exponential dichotomy and expansivity

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Digital Object Identifier (DOI) 10.1007/s10231-004-0141-5 Annali di Matematica 185, S171–S185 (2006) 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 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 systems. They say that the linear system (1) is uniformly noncritical if for all θ with 0 <θ< 1 there exists T = T(θ) > 0 such that if x(t) is any solution of (1), then |x(t)|≤ θ sup |ut |≤T |x(u )| holds for t T . Note it is clear that if this holds for one θ and T with 0 <θ< 1, then it holds for θ m with mT instead of T . Hence if it holds for one θ with 0 <θ< 1, it holds for all such θ . K.J. Palmer: Department of Mathematics, National Taiwan University, Taipei, Taiwan 106, e-mail: [email protected] This work was done with the support of grants 91-2115-M-002-020 and 92-2115-M- 002-022 from the National Research Council.

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Page 1: Exponential dichotomy and expansivity

Digital Object Identifier (DOI) 10.1007/s10231-004-0141-5

Annali di Matematica 185, S171–S185 (2006)

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, 2003Published online: March 22, 2005 – © Springer-Verlag 2005

Abstract. In this work we show if a linear nonautonomous system of ordinary differentialequations satisfies a bounded growth condition, then it is exponentially expansive (a slightlystronger 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 line if and only if it hasan exponential dichotomy on both half-lines and no nontrivial bounded solution. We relatethese theorems to Sacker and Sell’s work on linear skew-product flows and also examinethe 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 thelinear differential system

x = A(t)x, (1)

has bounded growth on [0,∞), then it has an exponential dichotomy if and only ifit is “uniformly noncritical”.

Krasovskii defined the concept of uniformly noncriticality for nonlinear sys-tems. Massera and Schäffer specialized it to linear systems. They say that thelinear system (1) is uniformly noncritical if for all θ with 0 < θ < 1 there existsT = T(θ) > 0 such that if x(t) is any solution of (1), then

|x(t)| ≤ θ sup|u−t|≤T

|x(u)|

holds for t ≥ T . Note it is clear that if this holds for one θ and T with 0 < θ < 1,then it holds for θm with mT instead of T . Hence if it holds for one θ with 0 < θ < 1,it holds for all such θ .

K.J. Palmer: Department of Mathematics, National Taiwan University, Taipei, Taiwan 106,e-mail: [email protected]

� This work was done with the support of grants 91-2115-M-002-020 and 92-2115-M-002-022 from the National Research Council.

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S172 K.J. Palmer

The main purpose of this article is to extend Massera and Schäffer’s result toequations (1) on (−∞,∞). We also define a concept of exponential expansivity(suggested by a similar concept from the theory of dynamical systems) and showthat when (1) satisfies a bounded growth condition both concepts are equivalent to(1) having an exponential dichotomy on both half-lines and no nontrivial boundedsolution. 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 hasa nontrivial bounded solution. This leads us to briefly consider the relationshipof our results to those of Sacker and Sell [7]. Finally we examine the robustnessof these properties under both linear and nonlinear perturbation. However, beforestudying systems on the whole line, we study systems on the half-line and showfor systems with bounded growth that exponential expansivity is equivalent toexponential dichotomy and nonuniform criticality. We also examine what happenswhen the bounded growth condition is removed.

2. Preliminary definitions

We consider a linear differential system (1), where A(t) is an n × n matrix func-tion continuous on an interval J , which is usually [0,∞), (−∞, 0] or (−∞,∞).Throughout the paper we denote by X(t) the fundamental matrix satisfyingX(0) = I .

Definition 1. We say that Eq. (1) has bounded growth on an interval J if thereexist constants C and M such that

|X(t)X−1(s)| ≤ CeM(t−s)

for all t, s in J with s ≤ t. (This is satisfied if A(t) is bounded.)

Definition 2. We say that Eq. (1) has bounded decay on an interval J if there existconstants C and M such that

|X(t)X−1(s)| ≤ CeM(s−t)

for all t, s in J with s ≥ t. (This is satisfied if A(t) is bounded.)

