Proceedings of the American Control Conference
San Diego, California
and $ ( t , $ ( h , x ) ) = $(t + h , z ) for all t , h 2 0 and x E
V. Given t > 0 and x E 2, it will often be convenient
to denote the map $(t,.) : V + V by $ t , and the map
x ) : [0,CO) + V by $".
$(e,
$t(U{= U for all t 2 0. The orbit 0"of a point x E V is
the set $"([O, 00)). The orbit of x is bounded if 0"is contained in a compact set. The positive limit set of x is the
set 0% of points z E 2) such that there exists a sequence
ti -+ 00 satisfying $(ti,%) + z. If Q c V is positively
invariant and closed, then it follows that 0% C 0 for all
x E 9. Also, limt+, $ ( t , x ) exists if and only if O& contains exactly one point.
A function V : Q + B is said t o be proper relative t o
Q 2)if V-'(K:) is a relatively compact subset of 0 for all
compact subsets K: of R
Given a continuous function V : Q + B defined on the
positively invariant set Q C V ,we define
1
V ( 4 = hlim
-[V($(t
- t ~h
+ h , x ) )- V($@,x))I,
(2)
whenever the limit on the right hand side exists. For completeness, we state the following well known criterion for
boundedness of the orbits of (1).
Proposition 2.1. Let Q C V be positively invariant
under (1). If there exists a continuous function V : Q + B
that is proper relative to Q and such that V is defined on
Q, V ( x ) 2 0 and V ( z )5 0 for all z E Q, then every orbit
in Q is bounded.
An equilibrium point of (1) is a point x E V satisfying
f ( z ) = 0 or, equivalently, $ ( t , x ) = x for all t 2 0. An
equilibrium point x E 8 G V is Lyapunov stable with respect to the positively invariant set Q if, for every relatively
open subset U, of B containing z , there exists a relatively
open subset U6 of 0 containing x such that $t(u6) U,for
all t 2 0.
The following lemmas will prove useful in the subsequent development.
0,00) + Bn be a continuLemma 2.1.
Let*
ously differentiable function. i is absolutely integrable on
[0,CO), then limt,,
y ( t ) exists. Conversely, if limt,,
y(t)
and limt,,
( l / ~ ~ y ( t$(t)
) ~ ~exist
)
and c(t)# 0 for all t 2 0,
then y is absolutely integrable.
Proof. First we note that limt+,y(t) exists if and
only if for every E > 0 there exists T,> 0 such that for all
h 2 0 and t > T,, Ilv(t h) - y(t)II < E For every t , h 2 0, we have
t+h
t+h
t
t
J lb(411dT - 6 J
2 1.
IIY(.>lldT
(6)
t+h
J 113i(.)ldT.
= (1 - 6)
(7)
t
Since limt,,
y ( t ) exists, equation (7) implies that p is ab0
solutely integrable.
Lemma 2.2.
Let Q G 2) be positively invariant
under (1) and let z E Q. If a point z E U& n Q is a
Lyapunov stable equilibrium point with respect to Q, then
z = limt,,
$ ( t , x ) and 02 = { z } .
Proof. Let U,be an open neighborhood of z. By Lyapunov stability of z, there exists a relatively open subset
U6 C 0 containing z such that $t(&)
U, for d l t >_ 0.
Since z E 0", there exists h 2 0 such that $(h, z ) E Ua.
Therefore, $Tt
h , x ) = $t($(h,x)) E qt(U6)
U, for
$(t,x ) . It immediately folall t > 0. Hence z = limt,,
lows that $ ( t i , x ) + z for all sequences ti + 00 and thus
0
: = {z}.
0
+
3.
Convergence and Semistability
In the remainder of the paper, we assume that 0 C 2)
is a closed set that is positively invariant under (1). The
$(t,x )
system (1) is Convergent with respect to Q if limt,,
exists for every x E Q.
If the system (1) is convergent with respect to 0, then
the dynamics give rise t o a function R : Q -+ 9 given by
n ( x ) = limt,,
$(t,x ) , x E B. It follows from continuity
h,x) =
that for every h 2 0, $h(R(z)) = limt+,$(t
R ( z ) . Thus R ( x ) is an equilibrium point for (1) for all
x E B*. Consequently, R o R = R. The following result
links the continuity of the function R at a point x to the
stability of the equilibrium point R ( x ) .
Proposition 3.1.
