Digital Object Identifier (DOI) 10.1007/s10231-004-0141-5
Kenneth J. Palmer
Exponential dichotomy and expansivity
⋆This article is dedicated to Prof. Roberto Conti on the occasion of his eightieth birthday
Received: December 6, 2003
Published online: March 22, 2005 –Springer-Verlag 2005
Abstract. In this work we show if a linear nonautonomous system of ordinary differential equations satisfies a bounded growth condition, then it is exponentially expansive (a slightly stronger condition than uniformly noncritical as defined by Massera and Schäffer) on a half-line if and only if it has an exponential dichotomy and on a whole half-line if and only if it has an exponential dichotomy on both half-lines and no nontrivial bounded solution. We relate these theorems to Sacker and Sell’s work on linear skew-product flows and also examine the robustness of exponential expansivity under perturbation.
Mathematics Subject Classification (2000).34D09
Key words.exponential dichotomy – uniformly noncritical – hull
1. Introduction
Inspired by work of Krasovskii [3], Massera and Schäffer [4, 5] showed that if the linear differential system
˙
x= A(t)x, (1)
has bounded growth on[0,∞), then it has an exponential dichotomy if and only if it is “uniformly noncritical”.
Krasovskii defined the concept of uniformly noncriticality for nonlinear sys-tems. Massera and Schäffer specialized it to linear syssys-tems. They say that the linear system (1) isuniformly noncriticalif for allθwith 0 < θ <1 there exists
T =T(θ) >0 such that ifx(t)is any solution of (1), then
|x(t)| ≤θ sup |u−t|≤T|
x(u)|
holds fort≥T. Note it is clear that if this holds for oneθandT with 0< θ <1, then it holds forθmwithmTinstead ofT. Hence if it holds for oneθwith 0< θ <1, it holds for all suchθ.
K.J. Palmer: Department of Mathematics, National Taiwan University, Taipei, Taiwan 106, e-mail:palmer@math.ntu.edu.tw
The main purpose of this article is to extend Massera and Schäffer’s result to equations (1) on(−∞,∞). We also define a concept of exponential expansivity (suggested by a similar concept from the theory of dynamical systems) and show that when (1) satisfies a bounded growth condition both concepts are equivalent to (1) having an exponential dichotomy on both half-lines and no nontrivial bounded solution. We also show that when A(t) is bounded and uniformly continuous, (1) has these three equivalent properties if and only if no equation in the hull has a nontrivial bounded solution. This leads us to briefly consider the relationship of our results to those of Sacker and Sell [7]. Finally we examine the robustness of these properties under both linear and nonlinear perturbation. However, before studying systems on the whole line, we study systems on the half-line and show for systems with bounded growth that exponential expansivity is equivalent to exponential dichotomy and nonuniform criticality. We also examine what happens when the bounded growth condition is removed.
2. Preliminary definitions
We consider a linear differential system (1), whereA(t)is ann×nmatrix func-tion continuous on an intervalJ, which is usually[0,∞),(−∞,0]or(−∞,∞). Throughout the paper we denote by X(t) the fundamental matrix satisfying
X(0)=I.
Definition 1. We say that Eq. (1) hasbounded growthon an interval J if there exist constantsCandMsuch that
|X(t)X−1(s)| ≤CeM(t−s)
for allt,sinJwiths≤t. (This is satisfied if A(t)is bounded.)
Definition 2. We say that Eq. (1) hasbounded decayon an intervalJif there exist constantsCandMsuch that
|X(t)X−1(s)| ≤CeM(s−t)
for allt,sinJwiths≥t. (This is satisfied if A(t)is bounded.)
Definition 3. We say Eq. (1) has anexponential dichotomyon an intervalJif there exist a projectionPand constantsK ≥1,α >0 such that
|X(t)PX−1(s)| ≤Ke−α(t−s) for s≤t inJ |X(t)(I−P)X−1(s)| ≤Ke−α(s−t) for t≤sin J.
