A Converse Sum of Squares Lyapunov Result with a Degree Bound

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by  Matthew M. Peet

1
A Converse Sum of Squares Lyapunov Result with
a Degree Bound
Matthew M. Peet and Antonis Papachristodoulou
Abstract—Sumof Squaresprogramminghasbeenusedexten- references on early work on optimization of polynomials,
sively over thepast decadeforthestabilityanalysis of nonlinear see[2],[3],and[4].Formorerecentworksee[5]and[6].For
systems butseveral questionsremain unanswered.In thispaper,
a recent review paper, see [7]. Today, there exist a number of
we show that exponential stability of a polynomial vector field
softwarepackagesforoptimizationoverpositivepolynomials,
on a bounded set implies the existence of a Lyapunov function
which is a sum-of-squares of polynomials. In particular, the e.g. SOSTOOLS [8] and GloptiPoly [9].
main result states that if a system is exponentially stable on At the same time, there are still a number of unanswered
a bounded nonempty set, then there exists an SOS Lyapunov questions regarding the use of sum of squares as a relaxation
function which is exponentially decreasing on that bounded set. to nonnegativity and its use for the analysis of nonlinear
TheproofisconstructiveandusesthePicarditeration.Abound
systems. Unanswered questions include, for example, a series
on the degree of this converse Lyapunov function is also given.
of questions on controller synthesis and the role of duality
This result implies that semidefinite programming can be used
to answer the question of stability of a polynomial vector field to convexify this problem, as well as estimating regions of
with a bound on complexity. attraction of equilibria. On the computation and optimization
side,itisunclearwhethermulti-corecomputingcouldbeused
for computation,as well as how to take advantage of sparsity
I. INTRODUCTION
in semidefinite programming.
Computational and numerical algorithms are extensively
Inthispaper,we donotconsidertheproblemofcomputing
used in control theory. A particular example is semidefinite
sum-of-squaresLyapunov functions. Such work can be found
programming conditions for addressing linear control prob-
in,e.g.[4],[10]–[12].Instead,weconcentrateontheproperties
lems, which are formulated as Linear Matrix Inequalities
of the converse Lyapunov functions for systems of the form
(LMIs).Usingsuchtools,severalquestionsontheanalysisand
synthesis of linear systems can be formulated and addressed x˙(t)= f(x(t)),
effectively. In fact, ever since the 1990s [1], LMIs have had
where f :Rn →Rn is polynomial. In particular, we address
a significant impactin the controlfield, to the point that once
the question of whether an exponentially stable nonlinear
the solution of a control problem has been formulated as the
system will have a sum-of-squares Lyapunov function which
solution to an LMI, it is considered solved.
establishes this property. This result adds to our previous
When it comes to nonlinear and infinite-dimensional sys-
work [13], where we were able to show that exponential sta-
tems,theequivalentproblemscanbeformulatedaspolynomial
bilityonaboundedsetimpliestheexistenceofaexponentially
non-negativity constraints under a Lyapunov framework, but
decreasing polynomial Lyapunov function on that set.
these are not, at first glance, as easy to solve. Polynomial
Work that is relevant to the one presented here includes
non-negativity is in fact NP hard. It is for this reason that
research on continuity properties, see e.g. [14], [15] and [16]
several researchers have looked at alternative tests for non-
and the overview in [17]. Infinitely-differentiable functions
negativity,thatarepolynomial-timecomplextotest,andwhich
were exploredin the work [18], [19].Other innovativeresults
imply non-negativity. One such relaxation is the existence
are found in [20] and [21]. The books [22] and [23] treat
of a sum of squares decomposition: the ability to optimize
further converse theorems of Lyapunov. Continuity of Lya-
over the set of positive polynomialsusing the sum-of-squares
punov functions is inherited from continuity of the solution
relaxationhasundoubtedlyopenedupnewwaysforaddressing
map with respect to initial condition. An excellent treatment
nonlinear control problems, in much the same way Linear
of this problem can be found in the text of Arnol’d [24].
Matrix Inequalities are used to address analysis questions
Unliketheworkin[13],thispaperiscloselytiedtosystems
for linear finite-dimensional systems. However, there remain
theory as opposed to approximation theory. Our method is to
several open questions about how these methods can be used
take a well-knownformof converseLyapunovfunctionbased
to search for Lyapunov functions for nonlinear systems. For
onthesolutionmapandusethePicarditerationtoapproximate
M.M.PeetiswiththeDepartmentofMechanical,Materials,andAerospace the solution map. The advantage of this approach is that if
Engineering, Illinois Institute ofTechnology, 10West32ndStreet, E1-252B, the vector field is polynomial, the Picard iteration will also
Chicago, IL60616,[email protected]
be polynomial. Furthermore, the Picard iteration inductively
A. Papachristodoulou is with the Department of Engineering Sci-
ence, University of Oxford, Parks Road, Oxford, OX1 3PJ, U.K. retainsalmostallthepropertiesofthesolutionmap.Theresult
[email protected]. isanewformofiterativeconverseLyapunovfunction,V .This
k
This material is based upon work supported by the National Science
function is discussed in Section VI.
Foundation under Grant No. CMMI 110036 and from EPSRC grants
EP/H03062X/1,EP/I031944/1 andEP/J012041/1 The first practical contribution of this paper is to give a
2
bound on the number of decision variables involved in the SOS decomposition, which can be tested using Semidefinite
question of exponential stability of polynomial vector fields programming.
on bounded sets. This is because SOS functions of bounded Consider, for example, the problem of ensuring that a
degree can be parameterized by the set of positive matrices polynomial p(x)∈R[x] satisfies p(x)≥0.Thisproblemarises
of fixed size. Furthermore, we note that the question of naturally when trying to construct Lyapunov functions for
existence of a Lyapunov function with negative derivative is the stability analysis of dynamical systems, which is the
convex.Therefore,if the question of polynomialpositivity on topic of this paper. Since ensuring non-negativityis hard [25]
a boundedset is decidable, we can conclude that the problem manyresearchershaveinvestigatedalternativewaystodothis.
of exponential stability of polynomial vector fields on that In[26],the existenceofa SumofSquaresdecompositionwas
set is decidable. The further complexity benefit of using SOS usedforthatpurpose,whichinvolvesthepresentationofother
Lyapunov functions is discussed in Section VIII. polynomials p(x) such that
i
The main result of the paper is stated and proven in
k
Section VI. Preceding the main result is a series of lemmas p(x)=(cid:229) p(x)2 (1)
i
thatareusedintheproofofthemaintheorem.InSubsectionV i=1
we show that the Picard iteration is contractive on a certain
Algorithmsforensuringthis haveappearedin the 1990’s[27]
metric space; and in Subsection V-A we propose a new way
butitwasnotuntiltheturnofthecenturythatthiswasrecog-
ofextendingthePicarditeration.InSectionV-Bweshowthat
nizedasbeingsolvableusingSemidefiniteProgramming[28].