Suppose (1) is convergent with
respect to Q. If R ( x ) is a Lyapunov stable equilibrium for
some x E Q, then R is continuous a t z.
Proof. Suppose R ( x ) is Lyapunov stable for some x E
0. Let U,be an open neighborhood of R ( x ) . Choose open
C U, and
neighborhoods U and 246 of R ( x ) such that
U for d l t 2 0. Let { x i } be a sequence in Q
$t(&)
converging t o x. There exists h > 0 such that $(h, x ) E &.
Since $(h,xi) + $(h,x) as i + 00, there exists N such
t+h
00
that for all i > N , $(h,xi) E 245. Therefore, $(t h , z i ) E
IlYO + h) - Y(t>II I
113i(~>Ild~
I l l Y ( m T - (3) $t(Us) C U for all t 2 0 and i > N and, consequently,
t
t
?(xi) E c U, for all i > N. Thus R(xi) + R(z) as
0
Suppose y is absolutely integrable. Then, for every E > 0, z + 00 and hence R is continuous at x .
there exists T, > 0 such that the second integral in the
The following result provides a Lyapunov-type test for
above equation is less than e for all t > T,. Hence we convergence.
y ( t ) exists.
conclude that limt,,
Theorem 3.1. Suppose 0"is bounded for all x E Q
i(t)/llP(t)ll and there exists a continuous function V : Q + such that
To prove the second part, let a = limt,,
and note that llall = 1. Let 6 E (0,l) and choose T6 suf- V is defined on Q and such that V ( x ) 2 0 and V ( x ) 5 0 for
ficiently large such that IIa - y ( t ) ( l / [ ~ ~ ( t )<[ ~6) for
~ ~ all all z E Q. Let M be the largest invariant set contained in
t > T6,that is, Ilalls(t)ll - y(t)II < dlllj(t)ll for all t > T6. V-' (0) and suppose that there exists a set U containing M
Then, for every t > T6 and h > 0, we have
that is relatively open in Q and an absolutely continuous,
nondecreasing, real-valued function g defined on a closed
interval containing V ( U )such that
(4)
+
:A
a
+
J
+
J
a
Ilf(x)ll 5 -g'(V(x))Wx)
1609
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(8)
almost everywhere on U. Then the system (1) is convergent
with respect to Q.
Proof. Let z E Q and denote y = $". Under the
hypotheses on V , every bounded solution converges to M .
Thus, there exists T > 0 such that y ( t ) E U for all t 2 T .
Consequently, IIY ( t )II = IIf (Y( t )1II I - 9 ' V (Y0))1V (Y( t )1
for almost every t > T and we have,
bo
00
J 113i(t>lldt I
-Js'(V(y(t)))V(v(t))dt
T
T
= 9(V(Y(T)))
- &$V(dt))).
V(y(t)l
is nonincreasing and bounded below,
Since V ( y ( t) exists. In addition, since g is continuous
limt+oo
on a closed interval containing V(U),it follows from the
above inequality that i is absolutely integrable. Lemma
2.1 now implies that limt+my(t) exists. Thus, (1) is con0
vergent with respect t o Q.
The following theorem provides a partial converse t o
Theorem 3.1.
Theorem 3.2. Suppose that for every connected comhas a continuponent C of Q\f-'(O), the function
ous extension to the closure of C. If (1) is convergent with
respect to 0, then the function V ( z )=
llf($~(t,x))lldt
is defined on Q and satisfies the inequality (8) on Q with
g : R + IR given by g(h) = h.
Proof. Consider x E Q\f-'(O)
and let C be the concontaining x. Then 0"C C
nected component of G\f-'(O)
and hence n ( x ) E E. Let U denote the continuous extension of the function
to ?
. and denote y = $".
Then limt-tw &y(t)
= u ( n ( z ) )exists. If y(T) = 0
for some T > 0, then i ( t ) = 0 for all t > T and y
is absolutely integrable. If $(t) # 0 for all t > 0, then
Lemma 2.1, implies that $ is absolutely integrable. Thus
V ( x )=
Ilf($(t,x))l(dtis defined for all x E Q. V can
be easily computed to be V ( z ) = -Ilf(x)11. The inequality (8) is clearly satisfied with the function g defined by
0
g(h) = h, h E R
Remark 3.1. In Lemma 2.1,
Ily(t)lldt represents
the Euclidean length of the image of the function y in R" .