Definition 4. We say Eq. (1) isuniformly noncriticalon an intervalJif there exist
T>0 and 0< θ <1 such that ifx(t)is a solution of (1) then
|x(t)| ≤θ sup |u−t|≤T|
x(u)|
Definition 5. We say Eq. (1) isexponentially expansiveon an interval Jif there exist positive constantsLandβsuch that ifx(t)is any solution of (1) and[a,b] ⊂J, then fora≤t≤b
|x(t)| ≤L[e−β(t−a)|x(a)| +e−β(b−t)|x(b)|].
Definition 5 originates in the theory of dynamical systems. LetSbe a compact invariant set for a diffeomorphism f : Rn → Rn. We say that f is expansive
on S if there exists a positive constantd such that wheneverx, yare in Sand
|fk(x)− fk(y)| ≤dfor all integerskthenx=y. IfSis a compact hyperbolic set for f, then f is expansive onS. In fact, f isexponentially expansiveonS(confer Palmer [6, p. 42]), that is, there exist positive constantsL andλ <1 such that if
{xk}bk=a and{yk}bk=a are orbits of f withxk ∈ Sand such that|xk−yk| ≤ dfor
a≤k≤b, then
|xk−yk| ≤Lλk−a|xa−ya| +Lλb−k|xb−yb| fora≤k≤b.
3. Characterisation of exponential dichotomy for equations on the half-line
Massera and Schäffer [4] (confer also Coppel [1]) showed that if (1) has bounded growth, then (1) has an exponential dichotomy on[0,∞)if and only if it is uni-formly noncritical. Our first theorem shows that uniform noncriticality is equivalent to exponential expansivity when the system has bounded growth and also examines what is true when this condition does not hold.
Theorem 1. Let A(t)be an n×n matrix function which is continuous on[0,∞)
such that Eq.(1)has bounded growth on[0,∞). Then the following three statements are equivalent:
(i) Eq.(1)has an exponential dichotomy on[0,∞);
(ii) Eq.(1)is exponentially expansive on[0,∞);
(iii) Eq.(1)is uniformly noncritical on[0,∞).
Without the assumption of bounded growth, it is still true that(i)⇒(ii)⇒(iii). Proof. First we assume no bounded growth. Suppose (i) holds. LetKandαbe the constant in the dichotomy. Then ifx(t)is a solution of (1),
x(t)=X(t)PX−1(a)x(a)+X(t)(I−P)X−1(b)x(b) and so if 0≤a≤t≤b,
|x(t)| ≤Ke−α(t−a)|x(a)| +Ke−α(b−t)|x(b)|.
Hence (ii) holds if (i) does. Next if (ii) holds as in the previous inequality, then if
t≥T,
|x(t)| ≤Ke−αT|x(t−T)| +Ke−αT|x(t+T)| ≤θ sup |u−t|≤T|
x(u)|,
Now we assume bounded growth. To prove the equivalence, we need only show that(iii)⇒(i). However, this is just Massera and Schäffer’s Theorem 3.5 in [4] (confer also Proposition 1 in Coppel [1, p. 14]). So the theorem follows.
Remark 1. Now we discuss the extent to which exponential expansivity implies dichotomy when bounded growth is not assumed. We show thatif(1)is exponen-tially expansive on[0,∞), there are positive constants K andβand a projection P such that for allξ
|X(t)Pξ| ≤Ke−β(t−s)|X(s)Pξ| for 0≤s≤t and
|X(t)(I−P)ξ| ≤Ke−β(s−t)|X(s)(I−P)ξ| for 0≤t≤s.
For suppose Eq. (1) is exponentially expansive on [0,∞). Then if x(t) is a solution of (1)
|x(t)| ≤L[e−β(t−a)|x(a)| +e−β(b−t)|x(b)|]
for 0≤a≤t≤b. Ifx(t)is bounded we may letb→ ∞to obtain for 0≤a≤t |x(t)| ≤Le−β(t−a)|x(a)|.
Now let P be any projection with R(P) = {ξ : supt≥0|X(t)ξ| < ∞}. Let ξ ∈N(P)with|ξ| =1.
We first show that|X(t)ξ| → ∞ast → ∞uniformly inξ. For if not, there existBand sequencestn → ∞andξnsuch that|xn(tn)| = |X(tn)ξn|<Bfor alln. Fixt≥0. Ifnis large enough,t ≤tnand then
|X(t)ξn| = |xn(t)| ≤Le−βt|xn(0)| +e−β(tn−t)|xn(tn)|≤L(1+B). Without loss of generality, we can assume thatξn→ξwithξ ∈N(P)and|ξ| =1. Then, lettingn→ ∞, we get|X(t)ξ| ≤L(1+B)and this holds for allt≥0. This meansξ ∈R(P), which is a contradiction.