the Picard iteration approximately retains the differentiability
In particular, (1) can be shown equivalent to the existence of
properties of the solution map, before we prove the main
a Q(cid:23)0 and a vector of monomials Z(x) of degree less than
result. The implications of the main result are then explored
or equal half the degree of p(x), such that
in Section VIII and Section VII. A detailed example is given
in Section IX. The paper is concluded in Section X. p(x)=Z(x)TQZ(x)
Intheaboverepresentation,thematrixQisnotunique,infact
II. MAIN RESULT it can be represented as
Before we begin the technical part of the paper, we give a Q=Q +(cid:229) l Q (2)
0 i i
simplified version of the main result. i
Theorem 1: Suppose that f is polynomial of degree q and whereQ satisfyZ(x)TQZ(x)=0.Thesearchforl suchthat
i i i
that solutions of x˙= f(x) satisfy Qin(2)issuchthatQ(cid:23)0isaLinearMatrixInequality,which
kx(t)k≤Kkx(0)ke−l t can be solved using Semidefinite Programming. Moreover,
if p(x) has unknown coefficients that enter affinely in the
for some l >0, K ≥1 and for any x(0)∈M, where M is representation (1), Semidefinite Programming can be used to
a bounded nonempty region of radius r. Then there exist find values for them so that the resulting polynomial is SOS.
a ,b ,g >0 and a sum-of-squares polynomial V(x) such that Thislatterobservationcanallowustosearchforpolynomi-
for any x∈M, alsthatsatisfySOSconditions:themostimportantexampleis
a kxk2≤V(x)≤b kxk2 intheconstructionofLyapunovfunctions,whichisthetopicof
thispaper.Formoredetails,pleasesee[10],[28].Thequestion
(cid:209) V(x)Tf(x)≤−g kxk2. that we address in this paper is whether Sum of Squares
Lyapunov functions always exist for locally exponentially
Further, the degree of V will be less than
2q(Nk−1), where k(L,l ,K) is any integer such that stable systems.
c(k):=(cid:229) N−1 eTL+K(TL)k iK2(TL)k<K, and
i=0
(cid:0)log2K2 (TL)(cid:1)k 1 IV. NOTATIONAND BACKGROUND
c(k)2+ K (1+c(k))(K+c(k))< ,
2l T 2 ThecoreconceptweuseinthispaperisthePicarditeration.
c(k)2< l (1−(2K2)−lL) We usethistoconstructanapproximationtothesolutionmap
KLlog2K2 and then use the approximate solution map to construct the
Lyapunov function. Construction of the Lyapunov function
and N(L,l ,K) is any integer such that NT > log22lK2 and T < will be discussed in more depth later on.
21L forsomeT andwhereLisaLipschitzboundon f onB4Kr. Denote the Euclidean ball centered at 0 of radius r by Br.
Consider an ordinary differential equation of the form
III. SUM-OF-SQUARES
x˙(t)= f(x(t)), x(0)=x , f(0)=0, (3)
0
Sum of squares (SOS) methods have been introduced over
wherex∈Rnand f satisfiesappropriatesmoothnessproperties
the past decade to allow for the algorithmic solution of
for local existence and uniqueness of solutions. The solution
problems that frequently arise in systems and control the-
map is a function f which satisfies
ory, many of which can be formulated as polynomial non-
¶
negativity constraints that are however difficult to solve. In f (t,x)= f(f (t,x)) and f (0,x)=x.
these methods, non-negativityis relaxed to the existence of a ¶ t
3
A. Lyapunov Stability We begin by showing that for any radius r, there exists a
The use of Lyapunov functions to prove stability of ordi- T such thatthe Picard iterationis contractiveon XT,2r forany
nary differential equations is well-established. The following x∈Br.
theorem illustrates the use of Lyapunov functions. Lemma 7: Given r >0, let T <min{Qr,L1} where f has
tioDnsefiinni(t3io)nar2e:exWpeonseanytitahlalytsthtaeblseysotnemX idfetfihneeredebxyistthge,Keq>ua0- LXTip,2srch→itzXTfa,2crtoarndLtohnereB2erxiastnsdsoQm=e sfup∈x∈XBT2,r2rf(sxu).chThtheant Pfor:
such that for any x0∈X, t∈[0,T] and x∈Br,
kx(t)k≤Kkx0ke−gt d f (t)= f(f (t)), f (0)=x
dt
for all t≥0.
and for any z∈X ,
T,2r
Theorem 3 (Lyapunov): Suppose there exist constants
a ,b ,g >0 and a continuously differentiable functionV such f −Pkz ≤(TL)kkf −zk.
that the following conditions are satisfied for all x∈U ⊂Rn. (cid:13) (cid:13)
(cid:13) (cid:13)
Proof: We fi(cid:13)rst show(cid:13)that for x∈Br, P:XT,2r→XT,2r. If
a kxk2≤V(x)≤b kxk2 q∈X , then sup kq(t)k≤2r and so
T,2r t∈[0,T]
(cid:209) V(x)Tf(x)≤−g kxk2
t
kPqk = sup x+ f(q(s)) ds
Then we have exponential stability of System (3) on X t∈[0,T](cid:13)(cid:13) Z0 (cid:13)(cid:13)
{x : {y:V(y)≤V(x)}⊂U}. (cid:13)(cid:13) T (cid:13)(cid:13)
≤kxk+ kf(q(s))kds
Z0
B. Fixed-Point Theorems T
≤r+ sup kf(y)kds
Definition 4: Let X be a metric space. A mappingF:X → Z0 y∈B2r
X is contractive with coefficient d∈[0,1) if
≤r+TQ<2r
kFx−Fyk≤dkx−yk x,y∈X.
Thus we conclude that Pq∈X . Furthermore, for q ,q ∈
T,2r 1 2
The following is a Fixed-Point Theorem. XT,2r,
Theorem 5 (Contraction Mapping Principle [29]): Let X
t
bweitahccoomefpfilceiteenmt det.riTchsepnactheearnedexleitstFs a:Xun→iquXebye∈aXcosnutcrahcttihoant kPq1−Pq2kX =t∈s[u0,pT](cid:13)(cid:13)Z0 (f(q1(s))−f(q2(s)))ds(cid:13)(cid:13)
(cid:13) (cid:13)
T (cid:13) (cid:13)
Fy=y. ≤ sup kf(q1(s))−f(q2(s))kds
Z0 t∈[0,T]
Furthermore, for any x ∈X,
0 ≤TL sup kq (s)−q (s)k=TLkq −q k
1 2 1 2 X
t∈[0,T]
Fkx −y ≤dkkx −yk.
