Báo cáo hóa học: "EXPONENTIAL DICHOTOMY OF DIFFERENCE EQUATIONS IN l p -PHASE SPACES ON THE HALF-LINE" docx

One can also use the results on exponential dichotomy of difference equations to obtain characterization of exponential dichotomy of evolution equations through the discretizing processes

IN lp-PHASE SPACES ON THE HALF-LINE

NGUYEN THIEU HUY AND VU THI NGOC HA

Received 14 November 2005; Revised 12 March 2006; Accepted 17 May 2006

For a sequence of bounded linear operators{ A n } ∞ n =0on a Banach spaceX, we investigate

the characterization of exponential dichotomy of the difference equations vn+1 = A n v n

We characterize the exponential dichotomy of difference equations in terms of the exis-tence of solutions to the equationsv n+1 = A n v n+f ninl p spaces (1≤ p < ∞) Then we

apply the results to study the robustness of exponential dichotomy of difference equa-tions

Copyright © 2006 N T Huy and V T N Ha This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis-tribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction and preliminaries

We consider the difference equation

x n+1 = A n x n, n ∈ N, (1.1) whereA n,n =0, 1, 2, , is a sequence of bounded linear operators on a given Banach

spaceX, x n ∈ X for n ∈ N.

One of the central interests in the asymptotic behavior of solutions to (1.1) is to find conditions for solutions to (1.1) to be stable, unstable, and especially to have an exponen-tial dichotomy (see, e.g., [1,5,7,12,16–20] and the references therein for more details

on the history of this problem) One can also use the results on exponential dichotomy

of difference equations to obtain characterization of exponential dichotomy of evolution equations through the discretizing processes (see, e.g., [4,7,9,18])

One can easily see that ifA n = A for all n ∈ N, then the asymptotic behavior of

solu-tions to (1.1) can be determined by the spectra of the operatorA However, the situation

becomes more complicated if{ A n } n ∈Nis not a constant sequence because, in this case, the spectra of each operatorA ncannot determine the asymptotic behavior of the solu-tions to (1.1) Therefore, in order to find the conditions for (1.1) to have an exponential dichotomy, one tries to relate the exponential dichotomy of (1.1) to the solvability of the

Hindawi Publishing Corporation

Advances in Di fference Equations

Volume 2006, Article ID 58453, Pages 1 14

DOI 10.1155/ADE/2006/58453

following inhomogeneous equation:

x n+1 = A n x n+f n, n ∈ N, (1.2)

in some certain sequence spaces for each given f = { f n } In other words, one wants to

relate the exponential dichotomy of (1.1) to the surjectiveness of the operatorT defined

by

(Tx) n:=x n+1 − A n x n forx =x n

belonging to a relevant sequence space (1.3)

In the infinite-dimensional case, in order to characterize the exponential dichotomy of (1.1) defined onN, beside the surjectiveness of the operatorT, one needs a priori

con-dition that the stable space is complemented (see, e.g., [5]) In our recent paper, we have replaced this condition by the spectral conditions of related operators (see [9, Corollary 3.3])

At this point, we would like to note that if one considers the difference equation (1.1) defined onZ, then the existence of exponential dichotomy of (1.1) is equivalent to the existence and uniqueness of the solution of (1.2) for a given f = { f n } n ∈Z, or, in other words, to the invertibility of the operatorT on suitable sequence spaces defined onZ This means that one can drop the above priori condition in the case that the difference equations are defined onZ(see [7, Theorem 3.3] for the original result and see also [2,

3,11,15] for recent results on the exponential dichotomy of difference equations defined

onZ)