Hence ifξ∈N(P)with|ξ| =1,|x(t)| = |X(t)ξ| → ∞ast→ ∞uniformly inξ. In particular, there existsT>0 such that|x(t)| ≥2 ift ≥T. We can assume
Tis chosen so thatLe−βT
≤1. Then if 0≤t≤s,
|x(t)| ≤L[e−βt|x(0)| +e−β(s−t)|x(s)|] =L[e−βt+e−β(s−t)|x(s)|]. So ifT ≤t≤s,
|x(t)| ≤1+Le−β(s−t)|x(s)| ≤1
2|x(t)| +Le −β(s−t)
|x(s)|.
Thus
|x(t)| ≤2Le−β(s−t)|x(s)| if T ≤t≤s.
If we letMbe a constant such that|X(t)X−1(s)| ≤Mfor 0≤s≤t≤T, a simple argument shows that
However, as shown by an example in the appendix, it does not in general follow thatX(t)PX−1(t)is bounded and so (1) need not have an exponential dichotomy. Thus (ii)⇒(i) does not hold in general. Note however (ii)⇒(i) does hold for scalar equations by the above argument. It follows from this that in general (iii) does not imply (ii) since Massera and Schäffer [4, p. 561] give an example of a scalar equation which is uniformly noncritical but does not have an exponential dichotomy. Of course, their example also shows that in general (iii) does not imply (i).
4. Equations on the whole line
Now we see what Theorem 1 looks like for systems on(−∞,∞).
Theorem 2. Let A(t)be an n×n matrix function which is continuous on(−∞,∞)
such that Eq.(1)has bounded growth on[0,∞)and bounded decay on(−∞,0]. Then the following three statements are equivalent:
(i) Eq. (1) has exponential dichotomies on[0,∞)and (−∞,0] and has no nontrivial bounded solution;
(ii) Eq.(1)is exponentially expansive on(−∞,∞);
(iii) Eq.(1)is uniformly noncritical on(−∞,∞).
Without the assumptions of bounded growth and decay, it is still true that
(i)⇒(ii)⇒(iii).
Note that if (1) has an exponential dichotomy on(−∞,∞), then it certainly has exponential dichotomies on[0,∞)and(−∞,0], and in addition has no nontrivial bounded solution. However the converse is not true. Consider, for example, the scalar equationx˙=a(t)x, wherea(t)=1 fort≥1 anda(t)= −1 fort≤ −1. This has exponential dichotomies on[0,∞)and(−∞,0], and no nontrivial bounded solution but does not have an exponential dichotomy on(−∞,∞).
In order to prove the theorem, we use the following lemmas.
Lemma 1. Let (1) have exponential dichotomies on both [0,∞) and
(−∞,0]. Then(1)has no nontrivial bounded solution if and only if we can choose the projections P and Q for the dichotomies on[0,∞)and(−∞,0]respectively such that
PQ=Q P=P.
Proof. Suppose (1) has no nontrivial bounded solution. LetPbe the projection for the dichotomy on[0,∞)andQthat on(−∞,0]. Then it follows from Coppel [1, p. 16] thatR(P)= {ξ :sup
t≥0|X(t)ξ|<∞}andN(Q)= {ξ:supt≤0|X(t)ξ|<∞}. The nullspace of P can be any subspace complementary toR(P)and the range of Q can be any subspace complementary toN(Q). However, since (1) has no nontrivial bounded solution,R(P)∩N(Q)= {0}and so there is a subspaceV such that Rn = R(P)⊕V ⊕N(Q). Thus we may takeN(P) = V ⊕N(Q), R(Q) = R(P)⊕V so that N(Q) ⊂ N(P), R(P) ⊂ R(Q). It follows that
Conversely, suppose we can choose the projectionsPandQfor the dichotomies on [0,∞) and (−∞,0] respectively such that PQ = Q P = P. Let x(t) be a bounded solution of (1). Thenx(0) ∈ R(P)∩N(Q)and sox(0)= Px(0)=
PQx(0)=P·0=0. Sox(t)is the trivial solution. Thus the lemma is proved.