0 0
(cid:13) (cid:13) Therefore, by the contraction mapping theorem, the Picard
(cid:13) (cid:13)
To apply these r(cid:13)esults to(cid:13)the existence of the solution map, iteration convergeson [0,T] with convergence rate (TL)k.
we use the Picard iteration.
A. Picard Extension Convergence Lemma
V. PICARD ITERATION
In this section we propose an extension to the Picard iter-
WebeginbyreviewingthePicarditeration.Thisisthebasic
ation approximation. We divide the interval into subintervals
mathematical tool we will use to define our approximation to
on which the Picard iteration is guaranteed to converge. On
the solution map and can be found in many texts, e.g. [30].
each interval, we apply the Picard iteration using the final
Definition 6: ForgivenT andr,definethecompletemetric
valueofthesolutionestimatefromthepreviousintervalasthe
space
initial condition,x. For a polynomialvectorfield, the result is
X := q(t): sup kq(t)k≤r, q is continuous. (4) a piece-wise polynomial approximation which is guaranteed
T,r t∈[0,T]
(cid:8) (cid:9) to converge on an arbitrary interval – see Figure 1 for an
with norm
illustration.
kqk = sup kq(t)k.
X Definition 8: Supposethatthesolutionmapf existsont∈
t∈[0,T]
[0,¥ ) and kf (t,x)k≤Kkxk for any x∈B . Suppose that f
r
For a fixed x∈B and q∈X , the Picard Iteration [31],
r T,r has Lipschitz factor L on B and is bounded on B with
4Kr 4Kr
is defined as bound Q. Given T <min{2Kr,1}, let z=0 and define
Q L
t
(Pq)(t),x+ f(q(s))ds. Gk(t,x):=(Pkz)(t,x)
Z0 0
In this paper, we also define the Picard iteration iteration on and for i>0, define the functions G recursively as
i
functions z(t,x) as
Gk (t,x):=(Pkz)(t,Gk(T,x)).
t i+1 i
(Pz)(x,t),x+ f(z(x,s))ds. The Gk are Picard iterations Pkz(t,x) defined on each
Z0 i
4
x(t) and suppose that Gk−f ≤c (k)kxk on interval [iT−
¥ i−1
1.0
k=1 T,iT]. Then (cid:13) (cid:13)
(cid:13) (cid:13)
k=2
0.9 k=3 sup Gk(s,x)−f (s,x)
k=4 s∈[iT,iT+T](cid:13) (cid:13)
0.8 (cid:13) (cid:13)
k=5 = sup(cid:13) Gk(s−iT,x)−(cid:13)f (s,x)
True i
0.7 s∈[iT,iT+T](cid:13) (cid:13)
(cid:13) (cid:13)
= sup (cid:13)Pkz(s−iT,Gk (T,x))(cid:13)−f (s−iT,f (iT,x))
0.6 i−1
s∈[iT,iT+T](cid:13) (cid:13)
(cid:13) (cid:13)
0.5 ≤ sup (cid:13)Pkz(s−iT,Gk (T,x))−f (s−iT,Gk (T,x)(cid:13))
i−1 i−1
[ 0, T ] [ T, 2T ] [ 2T, 3T ] s∈[iT,iT+T](cid:13) (cid:13)
(cid:13) (cid:13)
0.2 0.4 0.6 0.8 1.0 + sup (cid:13)f (s−iT,Gki−1(T,x))−f (s−iT,f (iT,x)) (cid:13)
s∈[iT,iT+T](cid:13) (cid:13)
(cid:13) (cid:13)
Fig. 1. The Solution map f and the functions Gk for k=1,2,3,4,5 and We treat these(cid:13)final two terms separately. First note tha(cid:13)t
i
i=1,2,3 for the system x˙(t)=−x(t)3. The interval of convergence of the
PicardIteration isT=13. Gki−1(T,x) ≤kf (iT,x)k+ f (iT,x)−Gki−1(T,x)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13)≤Kkxk+c ((cid:13)k)kxk (cid:13)
(cid:13) (cid:13) i−1(cid:13) (cid:13)
≤(K+c (k))kxk.
i−1
sub-interval where we substitute the initial condition x 7→ Since ci−1(k)≤c(k)<K and x∈Br, we have Gki−1(T,x) ≤
Gki−1(t,x). Define the concatenation of these Gki as (K+Kci−1(k))kxk≤2Kr. Hence (cid:13)(cid:13) (cid:13)(cid:13)
Gk(t,x):=Gk(t−iT,x) ∀ t∈[iT,iT+T] and i=1,···,¥ . sup Pkz(s−iT,Gk (T,x))−f (s−iT,Gk (T,x))
i i−1 i−1
s∈[iT,iT+T](cid:13) (cid:13)
(cid:13) (cid:13)
If f is polynomial, then the Gk are polynomial for any ≤ sup(cid:13) dk f (s−iT,Gk (T,x)) ≤Kdk Gk (T,x)(cid:13)
i i−1 i−1
i,k and Gk is continuously differentiable in x for any k. The s∈[iT,iT+T] (cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
following lemma providesseveral propertiesfor the functions ≤Kdk(K+c (cid:13)(k))kxk. (cid:13) (cid:13) (cid:13)
i−1
Gk.
Now,ifx∈B ,kf (s,x)k≤Krandsince Gk (T,x) ≤2Kr
r i−1
Lemma 9: Given d > 0, suppose that the solution map and f is Lipschitz on B4Kr, it is well-know(cid:13)n that (cid:13)
(cid:13) (cid:13)
f (t,x) exists on t∈[0,d ] and on x∈B . Further suppose that
kf (t,a,x)k≤Kkxkforanyx∈Br.Supprosethat f isLipschitz s∈[iTsu,iTp+T](cid:13)f (s−iT,Gki−1(T,x))−f (s−iT,f (iT,x))(cid:13)
on B with factor L and bounded with bound Q. Choose (cid:13) (cid:13)
T <m4Kirn{2QKr,L1} and integer N >d /T. Then let Gk and Gki ≤ sup(cid:13) eL(s−iT) Gki−1(T,x)−f (iT,x) ≤eTLci−(cid:13)1(k)kxk
be defined as above. s∈[iT,iT+T] (cid:13)(cid:13) (cid:13)(cid:13)
(cid:13) (cid:13)
Combining, we conclude that
Define the function
c(k)=N(cid:229)−1 eTL+K(TL)k iK2(TL)k. s∈[iTsu,iTp+T](cid:13)Gki(s−iT,x)−f (s,x)(cid:13)
(cid:13) (cid:13)
i=0(cid:16) (cid:17) ≤eTLc (cid:13)(k)kxk+Kdk(K+c (cid:13)(k))kxk
i−1 i−1
Given any k sufficiently large so that c(k)<K, then for any =((ed+Kdk)c (k)+K2dk)kxk=c(k)kxk.
i−1 i
x∈B ,
r
sup Gk(s,x)−f (s,x) ≤c(k)kxk. (5) Since ci(k)≤c(k), and d <NT, by induction, we conclude
s∈[0,d ](cid:13) (cid:13) that
(cid:13) (cid:13)
(cid:13) (cid:13) sup Gk(s,x)−f (s,x) ≤c(k)kxk.