Likewise, for the system ( l ) , the integral J" Ilf($(t, x))lldt
represents the Euclidean length of the orbit 0".
Thus the
first part of Lemma 2.1 implies that if every orbit in Q
has a finite Euclidean length, then (1) is convergent with
respect to 6. Indeed, the proof of Theorem 3.1 essentially
shows that if the inequality (8) is satisfied, then every orbit
in Q has finite length. Theorem 3.2 states that if the unit
vector field along f can be extended continuously in the
stated fashion, then convergence with respect to Q necessarily implies that every orbit in 0 has finite length.
An equilibrium x E V is semistable with respect t o Q if
there exists a set U containing x that is relatively open in Q
and such that limt_,bo$(t,z)exists for all x E U and every
equilibrium point in U is Lyapunov stable with respect t o Q.
The system (1) is said to be semistable with respect t o Q if
every equilibrium point in Q is semistable with respect to Q.
The following is a Lyapunov-type result for semistability.
Theorem 3.3. Suppose 8" is bounded for all x E Q
and there exists a continuous function V : Q + R such
c
hf(-)
hf(.)
S
F
that V is defined on Q, and V ( x ) 2 0 and V ( z ) 5 0 for
all x E Q. If every point in the largest invariant subset
M of V - ' ( O ) is a Lyapunov stable equilibrium point with
respect to 9, then (1) is semistable with respect to G.
Proof. Since every solution is bounded, it follows from
the hypotheses on V that, for every x E Q , 8 $ is nonempty
and contained in the largest invariant subset M of V - ' ( O ) .
Since every point in M is a Lyapunov stable equilibrium
with respect to Q, it follows from Lemma 2.2 that 82
contains a single point for every x E Q and (1) is convergent. Since n ( x ) E M is Lyapunov stable for every x E 0,
semistability follows.
0
The following example illustrates theorems 3.1 and 3.2.
Example 3.1. Consider the system (1) where f :
R2 + R2 is given by
f(z)= f,(x)sign(z? + xg - 1)1xf + xs - lla
fe(z)sign(zq xg - 1)Ix: xi - 11°,
+
+
+
where a,/3 2 1 and the vector fields f r and fe are given by
f,.(~) = [-xi - z Z ] ~ and fe(2) = [ 2 2 The vector fields f r and fe point in the radial and tangential directions, respectively, and thus the parameters a
and 0 decide the relative rates at which solutions move in
the radial and tangential directions. This can be seen more
clearly by rewriting equations (1) and (9) in terms of polar
coordinates r = ,/and 8 = tan-'(y/z) as
+
= -rsign(r2
6
= -sign(r2 - 1)lr2 - 110.
- 1)lr2 - lla,
(9)
(10)
It can be easily seen from equations (9) and (10) that
the set of equilibria f-'(O) consists of the origin x = 0 and
the unit circle S' = {x E R2 : x: xi = 1). All solutions
of (1) starting from nonzero initial conditions y(0) # 0 approach the unit circle. Solutions starting outside the unit
circle spiral in clockwise towards the unit circle while solutions starting inside the unit circle spiral out counterclockwise. Consequently, all solutions are bounded.
First assume a 2 /J 1. In this case, the system is not
convergent with respect to Q = R2. This can be seen by
using equations (9) and (10) to compute
+
+
It can be seen that the solution r ( - )of equation (11) can
approach r = 1 only as 0 + 00 if r(0) < 1 and as 8 + -m
if r(0) > 1. This implies that for a - /3 2 1, the solutions
of (9) - (10) that converge t o the unit circle spiral around
an infinite number of times, thus ruling out convergence.
Figure 1 shows the phaseportrait of this system for a =
2, p = 1.
Next, consider the case a - p < 1 and the Lya- l)2.With a slight
punov function V ( x )=
abuse of notation, we write V ( z ) = $(r - 1)2 and compute the derivative of V along the solutions of (9)-(10) as
V ( x ) = -r(r+ l ) a l r - lla+', which is seen to take nonpositive values everywhere. Moreover, the set V - ' ( O ) consists
solely of equilibrium points and is thus invariant. We also
compute
f(4-
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If a 5 p, then
Thus, for all z E { (T cose, T sine) : 0 5 T < fi,T # l},
Ilf(z)II I -V(z)g'(V(z)), where g : [O,oo) + R is given by
g(h) = 2 f i .