Remark 2. Elaydi and Hajek [2] consider a system (1) satisfying anexponential trichotomy. This holds if and only if (1) has exponential dichotomies on[0,∞) and(−∞,0]with respective projectionsPandQsatisfyingR(P)+N(Q)=Rn or equivalently if PandQcan be chosen so that PQ =Q P = Q. (This can be regarded as a kind ofstrong transversality condition.) They show that (1) has an exponential trichotomy if and only if its adjoint system has exponential dichotomies on[0,∞)and(−∞,0]and no nontrivial bounded solution.
Lemma 2. Let(1)have exponential dichotomies on both[0,∞)and(−∞,0]and no nontrivial bounded solution. Then(1)is exponentially expansive on(−∞,∞). Proof. From Lemma 1, we can choose the projections P and Q so that PQ = Q P = P. Let K and αbe the constants in both dichotomies. Suppose x(t)is a solution of (1). Then, as in the proof of Theorem 1, it follows for 0≤a≤t≤b
anda≤t≤b≤0 that
|x(t)| ≤Ke−α(t−a)|x(a)| +Ke−α(b−t)|x(b)|.
Next note that ifx(t)is a solution then
x(0)=Px(0)+(I−P)x(0)=PQx(0)+(I−P)x(0)
and so ifa≤0≤b
|x(0)| ≤K|Qx(0)| + |(I−P)x(0)|
=K|X(0)Q X−1(a)x(a)| + |X(0)(I−P)X−1(b)x(b)|
≤K2eαa|x(a)| +Ke−αb|x(b)|.
Next ifa≤0≤t≤b,
|x(t)| ≤Ke−αt|x(0)| +Ke−α(b−t)|x(b)|
≤Ke−αt(K2eαa|x(a)| +Ke−αb|x(b)|)+Ke−α(b−t)|x(b)|.
So ifa≤0≤t≤b,
|x(t)| ≤K3e−α(t−a)|x(a)| +(K2+K)e−α(b−t)|x(b)|.
Similarly ifa≤t≤0≤b,
Proof of Theorem 2.That (i) implies (ii) follows from Lemma 2. That (ii) implies (iii) is shown as in the proof of Theorem 1.
Finally we show that (iii) implies (i). So suppose (1) is uniformly noncritical on (−∞,∞). Then it follows from Theorem 3.5 in Massera and Schäffer [4] (confer also Proposition 1 in Coppel [1, p. 14]) and its analogue for(−∞,0]that (1) has exponential dichotomies on both half-lines. Next letx(t)be a bounded solution of (1) and setx =sup−∞<t<∞|x(t)|. Then for allt
|x(t)| ≤θ sup |u−t|≤T|
x(u)| ≤θx.
Hencex ≤θx. It follows thatx =0 and sox(t)=0 for allt. Thus (1) has no nontrivial bounded solution and the proof of the theorem is complete.
5. Hull considerations and relation with Sacker-Sell theory
Now supposeA(t)is bounded and uniformly continuous on(−∞,∞). Then the
hull H(A), which is defined as the set of matrix functions A˜(t)for which there exists a sequencetksuch that A(t+tk)→ ˜A(t)uniformly on compact intervals, is compact in the topology of uniform convergence on compact intervals.
Then we can show the following theorem, which is implicit in Coppel [1, Lecture 9].
Theorem 3. Suppose A(t) is bounded and uniformly continuous on
(−∞,∞). Then Eq.(1) is exponentially expansive on (−∞,∞) if and only if the equation
˙
x= ˜A(t)x (2)
has no nontrivial bounded solution for allA˜∈H(A).
Proof. Suppose (1) is exponentially expansive on(−∞,∞). Then, by Theorem 2, (1) has exponential dichotomies on both half-lines and no nontrivial bounded solution. Now if A˜∈ H(A), there are just three possibilities:A˜is a translate ofA,
˜
Ais in theω−limit set ofA(that is, the sequencetk→ ∞) orA˜is in theα−limit set ofA(that is, the sequencetk→ −∞). If A˜is a translate ofA, then clearly (2) has no nontrivial bounded solution. IfA˜is in theω−limit set orα−limit set of A, then it follows from Lemma 1 in Coppel [1, p. 70] and its analogue on(−∞,0]that (2) has an exponential dichotomy on(−∞,∞)and hence no nontrivial bounded solution.