Definition 3. We say Eq. (1) has an exponential dichotomy on an interval J if thereexist a projection P and constants K ≥ 1, α > 0 such that

|X(t)PX−1(s)| ≤ Ke−α(t−s) for s ≤ t in J

|X(t)(I − P)X−1(s)| ≤ Ke−α(s−t) for t ≤ s in J.

Definition 4. We say Eq. (1) is uniformly noncritical on an interval J if there existT > 0 and 0 < θ < 1 such that if x(t) is a solution of (1) then

|x(t)| ≤ θ sup|u−t|≤T

|x(u)|

for all t such that [t − T, t + T ] ⊂ J .

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Exponential dichotomy and expansivity S173

Definition 5. We say Eq. (1) is exponentially expansive on an interval J if thereexist positive constants L and β such that if x(t) is any solution of (1) and [a, b] ⊂ J ,then for a ≤ 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. Let S be a compact

invariant set for a diffeomorphism f : Rn → Rn . We say that f is expansive

on S if there exists a positive constant d such that whenever x, y are in S and| f k(x) − f k(y)| ≤ d for all integers k then x = y. If S is a compact hyperbolic setfor f , then f is expansive on S. In fact, f is exponentially expansive on S (conferPalmer [6, p. 42]), that is, there exist positive constants L and λ < 1 such that if{xk}b

k=a and {yk}bk=a are orbits of f with xk ∈ S and such that |xk − yk| ≤ d for

a ≤ k ≤ b, then

|xk − yk| ≤ Lλk−a |xa − ya| + Lλb−k |xb − yb|for a ≤ 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 boundedgrowth, 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 equivalentto exponential expansivity when the system has bounded growth and also examineswhat 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 statementsare 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. Let K and α be theconstant in the dichotomy. Then if x(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 ift ≥ T ,

|x(t)| ≤ Ke−αT |x(t − T )| + Ke−αT |x(t + T )| ≤ θ sup|u−t|≤T

|x(u)|,

where θ = 2Ke−αT < 1 if T is sufficiently large. Hence (iii) holds.

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S174 K.J. Palmer

Now we assume bounded growth. To prove the equivalence, we need only showthat (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 impliesdichotomy when bounded growth is not assumed. We show that if (1) is exponen-tially expansive on [0,∞), there are positive constants K and β and a projectionP such that for all ξ

|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) isa solution of (1)

|x(t)| ≤ L[e−β(t−a)|x(a)| + e−β(b−t)|x(b)|]for 0 ≤ a ≤ t ≤ b. If x(t) is bounded we may let b → ∞ 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| → ∞ as t → ∞ uniformly in ξ . For if not, thereexist B and sequences tn → ∞ and ξn such that |xn(tn)| = |X(tn)ξn| < B for all n.Fix t ≥ 0. If n is large enough, t ≤ tn and then

|X(t)ξn | = |xn(t)| ≤ L[e−βt |xn(0)| + e−β(tn−t)|xn(tn)|

] ≤ L(1 + B).

Without loss of generality, we can assume that ξn → ξ with ξ ∈ N (P) and |ξ| = 1.Then, letting n → ∞, we get |X(t)ξ| ≤ L(1 + B) and this holds for all t ≥ 0. Thismeans ξ ∈ R(P), which is a contradiction.

Hence if ξ ∈ N (P) with |ξ| = 1, |x(t)| = |X(t)ξ| → ∞ as t → ∞ uniformlyin ξ . In particular, there exists T > 0 such that |x(t)| ≥ 2 if t ≥ T . We can assumeT is chosen so that Le−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 if T ≤ 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 let M be a constant such that |X(t)X−1(s)| ≤ M for 0 ≤ s ≤ t ≤ T , a simpleargument shows that

|x(t)| ≤ 2L MeβT e−β(s−t)|x(s)| if 0 ≤ t ≤ s.

Thus our assertion is proved with K = 2L MeβT .

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Exponential dichotomy and expansivity S175

However, as shown by an example in the appendix, it does not in general followthat X(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 forscalar 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 ofa scalar equation which is uniformly noncritical but does not have an exponentialdichotomy. Of course, their example also shows that in general (iii) does notimply (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 nonontrivial 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 hasexponential dichotomies on [0,∞) and (−∞, 0], and in addition has no nontrivialbounded solution. However the converse is not true. Consider, for example, thescalar equation x = a(t)x, where a(t) = 1 for t ≥ 1 and a(t) = −1 for t ≤ −1. Thishas exponential dichotomies on [0,∞) and (−∞, 0], and no nontrivial boundedsolution 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 choosethe projections P and Q for the dichotomies on [0,∞) and (−∞, 0] respectivelysuch that

PQ = Q P = P.