If p < a < p 1, then
+
Figure 2: Phase portrait of (9) for a = /3 = 1
Next, we apply Theorm 3.1 to a chemical reaction network.
Example 3.2. Consider two species z1 and 2 2 whose
chemical interactions are described by the reactions
z]
Again, for all z E ((rcos8,rsinO) : 0 5 r e fi, r # l},
Ilf(z)ll I -V(z)g'(V(z)), where g : [0,CO) + R is given by
g(h) = &!(2h)(p+'-Q)/2/(p
1 - a). Hence, in both the
cases a 5 p and < a < p 1, Theorem 3.1 implies that
the system is convergent with respect to 9 = R2. Figure 2
shows the phase portrait for this system for a = /3 = 1.
+
+
+
3-22
1
9,
221,
(14)
2x1
z1 +z2,
(15)
where kl and k2 are the reaction rate constants of the respective reactions. Assuming that the dynamics of the reactions are governed by mass action kinetics, the dynamics
are modelled by the kinetic equations [3]
=
(16)
X2
= -kl~l~2+k2~:,
(17)
where we have, with a slight abuse of notation, used z1 and
2 2 to denote the concentrations of the respective species.
The nonnegative quadrant 9 = {z E R2 : z1 2 0,xz 2
0) is known t o be positively invariant under the equations
(16)-(17) [3], which is consistent with the physical fact that
the concentrations of the species cannot become negative
as the reactions proceed.
The function V(z) = z1 +z2 is easily seen to be proper
relative t o 9, since, for every U 2 0, V-l([O,u]) n 9 is a
closed subset of the compact set {z E 9 : z1 5 u,z2 5 U}.
Also,
= 0 for all z E 9 so that, by Proposition 2.1,
every orbit in 9 is bounded.
To show convergence, define V(z) = i ( k l z 2 - k 2 2 1 ) ~
and compute V ( z ) = -(k1
k2)zl(k122 - k 2 ~ 1 ) ~Also,
.
Ilf(z)I]= f i ~ i l k l ~ 2k 2 ~ 1 ) .Thus, V-'(O) = f-'(O) SO
that V-'(O) is invariant. Moreover,
Xi
k 1 2 1 ~ 2- k 2 ~ : ,
o(z)
+
Figure 1: Phase portrait of (9) for a = 2, /3 = 1
To illustrate Theorem 3.2, we let 9 = { (T cos 0, T sin e) :
T
2 f} and note that the closed set 9 is positively in-
are
variant. The two connected components of G\f-'(O)
= {z : 11z11 > 1) and C2 = {z : f 5 1 1 ~ 1 1< 1). The
= &f(z)
is defined everywhere
unit vector field U(.)
on 9 except on the unit circle. Moreover, the vector field U
points clockwise outside the unit circle and counterclockwise inside, and hence does not have a continuous extension
to all of 9. However, U can be continuously extended t o
C1 and c
2 separately. Thus, Theorem 3.2 implies that for
a < /3 1, Jooo Ilf(+(t1z))11dtexists for all z E 9, that is,
every orbit in 9 has finite length, while, for a 2 /3 1,
every nontrivial orbit is of infinite length. Figures 1 and 2
also reflect this fact.
C1
+
+
so that inequality (8) is satisfied almost everywhere on 9
with g : [O,oo) + P defined by g(h) = 2 a (kl k2).
Hence, by Theorem (3.1), the system (16) - (17 is convergent with respect to 9. In other words, starting from any
arbitrary initial conditions, the concentrations z1 and 2 2
approach steady state values.
I
4.
+
Semistability of Linear Systems
In this section, we apply the results of the previous
section t o study the semistability of the linear system
lxt) = AY(t)l
(18)
1611
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where A E Rnx" is an n x n real matrix. Equation (18) is
a special case of the system considered in Section 1, with
2> = R", f ( 2 ) = Ax and $ ( t , a:) = eAtx, where eAt is the
matrix exponential of At. The set of equilibria f-'(O) of
(18) is N ( A ) ,the nullspace of A.