Conversely, suppose (2) has no nontrivial bounded solution for allA˜∈ H(A). Then it follows from Lemma 2 in Coppel [1, p. 75] that (1) is uniformly noncritical and hence, by Theorem 2, exponentially expansive on(−∞,∞). Thus the proof of the theorem is complete.
Y is a compact Hausdorff space. They assume the no nontrivial bounded solution property, that is,
B= {(x,y): sup −∞<t<∞|
φ(x,y,t)|<∞} = {0} ×Y.
We claim that the linear skew product flowπ(x,y,t)satisfies the no nontrivial bounded solution property if and only if the linear skew-product flow is uni-formly noncritical, that is, there exist θ, 0 < θ < 1, and T > 0 such that |x| ≤θsup|t|≤T|φ(x,y,t)|for all(x,y).
Suppose the flow is uniformly noncritical and let(ξ,y)∈B. Then for allt,
|φ(ξ,y,t)| ≤θ sup |u|≤T|
φ(φ(ξ,y,t), σ(y,t),u)|
=θ sup
|u|≤T|
φ(ξ,y,u+t)|
≤θ sup
−∞<u<∞|
φ(ξ,y,u)|
and hence sup−∞<t<∞|φ(ξ,y,t)| ≤ θsup−∞<t<∞|φ(ξ,y,t)|. This implies that
|φ(ξ,y,t)| =0 for alltand soξ =0.
Conversely, suppose the no nontrivial bounded solution property holds but the flow is not uniformly noncritical. Then there exist sequencesξkand ykwith
|ξk| =1 such that 1> 12sup|t|≤k|φ(ξk,yk,t)|for allk. By compactness, we can assume without loss of generality thatξk→ξ andyk→ yask→ ∞. It follows that|ξ| =1 and sup|t|<∞|φ(ξ,y,t)| ≤2. This is a contradiction and so the claim follows.
Notice that Theorem 3 is a consequence of this claim. Also it raises the following question:
Question. Isuniform noncriticalityequivalent toexponential expansivity, that is, the existence of positiveLandαsuch that
|x| ≤L[eαa|φ(x,y,a)| +e−αb|φ(x,y,b)|] for allx,yanda≤0≤b?
Now Sacker and Sell show that a linear skew product flow with the no non-trivial bounded solution property does have a certain property which is implied by exponential expansivity as just defined. They defineS= {(x,y): |φ(x,y,t)| →0 as t → ∞}andU = {(x,y) : |φ(x,y,t)| → 0 as t → −∞}. In Lemma 5 in [7], they show there exist constantsK ≥1 andα >0 such that for all(x,y)∈S,
t ≥ 0 the inequality|φ(x,y,t)| ≤ K|x|e−αt holds and for all(x,y)∈ U,t ≤ 0 the inequality|φ(x,y,t)| ≤ K|x|eαt holds. An analogous result holds for single equations.
Theorem 4. Suppose(1)has exponential dichotomies on both[0,∞)and(−∞,0]
and no nontrivial bounded solution. Then there exist positive constants L andα
such that if x(t)is a solution bounded on t≥0then
and if x(t)is a solution bounded on t≤0then
|x(t)| ≤Le−α(s−t)|x(s)| for − ∞<t≤s<∞.
Proof. Let the constants in the exponential dichotomies beKandα. Then it follows from Lemma 2 and its proof that if−∞<a≤t≤b<∞, then
|x(t)| ≤L[e−α(t−a))|x(a)| +e−α(b−t)|x(b)|]
withL = K+K3. Supposex(t)is a solution bounded ont ≥ 0. Then we may letb→ ∞ in the previous inequality to get|x(t)| ≤ Le−α(t−a)|x(a)|fora ≤ t. Similarly, if x(t)is a solution bounded on t ≤ 0, we may leta → −∞in the inequality to get|x(t)| ≤Le−α(b−t)|x(b)|fort≤b. Thus the theorem is proved.