Proof. Suppose (1) has no nontrivial bounded solution. Let P be the projection forthe dichotomy on [0,∞) and Q that on (−∞, 0]. Then it follows from Coppel [1,p. 16] that R(P)={ξ : supt≥0 |X(t)ξ|<∞} and N (Q)={ξ : supt≤0 |X(t)ξ|<∞}.The nullspace of P can be any subspace complementary to R(P) and the rangeof Q can be any subspace complementary to N (Q). However, since (1) has nonontrivial bounded solution, R(P) ∩ N (Q) = {0} and so there is a subspace Vsuch that Rn = R(P) ⊕ V ⊕ N (Q). Thus we may take N (P) = V ⊕ N (Q),R(Q) = R(P) ⊕ V so that N (Q) ⊂ N (P), R(P) ⊂ R(Q). It follows thatPQ = Q P = P.

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S176 K.J. Palmer

Conversely, suppose we can choose the projections P and Q for the dichotomieson [0,∞) and (−∞, 0] respectively such that PQ = Q P = P. Let x(t) bea bounded solution of (1). Then x(0) ∈ R(P) ∩ N (Q) and so x(0) = Px(0) =PQx(0) = P · 0 = 0. So x(t) is the trivial solution. Thus the lemma is proved.

Remark 2. Elaydi and Hajek [2] consider a system (1) satisfying an exponentialtrichotomy. This holds if and only if (1) has exponential dichotomies on [0,∞)

and (−∞, 0] with respective projections P and Q satisfying R(P) + N (Q) = Rn

or equivalently if P and Q can be chosen so that PQ = Q P = Q. (This can beregarded as a kind of strong transversality condition.) They show that (1) has anexponential trichotomy if and only if its adjoint system has exponential dichotomieson [0,∞) and (−∞, 0] and no nontrivial bounded solution.

Lemma 2. Let (1) have exponential dichotomies on both [0,∞) and (−∞, 0] andno 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 α be the constants in both dichotomies. Suppose x(t) isa solution of (1). Then, as in the proof of Theorem 1, it follows for 0 ≤ a ≤ t ≤ band a ≤ t ≤ b ≤ 0 that

|x(t)| ≤ Ke−α(t−a)|x(a)| + Ke−α(b−t)|x(b)|.

Next note that if x(t) is a solution then

x(0) = Px(0) + (I − P)x(0) = PQx(0) + (I − P)x(0)

and so if a ≤ 0 ≤ b

|x(0)| ≤ K |Qx(0)| + |(I − P)x(0)|= K |X(0)QX−1(a)x(a)| + |X(0)(I − P)X−1(b)x(b)|≤ K2eαa|x(a)| + Ke−αb|x(b)|.

Next if a ≤ 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 if a ≤ 0 ≤ t ≤ b,

|x(t)| ≤ K3e−α(t−a)|x(a)| + (K2 + K )e−α(b−t)|x(b)|.

Similarly if a ≤ t ≤ 0 ≤ b,

|x(t)| ≤ (K + K3)e−α(t−a)|x(a)| + K2e−α(b−t)|x(b)|.

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Exponential dichotomy and expansivity S177

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] (conferalso Proposition 1 in Coppel [1, p. 14]) and its analogue for (−∞, 0] that (1) hasexponential dichotomies on both half-lines. Next let x(t) be a bounded solution of(1) and set ‖x‖ = sup−∞<t<∞ |x(t)|. Then for all t

|x(t)| ≤ θ sup|u−t|≤T

|x(u)| ≤ θ‖x‖.

Hence ‖x‖ ≤ θ‖x‖. It follows that ‖x‖ = 0 and so x(t) = 0 for all t. Thus (1) hasno nontrivial bounded solution and the proof of the theorem is complete.