Proof: Suppose x∈Br. By assumption, the conditionsof s∈[0,d ](cid:13)(cid:13) (cid:13)(cid:13)
Lemma 7 are satisfied using r′=2Kr. Let z(t,x)=0. Define (cid:13) (cid:13)
the convergencerate d=TL<1. By Lemma 7,
sup Gk(s,x)−f (s,x) = sup (Pkz)(s,x)−f (s,x)
0 B. Derivative Inequality Lemma
s∈[0,T](cid:13) (cid:13) s∈[0,T](cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13)≤dk su(cid:13)p kf (s,x)k≤Kdkkx(cid:13)k. In this critical lemma, we show that the Picard iteration
s∈[0,T] approximately retains the differentiability properties of the
ThusEquation(5)issatisfiedontheinterval[0,T].Weproceed solutionmap.Theproofisbasedoninduction,withakeystep
by induction. Define basedonanapproachin[32](ProofofThm4.14).Thislemma
is then adapted to the extended Picard iteration introduced in
i
c(k)= (cid:229) (ed+Kdk)jK2dk. the previous section.
i
Lemma 10: Suppose that the conditions of Lemma 7 are
j=1
5
satisfied. Then for any x∈B and any k≥0, satisfied. Then for any x∈B ,
r r
¶ ¶ (TL)k ¶ ¶ (TL)k
sup (Pkz)(t,x)Tf(x)− (Pkz)(t,x) ≤ kxk sup Gk(t,x)Tf(x)− Gk(t,x) ≤ (K+c(k))kxk
(cid:13)¶ x ¶ t (cid:13) T (cid:13)¶ x ¶ t (cid:13) T
t∈[0,T](cid:13) (cid:13) t∈[0,T](cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
Proo(cid:13)f: Begin with the identity for k≥1(cid:13) P(cid:13)roof: Recall that (cid:13)
(Pkz)(t,x)=x+ t f((Pk−1z)(s,x))ds Gk(t,x):=Gi(t−iT,x) ∀ t∈[iT,iT+T] and i=1,···,¥ .
Z0 and Gk (t,x)=Pkz(t,Gk(T,x)) where z=0. Then for t ∈
0 i+1 i
=x+ f((Pk−1z)(s+t,x))ds. [iT,iT+T],
Z−t
¶ ¶
Then, by differentiating the right-hand side, we get Gk(t,x)Tf(x)− Gk(t,x)
(cid:13)¶ x ¶ t (cid:13)
¶ (cid:13) (cid:13)
(Pkz)(t,x) (cid:13) ¶ ¶ (cid:13)
¶ t (cid:13)= Gk(t−iT,x)Tf(x)− (cid:13)Gk(t−iT,x)
(cid:13)¶ x ¶ t i (cid:13)
= f((Pk−1z)(0,x)) (cid:13) (cid:13)
(cid:13) ¶ ¶ (cid:13)
0 ¶ =(cid:13)− Pkz(t−iT,Gk(T,x))+ Pk(t−iT(cid:13),Gk(T,x))Tf(x)
+ (cid:209) f((Pk−1z)(s+t,x))T (Pk−1z)(s+t,x)ds (cid:13) ¶ t i ¶ x i (cid:13)
= f((Pk−1z)Z(−0t,x))+Z0t(cid:209) f((Pk−1z)(s,¶x1))T¶¶s(Pk−1z)(s,x)ds (cid:13)(cid:13)(cid:13) ≤ (TTL)k(cid:13)(cid:13)Gki(T,x)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13)
t ¶ As was shown(cid:13)in the pro(cid:13)of of Lemma 9, Gk(T,x) ≤(K+
= f(x)+ (cid:209) f((Pk−1z)(s,x))T (Pk−1z)(s,x)ds, i
Z0 ¶ s ci(k))kxk. Thus for t∈[iT,iT+T], (cid:13)(cid:13) (cid:13)(cid:13)
iwtsheitrhev¶¶airfiadbelenoatneds partial differentiation of f with respect to (cid:13)¶¶xGk(t,x)Tf(x)−¶¶tGk(t,x)(cid:13)≤ (TTL)k(K+ci(k))kxk
(cid:13) (cid:13)
¶ (Pkz)(t,x)=I+ t(cid:209) f((Pk−1z)(s,x))T ¶ (Pk−1z)(s,x)ds. Sin(cid:13)(cid:13)ce the ci are non-decreasing, (cid:13)(cid:13)
¶ x Z0 ¶ x ¶ ¶ (TL)k
Now define for k≥1, t∈s[u0p,d ](cid:13)(cid:13)¶ xGk(t,x)Tf(x)−¶ tGk(t,x)(cid:13)(cid:13)≤ T (K+c(k))kxk.
¶ ¶ (cid:13) (cid:13)
y (t,x):= (Pkz)(t,x)Tf(x)− (Pkz)(t,x). (cid:13) (cid:13)
k ¶ x ¶ t
For k≥2, we have
¶ ¶ VI. MAIN RESULT- A CONVERSE SOS LYAPUNOV
yk(t,x):= ¶ x(Pkz)(t,x)Tf(x)−¶ t(Pkz)(t,x) FUNCTION
t ¶ In this section, we combine the previous results to obtain
=− (cid:209) f((Pk−1z)(s,x))T (Pk−1z)(s,x)ds
Z0 ¶ t a converseLyapunovfunction which is also a sum-of-squares
t ¶ polynomial. Specifically, we use a standard form of converse
+ (cid:209) f((Pk−1z)(s,x))T (Pk−1z)(s,x)f(x)ds
Z0 ¶ x LyapunovfunctionandsubstituteourextendedPicarditeration
t for the solution map. Consider the system
= (cid:209) f((Pk−1z)(s,x))T·
Z0 x˙(t)= f(x(t)), x(0)=x . (6)
¶ ¶ 0
(Pk−1z)(s,x)f(x)− (Pk−1z)(s,x) ds
(cid:20)¶ x ¶ s (cid:21) Theorem 12: Supposethat f ispolynomialofdegreeqand
t that system (6) is exponentially stable on M with
= (cid:209) f((Pk−1z)(s,x))Ty (s,x)ds.