However, if one considers the difference equation (1.1) defined only onN, then the situation becomes more complicated, because for a given f = { f n } n ∈N, the solutions of the difference equation (1.2) on Nare not unique even in the case that the difference equation (1.1) has an exponential dichotomy Moreover, one does not have any informa-tion on the negative half-lineZ−:= {z ∈ Z:z ≤0}of the difference equations (1.1) and (1.2) (we refer the readers to [8] for more details on the differences between the expo-nential dichotomy of the differential equations defined on the half-line and on the whole line) Therefore, one needs new ideas and new techniques to handle the exponential di-chotomy of difference equations defined only onN For differential equations defined on the half-line, such ideas and techniques have appeared in [14] (see also [8,13]) Those ideas and techniques have been exploited to obtain the characterization of exponential dichotomy of difference equations defined onNwithl∞-phase space of sequences de-fined onN(see [9]) As a result, we have obtained a necessary and sufficient condition for difference equations to have an exponential dichotomy This conditions related to the solvability of (1.2) inl∞spaces of sequences defined onN In the present paper, we will characterize the exponential dichotomy of (1.1) by the solvability of (1.2) in l p spaces (1≤ p < ∞) of sequences defined onN Moreover, we also characterize the exponential dichotomy by invertibility of a certain appropriate difference operator derived from the operatorT Consequently, we will use this characterization to prove the robustness of

ex-ponential dichotomy under small perturbations Our results are contained in Theorems

3.2,3.6,3.7, andCorollary 3.3

To describe more detailedly our construction, we will use the following notation: in this paper X denotes a given complex Banach space endowed with the norm  ·  As

usual, we denote byN,R,R +, andCthe sets of natural, real, nonnegative real, and com-plex numbers, respectively Throughout this paper, for 1≤ p < ∞we will consider the following sequence spaces:

l p(N,X) : =



v =v n

n ∈N:v n ∈ X :∞

n =0

v np

< ∞

 :=l p,

l0

p(N,X) : =v =v n

:v ∈ l p;v0=0

:=l0

p

(1.4)

endowed with the norm v  p:=(∞

n =0 v n  p)1/p Let{ A n } n ∈Nbe a sequence of bounded linear operators fromX to X which is

uni-formly bounded This means that there existsM > 0 such that  A n x  ≤ M  x for all

n ∈ Nandx ∈ X Next we define a discrete evolution family ᐁ =(U n,m)n ≥ m ≥0associated with the sequence{ A n } n ∈Nas follows:

U m,m =Id (the identity operator inX)

U n,m = A n −1A n −2··· A m forn > m. (1.5)

The uniform boundedness of{ A n } yields the exponential boundedness of the evolu-tion family (U n,m)n ≥ m ≥0 That is, there exist positive constantsK, α such that  U n,m x  ≤

Ke α(n − m)  x ; x ∈ X; n ≥ m ≥0

Definition 1.1 Equation (1.1) is said to have an exponential dichotomy if there exist a

family of projections (P n)n ∈NonX and positive constants N, ν such that

(1)A n P n = P n+1 A n;

(2)A n: kerP n →kerP n+1is an isomorphism and its inverse is denoted byA −1

| n; (3) U n,m x  ≤ Ne − ν(n − m)  x ; n ≥ m ≥0;x ∈ P m X;

(4) denoteU| m,n = A −1

| m A −1

| m+1 ··· A −1

| n −1;n > m, and U| m,m =Id, then

U| m,n x  ≤ Ne − ν(n − m)  x , n ≥ m ≥0;x ∈kerP n (1.6) The above family of projections (P n)n ∈Nis called the family of dichotomy projections

We define a linear operatorT as follows:

Ifu =u n

∈ l p set (Tu) n = u n+1 − A n u n forn ∈ N (1.7) Foru = { u n } ∈ l p, we have( Tu) n  =  u n+1 − A n u n  ≤  u n+1 +M  u n , hence Tu ∈ l p

and Tu  p ≤(1 +M)  u  p This means thatT is a bounded linear operator from l pinto

l p We denote the restriction ofT on l0

pbyT0, this means thatD(T0)= l0

pandT0u = Tu

foru ∈ l0

p From the definition ofT, the following are obvious.