Since $ ( t , x ) = eAtx, it is clear that the system (18) is
semistable if and only if limt+oo eAt exists. It can now be
deduced from the structure of the matrix exponential that
the system (18) is semistable with respect to 0 = R" if and
only if it is convergent with respect to R" if and only if A
has all its eigenvalues in the open left half complex plane
with the possible exception of a semisimple eigenvalue at
0. We call such a matrix semistable. It can be seen that
if A is semistable, then the index of A is zero or one, and
thus A is group invertible. The group inverse A# of A is a
special case of the Drazin inverse AD in the case in which
A has index zero or one [4].
Using Theorem 3.3, we derive a sufficient condition for
the semistability of (18) in terms of the matrix Lyapunov
equation. In the following result, P 2 0 denotes a symmetric nonnegative definite matrix P , while in the sequel,
R ( A ) denotes the range of the matrix A.
Proposition 4.1. Suppose there exist matrices P
0 and Q 2 0 in RnXnsuch that
>
ATP
+P A = -Q,
1
NI
(19)
=N(A).
Then (18) is semistable with respect t o R".
The proof of the above result requires the following
lemma.
Lemma 4.1. If there exist matrices P 0 and Q 2 0
in RnXnsuch that equations (19 and (20) hold, then
i ) N ( P ) C_ n / ( A ) N ( Q ) ,ii)
( A )n R ( A ) = (0).
Proof. i) If Ax = 0, then, equation (19) implies
xTQx = x T ( A T P PA)x = 0, so that by the nonnegative definiteness of Q , &a: = 0. Thus N ( A ) G N ( Q ) . It
can be similarly shown that N ( P ) N ( Q ) .
Now consider x E N ( P ) N ( Q ) . If Akx E N ( P ) G
N ( Q ) for some k 2 0, then 0 = -QAkx = ( A T P +
P A ) A k x = PAk+lx, so that Ak+'x E N ( P ) E N ( Q ) .
Since Akx E N ( P ) N ( Q ) for k = 0, it follows by induction that x is contained in the nullspace on the left
hand side of equation (20). Equation (20) now implies that
x E N ( A ) . Thus N ( P ) 5 N ( A ) C N(&).
ii) Consider x E N ( A ) n R ( A ) . Then A z = 0 and
there exists z E R" such that x = A z . From the first
part of the lemma, we have Qx = QAz = 0 . Therefore,
0 = -zTQx = z T ( A T P P A ) x = xTPa:, so that Px = 0.
Finally, zTQz = zT(ATP P A ) z = x T P z + z T P x = 0 ,
so that Q z = 0. This implies that z is contained in the
nullspace on the left hand side of equation (20). Hence, by
0
equation (20),Az = x = 0 as required.
Proof of Proposition 4.1. Since, by Lemma 4.1,
N ( A )n R ( A ) = { 0 } , it follows from [l, p. 1191 that A is
group invertible. Thus let R = I - AA# which is idempotent, but not necessarily symmetric. Then R is the unique
matrix in Rnxnsatisfyin N ( R ) = R ( A ) , R ( R ) = N ( A )
and Rz = x for all x E 4 A ) .
J
+
+
+
>
Consider the function V ( x ) = xTPx + xTRTRx. If
V ( x ) = 0 for some x E R", then Pa: = 0 and Rx = 0.
Lemma 4.1 i) implies that x E N ( A ) ,while Rx = 0 implies
x E R ( A ) . Lemma 4.1 ii) now implies that a: = 0. Thus
the function V is a positive definite quadratic function and
hence proper relative to Rn.We compute V ( x ) = -xTQx+
2xTRTRAx = -xTQx, and note that V - ' ( O ) = N ( Q ) .
Since V ( x )_< 0 for all x E R" , Proposition 2.1 implies that
every orbit of (18) is bounded.
To find the largest invariant subset M of N ( Q ) , consider a solution y of (1) such that Q y ( t ) = 0 for all t 0.
Taking successive derivatives yields Q A k y ( t ) = 0 for all
t 2 0 and k = 1,2,. ... Equation (20) now implies that
y t ) E N ( A ) for all t 2 0. Thus M
N ( A ) . However,
J ( A ) consists of only equilibrium points and is itself invariant. Therefore, M = N ( A ) .
Now, let x , E N ( A ) be an equilibrium point of (18) and
consider the function V ( x ) = V ( x - x , ) , which is positive
definite about the point a:,. Since Ax, = 0 , it follows that
U ( x )= V ( x - x , ) , which is negative semidefinite about 2,.
Hence it follows that x , is Lyapunov stable. Theorem 3.3
now implies that the system (18) is semistable.
0
>
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