Z0 k−1 kx(t)k≤Kkx(0)ke−l t,
This means that since (Pk−1z)(t,x)∈B , by induction
2r whereMisaboundednonemptyregionofradiusr.Thenthere
supky (t)k≤T sup (cid:209) f((Pk−1z)(t,x)) sup ky (t,x)k exist a ,b ,g >0 and a sum-of-squares polynomialV(x) such
k k−1
[0,T] t∈[0,T](cid:13) (cid:13)t∈[0,T] that for any x∈M,
(cid:13) (cid:13)
≤TL sup(cid:13)kyk−1(t,x)k≤(TL(cid:13))(k−1) sup ky1(t,x)k a kxk2≤V(x)≤b kxk2 (7)
t∈[0,T] t∈[0,T]
(cid:209) V(x)Tf(x)≤−g kxk2. (8)
Fork=1,(Pz)(t,x)=x,soy (t)= f(x)andsup ky (t)k≤
1 [0,T] 1
Lkxk. Thus Further, the degree of V will be less than 2q(Nk−1), where
(TL)k k(L,l ,K) is any integer such that c(k)<K,
sup ky (t)k≤ kxk.
k
T
t∈[0,T] log2K2 (TL)k 1
c(k)2+ K (1+c(k))(K+c(k))< , (9)
2l T 2
l
We now adapt this lemma to the extended Picard iteration. c(k)2< (1−(2K2)−lL). (10)
Lemma 11: Suppose that the conditions of Lemma 9 are KLlog2K2
6
where c(k) is defined as Negativity of the Derivative: Next, we prove the derivative
condition. Recall
c(k)=N(cid:229)−1 eTL+K(TL)k iK2(TL)k, (11) d
i=0(cid:16) (cid:17) Vk(x):= Gk(s,x)TGk(s,x)ds
Z0
and N(L,l ,K) is any integer such that NT > log22lK2 and T < = t+d Gk(s−t,x)TGk(s−t,x)ds
21L forsomeT andwhereLisaLipschitzboundon f onB4Kr. Zt
Proof: Define d = log22lK2 and d =TL. By assumption then since (cid:209) V(x(t))Tf(x(t)) = ddtV(x(t)), we have by the
d Leibnitz rule for differentiation of integrals,
N> .Next,wenotethatsincestabilityimplies f(0)=0, f is
T
boundedonanyB withboundQ=Lr.ThusforB ,wehave d
the bound Q=4KrrL. By assumption, T < 1 =4K2Krr = 2Kr. dtVk(x(t))= Gk(d ,x(t))TGk(d ,x(t)) − Gk(0,x(t))TGk(0,x(t))
2L 4KrL Q h i h i
Therefore,ifkisdefinedasabove,theconditionsofLemma9 t+d ¶
are satisfied. Define Gk as in Lemma 9. By Lemma 9, if k −Zt 2Gk(s−t,x(t))T¶ 1Gk(s−t,x(t))ds
is defined as above, Gk(s,x)−f (s,x) ≤c(k)kf (s,x)k on t+d ¶
s∈[0,d ] and x∈Br. (cid:13)(cid:13) (cid:13)(cid:13) +Zt 2Gk(s−t,x(t))T¶ 2Gk(s−t,x(t))f(x(t))ds
We propose the following Lyapunov functions, indexed by 2
= Gk(d ,x(t)) −kx(t)k2
k.
d (cid:13) (cid:13)
Vk(x):=Z0 Gk(s,x)TGk(s,x)ds +(cid:13)(cid:13)Z0d 2Gk(s,x((cid:13)(cid:13)t))T(cid:20)¶¶2Gk(s,x(t))f(x(t))−¶¶sGk(s,x(t))(cid:21)ds
We will show that for any k which satisfies Inequali-
¶
where recall f denotes partial differentiation of f with
ties (9), (10) and (11), then if we define V(x)=Vk(x), we ¶ i
respect to its ith variable. As per Lemma 11, we have
havethatV satisfiestheLyapunovInequalities(7)and(8)and
has degree less than 2q(Nk−1). The proof is divided into four ¶ ¶ dk
parts: (cid:13)¶ 2Gk(t,x(t))Tf(x(t))−¶ 1Gk(t,x(t))(cid:13)≤ T (K+c(k))kx(t)k
(cid:13) (cid:13)
LUypappeurnoavnfdunLctoiwone,rfirBsotucnondseidd:erTuoppperrovbeoutnhdaetdVnkesiss.Iaf xva∈liBd a(cid:13)(cid:13)nd as previously noted Gk(d ,x(t(cid:13)(cid:13))) 2 ≤ (K2e−2l (s−t)+
and s∈[0,d ]. Then c(k)2)kx(t)k2. Also, Gk(s,(cid:13)x(t)) ≤ K((cid:13)1+c(k))kx(t)k. We
(cid:13) (cid:13)
conclude that (cid:13) (cid:13)
2 2 (cid:13) (cid:13)
Gk(s,x) = f (s,x)+ Gk(s,x)T−f (s,x) d
(cid:13) (cid:13) (cid:13) h i(cid:13) Vk(x(t))≤(K2e−2ld +c(k)2)kx(t)k2−kx(t)k2
(cid:13) (cid:13) (cid:13) (cid:13) 2 dt
(cid:13) (cid:13) ≤(cid:13)kf (s,x)k2+ Gk(s,x)T−f (s,(cid:13)x) dk
(cid:13)h i(cid:13) +2d K(1+c(k))(K+c(k))kx(t)k2
As per Lemma 9, Gk(s,x)−(cid:13)(cid:13)f (s,x) ≤ c(k)kf (s(cid:13)(cid:13),x)k ≤ T
dk
Kc(k)kxk. From stabil(cid:13)ity we have kf (s,(cid:13)x)k≤Kkxk. Hence, ≤ K2e−2ld +c(k)2−1+2d K (1+c(k))(K+c(k))kx(t)k2.