5. Hull considerations and relation with Sacker-Sell theory

Now suppose A(t) is bounded and uniformly continuous on (−∞,∞). Then thehull H(A), which is defined as the set of matrix functions A(t) for which thereexists a sequence tk such that A(t + tk) → A(t) uniformly on compact intervals, iscompact 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 ifthe equation

x = A(t)x (2)

has no nontrivial bounded solution for all A ∈ H(A).

Proof. Suppose (1) is exponentially expansive on (−∞,∞). Then, by Theorem 2,(1) has exponential dichotomies on both half-lines and no nontrivial boundedsolution. Now if A ∈ H(A), there are just three possibilities: A is a translate of A,A is in the ω−limit set of A (that is, the sequence tk → ∞) or A is in the α−limitset of A (that is, the sequence tk → −∞). If A is a translate of A, then clearly (2)has no nontrivial bounded solution. If A is in the ω−limit set or α−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 boundedsolution.

Conversely, suppose (2) has no nontrivial bounded solution for all A ∈ H(A).Then it follows from Lemma 2 in Coppel [1, p. 75] that (1) is uniformly noncriticaland hence, by Theorem 2, exponentially expansive on (−∞,∞). Thus the proofof the theorem is complete.

Sacker and Sell [7] consider a linear skew product flow π(x, y, t) = (φ(x, y, t),σ(y, t)) on a product space X ×Y , where X is a finite-dimensional vector space and

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S178 K.J. Palmer

Y is a compact Hausdorff space. They assume the no nontrivial bounded solutionproperty, that is,

B = {(x, y) : sup−∞<t<∞

|φ(x, y, t)| < ∞} = {0} × Y.

We claim that the linear skew product flow π(x, y, t) satisfies the no nontrivialbounded 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| ≤ θ sup|t|≤T |φ(x, y, t)| for all (x, y).

Suppose the flow is uniformly noncritical and let (ξ, y) ∈ B. Then for all t,

|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)| ≤ θ sup−∞<t<∞ |φ(ξ, y, t)|. This implies that|φ(ξ, y, t)| = 0 for all t and so ξ = 0.

Conversely, suppose the no nontrivial bounded solution property holds butthe flow is not uniformly noncritical. Then there exist sequences ξk and yk with|ξk| = 1 such that 1 > 1

2 sup|t|≤k |φ(ξk, yk, t)| for all k. By compactness, we canassume without loss of generality that ξk → ξ and yk → y as k → ∞. It followsthat |ξ| = 1 and sup|t|<∞ |φ(ξ, y, t)| ≤ 2. This is a contradiction and so the claimfollows.

Notice that Theorem 3 is a consequence of this claim. Also it raises the followingquestion:

Question. Is uniform noncriticality equivalent to exponential expansivity, that is,the existence of positive L and α such that

|x| ≤ L[eαa|φ(x, y, a)| + e−αb|φ(x, y, b)|]for all x, y and a ≤ 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 byexponential expansivity as just defined. They define S = {(x, y) : |φ(x, y, t)| → 0as t → ∞} and U = {(x, y) : |φ(x, y, t)| → 0 as t → −∞}. In Lemma 5 in[7], they show there exist constants K ≥ 1 and α > 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 ≤ 0the inequality |φ(x, y, t)| ≤ K |x|eαt holds. An analogous result holds for singleequations.

Theorem 4. Suppose (1) has exponential dichotomies on both [0,∞) and (−∞, 0]and no nontrivial bounded solution. Then there exist positive constants L and α

such that if x(t) is a solution bounded on t ≥ 0 then

|x(t)| ≤ Le−α(t−s)|x(s)| for − ∞ < s ≤ t < ∞

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Exponential dichotomy and expansivity S179

and if x(t) is a solution bounded on t ≤ 0 then

|x(t)| ≤ Le−α(s−t)|x(s)| for − ∞ < t ≤ s < ∞.

Proof. Let the constants in the exponential dichotomies be K and α. Then it followsfrom 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)|]with L = K + K3. Suppose x(t) is a solution bounded on t ≥ 0. Then we maylet b → ∞ in the previous inequality to get |x(t)| ≤ Le−α(t−a)|x(a)| for a ≤ t.Similarly, if x(t) is a solution bounded on t ≤ 0, we may let a → −∞ in theinequality to get |x(t)| ≤ Le−α(b−t)|x(b)| for t ≤ 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 boundedsolution. 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 and α, then the projections Pand Q for (3) can be chosen to satisfy P Q = Q P = P and there are constants N,N1 depending only on K and α such that

|P − P| ≤ Nδ, |Q − Q| ≤ Nδ

and the constants in the dichotomies for (3) are 5K2[2 + 5K2 N1δ]/4, α − 2Kδ.