6. Perturbations
First we prove a roughness theorem for linear systems.
Theorem 5. Let A(t)be an n×n matrix function which is continuous on(−∞,∞)
such that Eq.(1)has exponential dichotomies on both half-lines and no bounded solution. Then if B(t)is an n×n matrix function which is continuous on(−∞,∞)
such that
|B(t)| ≤δ≤α/48√2K4 for all t, then the perturbed system
˙
x= [A(t)+B(t)]x (3)
has exponential dichotomies on both half-lines and no bounded solution. Moreover, if P and Q are projections for the dichotomy of (1) on [0,∞) and(−∞,0] satisfying PQ=Q P= P and the constants are K andα, then the projections P¯ andQ for¯ (3)can be chosen to satisfyP¯Q¯ = ¯QP¯= ¯P and there are constants N, N1depending only on K andαsuch that
| ¯P−P| ≤Nδ, | ¯Q−Q| ≤Nδ
and the constants in the dichotomies for(3)are5K2[2+5K2N
1δ]/4,α−2Kδ.
Proof. By Lemma 1, (1) has exponential dichotomies on[0,∞)and(−∞,0]with respective projectionsP andQ satisfying PQ = Q P = P. Let the constants in both dichotomies beKandα. Then, sinceδ < α/4K2, it follows from Proposition 1 in Coppel [1, p. 34] that (3) has an exponential dichotomy on[0,∞)with constants 5K2/2,α−2Kδand projectionP˜satisfying for allt≥0,
|Y(t)PY˜ −1(t)−X(t)PX−1(t)| ≤δ1=4α−1K3δ,
whereY(t)is the fundamental matrix for (3) satisfyingY(0)=I. Similarly, (3) has an exponential dichotomy on(−∞,0]with constants 5K2/2,α−2Kδand projec-tionQ˜satisfying
Now it may not be true that P˜Q˜ = ˜QP˜ = ˜P. We modify the projections so that it is true. To this end, write
P1=P, P2 =Q−P, P3=I−Q
and let ki be the rank of Pi. Note that P1, P2 and P3 are mutually orthogonal projections such thatP1+P2+P3=I. Then, provided we use the Euclidean norm inRn, it follows from an obvious extension of Lemma 1 in Coppel [1, p. 39] that there is an invertible matrixSsuch thatSPi0S−1=Pifori =1,2,3, where
P10=diagIk1,0,0
, P20 =diag0,Ik2,0
, P30=diag0,0,Ik3
and
|S| ≤√3, |S−1| ≤ [|P1|2+ |P2|2+ |P3|2]1/2≤
√
6K.
Write S = [S1 S2 S3], where Si has ki columns. The columns of S1 form a basis for R(P), those of S2 a basis for V = N(P)∩R(Q) and those of
S3 a basis for N(Q). We define S¯ = [ ˜PS1 S2 (I− ˜Q)S3]. Then S¯− S =
[(P˜−P)S1 0(Q− ˜Q)S3]and so
| ¯S−S| ≤ |S|[| ˜P−P| + |Q− ˜Q|] ≤2√3δ1.
Then since 6√2Kδ1≤ 12,S¯is invertible and
| ¯S−1| ≤2|S−1| ≤2√6K.
Now we define the projections
¯
P= ¯S P10S¯−1, Q¯ = ¯SI−P30S¯−1.
Then P¯Q¯ = ¯QP¯ = ¯Pand since P¯S¯= ¯S P10 = [ ˜P S10 0],(I− ¯Q)S¯ = ¯S P30 =
[0 0(I− ˜Q)S3], it follows thatR(P¯)=R(P˜)andN(Q¯)=N(Q˜). Moreover
| ¯P−P| =S P¯ 10S¯−1−SP10S−1
≤ | ¯S−S|P10S¯−1 +
SP10
| ¯S−1−S−1|
≤ | ¯S−S|| ¯S−1| + |S|| ¯S−1||S− ¯S||S−1| = | ¯S−S|| ¯S−1|[1+ |S||S−1|]
≤12√2K[1+3√2K]δ1=Nδ.
Similarly,| ¯Q−Q| ≤Nδ.