(cid:13) (cid:13) (cid:18) T (cid:19)
d 2
V (x)= Gk(s,x) ds≤d K2 1+c(k)2 kxk2. Therefore, we have strict negativity of the derivative since
k
Z0 (cid:13) (cid:13)
(cid:13) (cid:13) (cid:0) (cid:1) dk
Thereforetheupp(cid:13)erbound(cid:13)ednessconditionissatisfiedforany K2e−2ld +c(k)2+2d (1+c(k))(K+c(k))
k≥0 with b =d K2(1+c(k)2)>0. T
1 Klog2K2dk
Next we consider the strict positivity condition. First we = +c(k)2+2 (1+c(k))(K+c(k))<1
2 2l T
note
kf (s,x)k2= Gk(s,x)+ f (s,x)−Gk(s,x) 2 Thus ddtVk(x(t))≤−g kx(t)k2 for some g >0.
(cid:13) h i(cid:13)
≤(cid:13)(cid:13)Gk(s,x) 2+ f (s,x)−Gk(s,x(cid:13)(cid:13)) 2 Sum of Squares: Since f is polynomial and z is trivially
(cid:13) (cid:13) (cid:13) (cid:13) polynomial, (Pkz)(s,x) is a polynomialin x and s. Therefore,
(cid:13) (cid:13) (cid:13) (cid:13)
which implies (cid:13) (cid:13) (cid:13) (cid:13) Vk(x) is a polynomial for any k>0. To show thatVk is sum-
of-squares, we first rewrite the function
2 2
Gk(s,x) ≥kf (s,x)k2− f (s,x)−Gk(s,x)
N iT
By Lips(cid:13)(cid:13)(cid:13)chitz con(cid:13)(cid:13)(cid:13)tinuity of f, kf (s(cid:13)(cid:13)(cid:13),x)k2≥e−2Lskxk(cid:13)(cid:13)(cid:13)2 and Vk(x)=i(cid:229)=1ZiT−ThGki(s−iT,x)TGki(s−iT,x)ids.
Gk(s,x)−f (s,x) ≤Kc(k)kxk. Thus
Since Gkz is a polynomial in all of its arguments, Gk(s−
(cid:13) (cid:13) i i
V(cid:13)k(x)=Z0d (cid:13)Gk(s,(cid:13)x)(cid:13)2ds≥(cid:18)21L(1−e−2Ld )−d Kc(k)2(cid:19)kxk2. irTep,xre)sTeGnkite(ds−asiTR,xi()x)iTsZis(usm)T-Zoif(-ss)qRuia(rxe)s.foIrt scoamnethpeorelyfonroemibael
(cid:13) (cid:13) vector R and matrix of monomial bases Z. Then
Therefore f(cid:13)or k as(cid:13) defined previously, 1 (1−e−2Ld )− i i
d Kc(k)2 >0 and so the positivity condition2Lholds for some V (x)=(cid:229)N R(x)T iT Z(s)TZ(s)dsR(x)=(cid:229)N R(x)TMR(x)
a >0. k i=1 i ZiT−T i i i i=1 i i i
7
WhereM = iT Z(s)TZ(s)ds≥0isaconstantmatrix.This 1060
i iT−T i i
provesthatVR is sum-of-squaressince it is a sum of sums-of-
k 1050
squares.
We conclude that V =Vk satisfies the conditions of the d1040
n
theorem for any k which satisfies Inequalities (9) and (10). ou
Degree Bound: Given a k which satisfies the inequality e B1030
e
conditions on c(k), we consider the resulting degree of Gk, Degr1020
and hence, of V . If f is a polynomial of degree q, and y is
k
a polynomial of degree d in x, then Py will be a polynomial 1010
of degree max{1,dq} in x. Thus since z=0, the degree of
Pkz will be qk−1. If N >1, then the degree of Gk will be 1000.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7
i
qNk−1. Thus the maximum degree of the Lyapunov function Exponential Decay Rate
is 2q(Nk−1). 103
In the proof of Theorem 12, the integration interval, d
was chosen such that the conditions will always be feasible d 102
n
u
for some k >0. However, this choice may not be optimal. Bo
e
Numerical experimentation has shown us that a better degree e
gr
e
boundmaybeobtainedbyvaryingthisparameterintheproof. D 101
However,thegivenvalueisonewhichwehavefoundtowork
well in the vast majority of cases.
Weconcludethissectionbycommentingontheformofthe
100
converse Lyapunov function, .7 1 2 3 4 5
Exponential Decay Rate
d
V (x):= Gk(s,x)TGk(s,x)ds. Fig. 2. Degree bound vs. Exponential Convergence Rate for K=1.2, r=
k Z0 L=1,q=5.Domains l <.7andl >.7areplotted separately forclarity.
Our Lyapunov function is defined using an approximation of
thesolutionmap.AdualapproachtosolutionoftheHamilton-
themonomialtermsinthevectorfield.Ifthecomplexityisstill
Jacobi-BellmandEquationwastakenin [33]usingoccupation
unacceptably high, then one can consider the use of parallel
measuresinsteadofPicarditeration.Indeed,the dualspaceof
computing: unlike single-core processing, parallel computing
the Sum of SquaresLyapunovfunctionscan be understoodin
power continues to increase exponentially. For a discussion
terms of moments of such occupation measures [34].
on using parallelcomputing to solve polynomialoptimization
Asafinalnote,theproofofTheorem12alsoholdsfortime-
problems, we refer to [35].
varying systems. Indeed the original proof was for this case.
However, because Sum-of-Squares is rarely used for time-
varying systems, the result has been simplified to improve
VII. QUADRATIC LYAPUNOV FUNCTIONS
clarity of presentation.
In this section, we briefly explore the implications of our
resultforthe existence of quadraticLyapunovfunctionsprov-
A. Numerical Illustration
ingexponentialstabilityofnonlinearsystems.Specifically,we
Toillustratethe degreeboundandhencethecomplexityof look at when the theorem predicts the existence of a degree
analyzinganonlinearsystem,weplotthedegreeboundversus bound of 2. We first note that when the vector field is linear,
the exponential convergence rate of the system. For given then q=1, which implies that 2q(Nk−1)=2 independentof N
parameters, this bound is obtained by numerically searching and k. Recall N is the number of Picard iterations, k is the
forthesmallestkwhichsatisfiestheconditionsofTheorem12. number of extensions and q is the degree of the polynomial
The convergence rate parameter can be viewed as a metric vector field, f. Hence an exponentially stable linear system
for the accuracy of the sum-of-squares approach: suppose has a quadratic Lyapunov function - which is not surprising.
we have a degree bound as a function of convergence rate, Instead we consider the case when q 6= 1. In this case,
d(g ). If it is not possible to find a sum-of-squares Lyapunov for a quadratic Lyapunov function, we require N =k =1 -
function of degree d(g ) proving stability, then we know that a single Picard iteration and no extensions. By examining
the convergencerate of the system must be less than g . the proof of Theorem 12, we see that if the conditions of
As can be seen, as the convergence rate increases, the the theorem are satisfied with N =k=1 then V(x)=xTx is
degreebounddecreasessuper-exponentially,sothatatg =2.4, a Lyapunov function which establishes exponential stability
only a quadratic Lyapunov function is required to prove of the system. Since this is perhaps the most commonly
stability.Forcaseswherehighaccuracyisrequired,thedegree used form of Lyapunov function, it is worth considering how
1
bound increases quickly; scaling approximately as eg . To conservative it is when applied to nonlinear systems of the
reduce the complexity of the problem, in come cases less form
conservativeboundsonthedegreecanbefoundbyconsidering x˙(t)= f(x(t)).