Remark 1.2 (i) kerT = { u = { u n } ∈ l p:u n = U n,0 u0,n ∈ N}.

(ii) It is easy to verify thatT0is injective Indeed, letu = { u n }, v = { v n } ∈ l0

pandT0u =

T0v Then we have u0= v0=0,u1=(T0v)0= v1,u2= A1u1+ (T0u)1= A1v1+ (T0v)1=

v2, , u n+1 = A n u n+ (T0u) n = A n v n+ (T0v) n = v n+1, for alln ∈ N Hence, u = v.

Recall that for an operatorB on a Banach space Y, the approximate point spectrum Aσ(B) of B is the set of all complex numbers λ such that for every  > 0, there exists

y ∈ D(B) with  y  =1 and( λ − B)y  ≤ .

In order to characterize the exponential stability and dichotomy of an evolution family,

we need the concept ofl p-stable spaces defined as follows

Definition 1.3 For a discrete evolution familyᐁ=(U m,n)m ≥ n ≥0, m,n ∈ N, on Banach

spaceX and n0∈ N, define the l p -stable space X0(n0) by

X0

n0 :=

x ∈ X : ∞

n = n0

U n,n

0xp < ∞

An orbitU m,n0x for m ≥ n0≥0 andx ∈ X0(n0) is called anl p-stable orbit

2 Exponential stability

In this section we will give a sufficient condition for stability of l p-stable orbits of a dis-crete evolution familyᐁ The obtained results will be used in the next section to charac-terize the exponential dichotomy of (1.1)

Theorem 2.1 Let the operator T0defined as above satisfy the condition 0 ∈ Aσ(T0) Then

every l p -stable orbit of ᐁ is exponentially stable Precisely, there exist positive constants N, ν

such that for any n0∈ N and x ∈ X0(n0),

U n,n

0x  ≤ Ne − ν(n − s)U s,n

0x, n ≥ s ≥ n0. (2.1)

Proof Since 0 ∈ Aσ(T0), we have that there exists a constantη > 0 such that

ηT0v

p ≥  v  p forv ∈ l0

To prove (2.1), we first prove that there is a positive constantl such that for any n0∈ N

andx ∈ X0(n0),

U n,n

0x  ≤ lU s,n

0x, n ≥ s ≥ n0≥0. (2.3) Fixn0∈ N, x ∈ X0(n0), ands ≥ n0 Taking

v =v n

withv n:=

⎧

⎨

⎩U n,n0x for n > s,

we havev ∈ l0

p By definition ofT0, we have (T0v) n = v n+1 − A n v n This yields

T0v n =

⎧

⎪

⎨

⎪

⎩

0 forn ≤ s −1,

U s+1,n0x for n = s,

0 forn > s.

(2.5)

By inequality (2.2), we have

ηU s+1,n

0x  ≥∞

k = s

U k,n

0xp 1/p

≥U n,n

0x forn > s ≥ n0. (2.6)

Hence,

U n,n

0x  ≤ ηU s+1,n

0x  = ηU s+1,s U s,n

0x  ≤ ηMU s,n

0x forn > s ≥ n0. (2.7) Puttingl =max{1,ηM }, we obtain (2.3)

We now show that there is a numberK = K(η,l) > 0 such that for any n0∈ Nand

x ∈ X0(n0),

U s+n,n

0x  ≤1

2U s,n

0x forn ≥ K, s ≥ n0. (2.8)

To prove (2.8), putu n:=U n,n0x, n ≥ n0, and leta < b be two natural numbers with a ≥ n0

such that u b  > 1/2  u a  From (2.3), we obtain that

lu a  ≥  u n> 1

2lu a fora ≤ n ≤ b. (2.9) Put now

v =v n

withv n =

⎧

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎩

0 for 0≤ n ≤ a,

u n n



k = a+1

1

u k fora + 1 ≤ n ≤ b,

u n b+1

k = a+1

1

u k forn ≥ b + 1.