Remark 3. We can also show thatgivenε > 0, then ifδis sufficiently small the inequalities
|Y(t)PY¯ −1(t)−X(t)PX−1(t)| ≤ε for t≥0
and
|Y(t)QY¯ −1(t)−X(t)Q X−1(t)| ≤ε for t≤0
hold.
To see this, using the same notation as in the proof of Theorem 5, first write
˜
P(t)=Y(t)PY˜ −1(t), P¯(t)=Y(t)PY¯ −1(t),P(t)= X(t)PX−1(t). Note thatP˜and
¯
P have the same range which implies that P˜P¯ = ¯P, P¯P˜ = ˜P and hence that
˜
P(t)P¯(t)= ¯P(t),P¯(t)P˜(t)= ˜P(t). Then fort≥0
| ¯P(t)− ˜P(t)| = | ¯P(t)(I− ˜P(t))|
= |Y(t)PY¯ −1(0)Y(0)(I− ˜P)Y−1(t)|
≤K1e−2(α−2Kδ)t,
where K1 = 25K 4 4 [1+
5K2
2 N1δ]2. Then| ¯P(t)− ˜P(t)| ≤ ε/2 fort ≥ T, where
T =log(2K1/ε)/[2α−4Kδ].
Next it follows from Proposition 2 in Coppel [1, p. 3] that|Y(t)| ≤ MeMTδ for 0 ≤ t ≤ T, where M = sup{|X(t)X−1(s)| : 0 ≤ s,t ≤ T}. By a similar argument applied to the adjoint system,|Y−1(t)| ≤MeMTδfor 0
≤t≤T. Thus for 0≤t≤T
| ¯P(t)− ˜P(t)| ≤ |Y(t)|| ¯P− ˜P||Y−1(t)| ≤M2e2MTδN1δ < ε/2 ifδis sufficiently small.
Finally fort≥0
| ¯P(t)−P(t)| ≤ | ¯P(t)− ˜P(t)| + | ˜P(t)−P(t)|< ε/2+δ1< ε ifδis sufficiently small. The inequality fort≤0 is proved similarly.
Now we look at nonlinear systems. We show that if the variational equation along a solution has an exponential dichotomy on both half-lines and no nontriv-ial bounded solution, then the nonlinear system has an exponentnontriv-ial expansivity property in the neighbourhood of the solution.
Theorem 6. Suppose f : R×Rn
→ Rn is continuous with continuous partial
derivative fx(t,x). Let x(t)be a solution of the system
˙
x= f(t,x) (4)
such that the variational system
˙
has exponential dichotomies on both half-lines and no nontrivial bounded solution. Suppose also that asδ→0
ω(δ)=sup{|fx(t,y)− fx(t,x(t))| : −∞<t<∞,|y−x(t)| ≤δ} →0.
Then there exist positive constants L,βand d such that if y(t)is a solution of(4)
satisfying
|y(t)−x(t)| ≤d for a≤t ≤b,
then for a≤t≤b,
|y(t)−x(t)| ≤L[e−β(t−a)|y(a)−x(a)| +e−β(b−t)|y(b)−x(b)|].
Proof. Write
z(t)=y(t)−x(t).
Thenz(t)satisfies
˙
z(t)= fx(t,x(t))z(t)+ f(t,x(t)+z(t))− f(t,x(t))− fx(t,x(t))z(t)
so thatz(t)can be thought of as a solution of the linear system
˙
z= [A(t)+B(t)]z, (6)
where
A(t)= fx(t,x(t)), B(t)= 1
0
fx(t,x(t)+θz(t))− fx(t,x(t))dθ.
We note that fora≤t ≤b,
|B(t)| ≤ω(d).
We supposedis so small thatδ=ω(d)satisfies the conditions of Theorem 5. Now we define
B(t)=
B(a) if t<a B(b) if t>b.
Then we can apply Theorem 5 to deduce that (6) has exponential dichotomies on both half-lines and no bounded solution. Also we can choose the projectionsP,Q
to satisfy PQ =Q P = Pand the associated dichotomy constants K1 andβcan be chosen to depend only on the constants associated with the dichotomies for (5). Then it follows from Lemma 2 and its proof that fora≤t≤b,
|z(t)| ≤L[e−β(t−a)|z(a)| +e−β(b−t)|z(b)|],
7. Appendix
In this appendix we give an example of a system on[0,∞)which is exponentially expansive but does not have an exponential dichotomy.