8
5
VIII. IMPLICATIONSFOR SUM-OF-SQUARES
4.5 PROGRAMMING
e 4 In this section we consider the implications that the above
Rat3.5 results have on Sum of Squares programming.
y
a 3
c
e
al D2.5 A. Bounding the number of decision variables
nti 2 Because the set of continuously differentiable functions is
e
n an infinite-dimensional vector space, the general problem of
o1.5
p
x finding a Lyapunov function is an infinite-dimensional feasi-
E 1
bility problem. However, the set of sum-of-squaresLyapunov
0.5 functionswithboundeddegreeisfinite-dimensional.Themost
0 significant implication of our theorem is a bound on the
0 01. 02. 03. 04. 05. 06. 07. 08. 09. 1
Lipschitz Factor numberofvariablesintheproblemofdeterminingstabilityof
a nonlinear vector field. The nonlinear stability problem can
Fig.3. RequireddecayrateforaquadraticLyapunovfunctionvs.Lipschitz now be expressed as a feasibility problem of the following
boundforK=1.2
form.
Theorem 13: Fora givenl , let2d bethe degreeboundas-
sociatedwithTheorem12anddefineN=(n+d)!.IfSystem(6)
n!d!
is exponentiallystable on M with decay rate l or greater,the
Inthe followingcorollarywe givesufficientconditionson the
vector field and decay rate for the Lyapunov function xTx to following is feasible for some a ,b ,g >0.
prove exponential stability. Find: P∈SN :
Corollary 1: Suppose that system (6) is exponentially sta- P≥0
ble with a kxk2≤Z(x)TPZ(x)≤b kxk2 for all x∈M
kx(t)k≤Kkx(0)ke−l t
(cid:209) Z(x)TPZ(x) T f(x)≤−g kxk2 for all x∈M
for some l >0, K≥1 and for any x(0)∈M, where M is a
(cid:0) (cid:1)
bounded nonempty region of radius r. Let L be a Lipschitz where Z(x) be the vector of monomials in x of degree d or
bound for f on B . Suppose that there exists some 1 > less.
4Kr 2L
d >0 such that Proof: The proof follows immediately from the fact that
apolynomialV ofdegree2d isSOSifandonlyifthereexists
K2e−2ld +c21+2Kd L(1+c1)(K+c1)<1 a P≥0 such thatV(x)=Z(x)TPZ(x).
and Kd L<1, where c =K2d L. Let V(x)=xTx. Then for Ourconditionboundsthenumberofvariablesinthefeasibility
1
problem associated with Theorem 13. If M is semialgebraic,
any x∈M,
thentheconditionsinTheorem13canbeenforcedusingsum-
V˙(x)=(cid:209) xTf(x)≤−b kxk2. of-squares and the Positivstellensatz [36]. The complexity of
for some b >0. solvingtheoptimizationproblemwilldependonthecomplex-
ityofthePositivstellensatztest.Ifpositivityonasemialgebraic
Proof:WereconsidertheproofofTheorem12.Thistime, set is decidable,as indicatedin [37],this implies the question
we set N =k =1 and T =d and determine if there exists of exponential stability on a bounded set is decidable.
a d =T < 1 which satisfies the upper-boundedness, lower-
2L
boundednessandderivativeconditions.BecauseV(x)=d xTx,
B. Local Positivity
the upper and lower boundedness conditions are immediately
satisfied. The derivative negativity condition is Another implication of our result is that it reduces the
K2e−2ld +c(1)2+2Kd L(1+c(1))(K+c(1))<1 complexityofenforcingthepositivityconstraint.Asdiscussed
in Section III, semidefinite programming is used to optimize
where c(1)=c =K2d L. This is satisfied by the statement of over the cone of sums-of-squares of polynomials. There are
1
the theorem. severaldifferentwaysthestabilityconditionscanbeenforced.
For example, we have the following theorem.
Note that neither the size of the region we consider nor
Theorem 14: Suppose there exist polynomial V and sum-
the degree of the vector field plays any role in determining
of-squarespolynomialss ,s ,s ands suchthatthefollowing
the degree bound. To illustrate the conditions for existence 1 2 3 4
conditions are satisfied for a ,g >0.
of a quadratic Lyapunovfunction, we plot the required decay
rate vs. the Lipschitz continuity factor in Figure 3 for K = V(x)−a kxk2=s (x)+g(x)s (x)
1 2
1.2. This plot shows that as the Lipschitz continuity of the −(cid:209) V(s)Tf(x)−g kxk2=s (x)+g(x)s (x)
3 4
vectorfield increases(andthe field becomeslesssmooth),the
conservatism of using the quadratic Lyapunov function xTx Then we have exponential stability of System (6) on
increases. {x : {y:V(y)≤V(x)}⊂U}.
9
The complexity of the conditions associated with Theo- 1
rem 14 is determined by the four sum-of-squares variables,
0.8
s. Theorem 14 uses the Positivstellensatz multipliers s and
i 2
s to ensure that the Lyapunovfunctionneed only be positive
4
0.6
and decreasing on the region X ={x:g(x)≥0}. However,
as we now know that the Lyapunov function can be assumed
0.4
SOS, we can eliminate the multiplier s , reducing complexity
2
of the problem.
0.2
Theorem 15: Suppose there exist polynomial V and sum-
of-squares polynomials s1, s2 and s3 such that the following 0
conditions are satisfied for a ,g >0.
−0.2
V(x)−a kxk2=s (x)
1
−(cid:209) (V(x)+a kxk2)Tf(x)−g kxk2=s (x)+g(x)s (x) −0.4
2 3
Thenwe haveexponentialstabilityofSystem(6)foranyx(0) −0.6
suchthat{y:V(y)≤V(x(0))}⊂X whereX:={x:g(x)≥0}.