Proof. By Lemma 1, (1) has exponential dichotomies on [0,∞) and (−∞, 0] withrespective projections P and Q satisfying PQ = Q P = P. Let the constants inboth dichotomies be K and α. Then, since δ < α/4K2, it follows from Proposition 1in Coppel [1, p. 34] that (3) has an exponential dichotomy on [0,∞) with constants5K2/2, α − 2Kδ and projection P satisfying for all t ≥ 0,

|Y(t)PY−1(t) − X(t)PX−1(t)| ≤ δ1 = 4α−1K3δ,

where Y(t) is the fundamental matrix for (3) satisfying Y(0) = I . Similarly, (3) hasan exponential dichotomy on (−∞, 0] with constants 5K2/2, α− 2Kδ and projec-tion Q satisfying

|Y(t)QY−1(t) − X(t)QX−1(t)| ≤ δ1

for all t ≤ 0.

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S180 K.J. Palmer

Now it may not be true that P Q = Q P = P. We modify the projections sothat 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 orthogonalprojections such that P1 + P2 + P3 = I . Then, provided we use the Euclidean normin Rn , it follows from an obvious extension of Lemma 1 in Coppel [1, p. 39] thatthere is an invertible matrix S such that SP0

i S−1 = Pi for i = 1, 2, 3, where

P01 = diag

(Ik1 , 0, 0

), P0

2 = diag(0, Ik2 , 0

), P0

3 = diag(0, 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 forma basis for R(P), those of S2 a basis for V = N (P) ∩ R(Q) and those ofS3 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 = SP01 S−1, Q = S

(I − P0

3

)S−1.

Then P Q = Q P = P and since P S = SP01 = [PS1 0 0], (I − Q)S = SP0

3 =[0 0 (I − Q)S3], it follows that R(P) = R(P) and N (Q) = N (Q). Moreover

|P − P| = ∣∣SP01 S−1 − SP0

1 S−1∣∣

≤ |S − S|∣∣P01 S−1

∣∣ + ∣∣SP01

∣∣|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.It follows that |P − P| ≤ Nδ + δ1 = N1δ, |Q − Q| ≤ N1δ and then from

Coppel [1, p. 17] that (3) has exponential dichotomies on [0,∞) and (−∞, 0] withprojections P, Q respectively and constants 5K2[1 + (5K2/2)N1δ]/2, α − 2Kδ.

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Exponential dichotomy and expansivity S181

Remark 3. We can also show that given ε > 0, then if δ is sufficiently small theinequalities

|Y(t)PY−1(t) − X(t)PX−1(t)| ≤ ε for t ≥ 0

and

|Y(t)QY−1(t) − X(t)QX−1(t)| ≤ ε for t ≤ 0

hold.

To see this, using the same notation as in the proof of Theorem 5, first writeP(t) = Y(t)PY−1(t), P(t) = Y(t)PY−1(t), P(t) = X(t)PX−1(t). Note that P andP have the same range which implies that P P = P, P P = P and hence thatP(t)P(t) = P(t), P(t)P(t) = P(t). Then for t ≥ 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 = 25K4

4 [1 + 5K2

2 N1δ]2. Then |P(t) − P(t)| ≤ ε/2 for t ≥ T , whereT = 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 similarargument applied to the adjoint system, |Y−1(t)| ≤ MeMTδ for 0 ≤ t ≤ T . Thus for0 ≤ t ≤ T

|P(t) − P(t)| ≤ |Y(t)||P − P||Y−1(t)| ≤ M2e2MTδ N1δ < ε/2

if δ is sufficiently small.Finally for t ≥ 0

|P(t) − P(t)| ≤ |P(t) − P(t)| + |P(t) − P(t)| < ε/2 + δ1 < ε

if δ is sufficiently small. The inequality for t ≤ 0 is proved similarly.