The solution(x1(t),x2(t))of the system
˙
x1= −x1+2etx2, x˙2=0 withx1(0)=ξ1,x2(0)=ξ2is
x1(t)=ξ1e−t+ξ2(et−e−t), x2(t)=ξ2.
One solution is x(t) = (e−t,0)and another is y(t)
= (et,1). So if the system has an exponential dichotomy on[0,∞), the projection will have rank 1 withx(t) spanning the stable subspace and can be chosen so that y(t)spans the unstable subspace. However, it is clear that the angle betweenx(t)andy(t)tends to zero as t → ∞. So this system does not have an exponential dichotomy on[0,∞). Nevertheless we will show that it is exponentially expansive.
The general solution of this system can be written asx(t)=(Aet
+Be−t,A). We will show that if 0≤a≤t≤b, then
|x(t)| ≤e−(t−a)|x(a)| +e−(b−t)|x(b)|.
Clearly this holds if A =0. So we can assume A =0. Then, dividing by A, we see that we need to show that for 0≤a≤t≤band allλ
f(t)≤e−(t−a)f(a)+e−(b−t)f(b),
where, if we use the maximum norm, f(t)=max{|et+λe−t|,1}.
First supposeλ≥0. Then for allt≥0, f(t)=et+λe−tand if 0≤a≤t≤b e−(t−a)f(a)+e−(b−t)f(b)=e2a−t+λe−t+et+λet−2b ≥et+λe−t = f(t).
Next suppose λ < 0. Then et
+λe−t is a strictly increasing function. We consider three cases:
Case 1. et+λe−t≥1.
Then we need to show that if 0≤a≤t≤bandλ <0
et+λe−t≤e−(t−a)f(a)+e−(b−t)(eb+λe−b),
where f(a)≥1. So it suffices to show
et+λe−t ≤e−(t−a)+e−(b−t)(eb+λe−b).
This is clear since
e−(t−a)+e−(b−t)(eb+λe−b)=e−(t−a)+et+λet−2b
Case 2. −1≤et
+λe−t
≤1. Supposeeb
+λe−b
≥1. Then we need to show that if 0≤a≤t ≤bandλ <0 1≤e−(t−a)f(a)+e−(b−t)(eb+λe−b),
that is,
1≤e−(t−a)f(a)+et+λet−2b.
The derivative of the right side with respect tobis−2λet−2b >0. So if we can prove the inequality forbsuch thateb
+λe−b
=1, it will follow for largerb. Hence we may assume that
−1≤et+λe−t ≤eb+λe−b ≤1 and need to prove that if 0≤a≤t ≤bandλ <0,
1≤e−(t−a)f(a)+e−(b−t). Next supposeea
+λe−a
≤ −1. Then we need to show that if 0≤a≤t ≤band λ <0
1≤e−(t−a)(−ea−λe−a)+e−(b−t), that is,
1≤ −e2a−t−λe−t+e−(b−t).
The derivative of the right side with respect toais−2e2a−t <0. So if we can prove the inequality forasuch thatea
+λe−a
= −1, it will follow for smallera. Hence we may assume that
−1≤ea+λe−a ≤eb+λe−b ≤1 and need to prove that if 0≤a≤t ≤bandλ <0,
1≤e−(t−a)+e−(b−t).
Now as a function oftin[a,b], the minimum value of the function on the right hand side is 2e−(b−a)/2. So we need to show thateb−a
≤4. However
−ea−e2a ≤λ≤eb−e2b
and so
(eb+ea)(eb−1−ea)=e2b−eb−ea−e2a≤0. Thuseb ≤1+eaand soeb−a≤2.
Case 3. et+λe−t≤ −1.
Then we need to show that if 0≤a≤t≤bandλ <0
−et−λe−t≤e−(t−a)(−ea−λe−a)+e−(b−t)f(b). Since f(b)≥1, it suffices to prove that
−et−λe−t ≤e−(t−a)(−ea−λe−a)+e−(b−t). This holds since
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