ThissimplificationreducesthesizeoftheSOSvariablesby −0.8
25%(from4 to 3).If the semialgebraicset X is definedusing
several polynomials (e.g. a hypercube), then the reduction in −1
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
the number of variables can approach 50% . SDP solvers are
Fig. 4. Plot of trajectories of the Van-der-Pol Oscillator. We estimate the
typicallyofcomplexityO(n6),wherenisthedimensionofthe overshootparameter asK∼=1
symmetric matrix variable. In the above example we reduced
n=4N to n= 3N. Thus this simplification can potentially
decrease computation by a factor of 82%.
IX. NUMERICAL EXAMPLE
101
Inthissection,weusetheVan-der-Poloscillatortoillustrate
htiesosuwt.nsTthtahebeldeze.egIrrnoeereeqbvuoeiurlsinbed-rtiiiumnmfleu,pehonoicnwetesovtfehret,htaehcicVsuaernqa-ucdyielirob-frPiouthlmeoisssctaislbtlaialbtiotlyer Degree Bound
with a domain of attraction bounded by the well-known
forward-time limit-cycle. The reverse-time dynamics are as
follows.
x˙ (t)=−x (t)
1 2
x˙2(t)=−m (1−x1(t)2)x2(t)+x1(t) 10100−2 10−1 100 101
Exponential Decay Rate
For simplicity, we choose m =1. On a ball of radius r, the
Lipschitz constant can be found from L=sup kDf(x)k, Fig.5. DegreeBoundfortheVan-der-PolOscillatorasaFunctionofDecay
x∈Br Rate
where k·k is the maximum singular value norm. We find a
Lipschitz constant for the Van-der-Pol oscillator on radius
100
r = 1 to be 2.1. Numerical simulations indicate K ∼= 1, as
illustrated in Figure 4. Given these parameters, the degree
10−1
bound plot is illustrated in Figure 5. Note that the choice
of K=1 dramatically improves the degree bound. Numerical
simulation shows the decay rate to be a relatively constant 10−2
l = .542 throughout the unit ball. This is illustrated in
Figure 6. This gives us an estimate of the degree bound as 10−3
d=6.
TofindtheconverseLyapunovfunctionassociatedwiththis
10−4
degree bound we construct the Picard iteration.
t
(Pz)(t,x)=x+ f(0)ds=x. 10−5
Z0
t
(P2z)(t,x)=x+ f(Pz(s,x))ds 10−6
Z0 0 2 4 6 8 10 12 14 16 18 20
t
=x+ f(x)ds=x+f(x)t Fig.6. Asemi-logplotofkxkforthreetrajectories. Weestimate l =.542
Z0 fortheVan-der-Poloscillator
10
0.8
The converse Lyapunov function is
d
V(x)= (P2z(s,x))T(P2z(s,x))ds 0.6
Z0
d 0.05
= (x+f(x)s)T(x+f(x)s)ds 0.4
Z0
d T
=Z0 (cid:20)f(xx)(cid:21) (cid:20)sII(cid:21) I sI (cid:20)f(xx)(cid:21)ds 0.2 0.05 0.03 0.01 0.005 0.03
(cid:2) (cid:3)
T d
x I sI x
= ds 0
(cid:20)f(x)(cid:21) Z0 (cid:20)sI s2I(cid:21) (cid:20)f(x)(cid:21)
= x T d I d 2/2I x −0.2 0.01 0.05
(cid:20)f(x)(cid:21) (cid:20)d 2/2I d 3/3I(cid:21)(cid:20)f(x)(cid:21)
If d =T = 1 = 1, for the Van-der-Pol Oscillator, we get the −0.4 0.03
2L 4
SOS Lyapunov function. 0.05
T −0.6
x 48I 6I x
192·V(x)=
(cid:20)f(x)(cid:21) (cid:20)6I I(cid:21)(cid:20)f(x)(cid:21)
T 2 −0.8
x 6.93I 2.45I x −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8
=
(cid:20)f(x)(cid:21) (cid:20)2.45I I (cid:21) (cid:20)f(x)(cid:21)
Fig. 7. Level Sets of the converse Lyapunov function, with Ball of radius
T
6.93x+2.45f(x) 6.93x+2.45f(x) r=.25
=
(cid:20) 2.45x+f(x) (cid:21) (cid:20) 2.45x+f(x) (cid:21)
=(6.93x −2.45x )2+ 2.45(x +x2x )+4.48x 2 3
1 2 1 1 2 2
+(2.45x−x )2+ (cid:0)x +x2x +1.45x 2 (cid:1) d=10
2 1 1 2 2
(cid:0) (cid:1) 2 d=8
As per the previous discussion, we use SOSTOOLS to
d=4
verify that this Lyapunov function proves stability. Note that d=6
we must show the function is decreasing on the ball of
1
radius r=.25, as the Lipschitz bound used in the theorem
is for the ball of radius B . We are able to verify that the
4r
Lyapunovfunctionis decreasingon the ball of radiusr=.25.
0
Some level sets of this Lyapunov function are illustrated in
Figure 7. Through experimentation, we find that when we
d=2
increase the ball to radius r = 1, the Lyapunov function
-1
is no longer decreasing. We also found that the quadratic
LyapunovfunctionV(x)=xTxisnotdecreasingontheballof
radius r=.25. Although we believe that our degree bound is
-2
somewhatconservative,theseresultsindicatetheconservatism
is not excessive.
ToexplorethelimitsoftheSOSapproach,fordegreebound
-3
2, 4, 6, 8 and 10, we find the maximum unit ball on which -3 -2 -1 0 1 2 3
we are able to find a sum-of-squares Lyapunov function. We
Fig.8. BestInvariant Regionvs.DegreeBoundwithLimitCycle
then use the largest sublevelset of this Lyapunovfunction on
which the trajectories decrease as an estimate for the domain
of attraction of the system. These level sets are illustrated
bounded set may be decidable. Furthermore, the converse
in Figure 8. We see that as the degree bound increases, our
Lyapunovfunctionwehaveusedinthispaperisrelativelyeasy
estimate of the domain of attraction improves.
to construct given the vector field and may find applications
in other areas of control. The main result also holds for time-
X. CONCLUSION varying systems.
In thispaper,we have usedthe Picarditeration to construct Recently, there has been interest in using semidefinite
an approximation to the solution map on arbitrarily long programmingfortheanalysisonnonlinearsystemsusingsum-
intervals. We have used this approximation to prove that of-squares.This paperclarifies severalquestionson the appli-
exponentialstabilityofapolynomialvectorfieldonabounded cationofthismethod.We nowknowthatexponentialstability
set implies the existence of a Lyapunov function which is a on a bounded set implies the existence of an SOS Lyapunov
sum-of-squares of polynomials with a bound on the degree. function and we know how complex this function may be. It
This implies that the question of exponential stability on a has been recently shown that globally asymptotically stable