Now we look at nonlinear systems. We show that if the variational equationalong a solution has an exponential dichotomy on both half-lines and no nontriv-ial bounded solution, then the nonlinear system has an exponential expansivityproperty 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

x = fx(t, x(t))x (5)

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S182 K.J. Palmer

has exponential dichotomies on both half-lines and no nontrivial bounded solution.Suppose also that as δ → 0

ω(δ) = sup{| fx(t, y) − fx(t, x(t))| : −∞ < t < ∞, |y − x(t)| ≤ δ} → 0.

Then there exist positive constants L, β and 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).

Then z(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 that z(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

0fx(t, x(t) + θz(t)) − fx(t, x(t)) dθ.

We note that for a ≤ t ≤ b,

|B(t)| ≤ ω(d).

We suppose d is so small that δ = ω(d) satisfies the conditions of Theorem 5. Nowwe 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 onboth half-lines and no bounded solution. Also we can choose the projections P, Qto satisfy PQ = Q P = P and the associated dichotomy constants K1 and β canbe chosen to depend only on the constants associated with the dichotomies for (5).Then it follows from Lemma 2 and its proof that for a ≤ t ≤ b,

|z(t)| ≤ L[e−β(t−a)|z(a)| + e−β(b−t)|z(b)|],

where L = K1 + K31 .

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Exponential dichotomy and expansivity S183

7. Appendix

In this appendix we give an example of a system on [0,∞) which is exponentiallyexpansive but does not have an exponential dichotomy.

The solution (x1(t), x2(t)) of the system

x1 = −x1 + 2etx2, x2 = 0

with x1(0) = ξ1, x2(0) = ξ2 is

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 systemhas an exponential dichotomy on [0,∞), the projection will have rank 1 with x(t)spanning the stable subspace and can be chosen so that y(t) spans the unstablesubspace. However, it is clear that the angle between x(t) and y(t) tends to zeroas 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 as x(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, wesee that we need to show that for 0 ≤ a ≤ t ≤ b and all λ

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 suppose λ ≥ 0. Then for all t ≥ 0, f(t) = et + λe−t and 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. Weconsider three cases:

Case 1. et + λe−t ≥ 1.Then we need to show that if 0 ≤ a ≤ t ≤ b and λ < 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

> et + λe−t e−2(b−t)

≥ et + λe−t .

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S184 K.J. Palmer

Case 2. −1 ≤ et + λe−t ≤ 1.Suppose eb +λe−b ≥ 1. Then we need to show that if 0 ≤ a ≤ t ≤ b and λ < 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 to b is −2λet−2b > 0. So if we canprove the inequality for b such that eb +λe−b = 1, it will follow for larger b. Hencewe may assume that

−1 ≤ et + λe−t ≤ eb + λe−b ≤ 1

and need to prove that if 0 ≤ a ≤ t ≤ b and λ < 0,

1 ≤ e−(t−a) f(a) + e−(b−t).

Next suppose ea + λe−a ≤ −1. Then we need to show that if 0 ≤ a ≤ t ≤ b andλ < 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 to a is −2e2a−t < 0. So if we can provethe inequality for a such that ea + λe−a = −1, it will follow for smaller a. Hencewe may assume that

−1 ≤ ea + λe−a ≤ eb + λe−b ≤ 1

and need to prove that if 0 ≤ a ≤ t ≤ b and λ < 0,

1 ≤ e−(t−a) + e−(b−t).

Now as a function of t in [a, b], the minimum value of the function on the righthand side is 2e−(b−a)/2. So we need to show that eb−a ≤ 4. However

−ea − e2a ≤ λ ≤ eb − e2b

and so(eb + ea)(eb − 1 − ea) = e2b − eb − ea − e2a ≤ 0.

Thus eb ≤ 1 + ea and so eb−a ≤ 2.

Case 3. et + λe−t ≤ −1.Then we need to show that if 0 ≤ a ≤ t ≤ b and λ < 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

−e2a−t − λe−t + e−(b−t) > −ete−2(t−a) − λe−t ≥ −et − λe−t .

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5. Massera, J.L., Schäffer, J.J.: Linear Differential Equations and Function Spaces. NewYork: Academic Press 1966

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