By symmetry, the same conclusion holds if fl(t) 2 C
6
(L). So
we have proved:
(IV) assume that ff; fi; `; ffi
0
; ffi
1
; ffi
2
; A; B; C are pos