(2.10)

Thenv ∈ l0

p By definition ofT0, we have

T0v =

T0v n with

T0v n =

⎧

⎪

⎪

⎪

⎪

0, for 0≤ n < a,

u n+1

u n+1 fora ≤ n ≤ b −1,

0 forn ≥ b.

(2.11)

By inequality (2.2), we obtain

η(b − a)1/p ≥  v  p (2.12) Using H¨older inequality forv and χ[a+1,b], where

χ[a+1,b] n =

⎧

⎨

⎩1 fora + 1 ≤ n ≤ b,

we have that

b



n = a+1

v n  ≤(b − a)1−1/p  v  p (2.14) Substituting this into inequality (2.12), we obtain that

η(b − a) ≥

b



n = a+1

Using now the estimates (2.9), we have

η(b − a) ≥ b

n = a+1

v n  = b

n = a+1

n



k = a+1

u n

u k

≥ b



n = a+1

1

2lu a n

k = a+1

1

lu a =(b − a)(b4l2− a + 1) >(b − a)2

4l2 .

(2.16)

This yieldsb − a < 4ηl2 PuttingK : =4ηl2, the inequality (2.1) follows

We finish by proving (2.1) Indeed, ifn ≥ s ≥ n0∈ Nwritingn − s = n1K + r for 0 ≤

r < K, and n1∈ N, we have

U n,n

0x  =  U n − s+s,n0x  =  U n1K+r+s,n0x

by ( 2.8 )

2n1

U r+s,n

0x by ( 2.3 )

2n1

U s,n

0x  ≤2le −((n − s)/K)ln2U s,n

0x. (2.17) TakingN : =2l and ν : =ln 2/K, the inequality (2.1) follows  From this theorem, we obtain the following corollary

Corollary 2.2 Under the conditions of Theorem 2.1, the space X0(n0) can be expressed as

X0

n0 =x ∈ X :U n,n

0x  ≤ Ne − ν(n − n0 ) x ; n ≥ n0≥0

for certain positive constants N, ν Hence, X0(n0) is a closed linear subspace of X.

3 Exponential dichotomy and perturbations

We will characterize the exponential dichotomy of (1.1) by using the operatorsT0,T In

particular, we will also get necessary and sufficient conditions for exponential dichotomy

in Hilbert spaces and finite-dimensional spaces Moreover, using our characterization of the exponential dichotomy, we can prove the robustness of the exponential dichotomy

of (1.1) under small perturbations Then we start with the following lemma which has a history that can be traced back to [14, Lemma 4.2] and to [6] and beyond

Lemma 3.1 Assume that ( 1.1) has an exponential dichotomy with corresponding family of projections P n , n ≥ 0, and constants N > 0, ν > 0, then M : =supn ≥0 P n  < ∞

Proof The proof is done in [9, Lemma 3.1] We present it here for sake of completeness Fixn0> 0, and set P0:=P n0;P1:=Id−P n0,X k = P k X, k =0, 1 Setγ0:=inf{x0+x1:

x k ∈ X k, x0 =  x1 =1} Ifx ∈ X and P k x =0,k =0, 1, then

γ n0≤



 P0x

P0x+ P1x

P1x ≤P10xP0x +P0x

P1xP1x





≤P10xx +P0x − P1x

P1x P1x



x

P0(x).

(3.1)

Hence, P0 ≤2/γ n0 It remains to show that there is a constantc > 0 (independent of n0) such thatγ n0≥ c For this, fix x k ∈ X k,k =0, 1, with x k  =1 By the exponential bound-edness ofᐁ, we have U n,n0(x0+x1) ≤Ke α(n − n0 ) x0+x1forn ≥ n0and constantsK,

α ≥0 Thus,

x0+x1 ≥ K −1e − α(n − n0 ) U n,n

0x0+U n,n0x1 

≥ K −1e − α(n − n0 )

N −1e ν(n − n0 )− Ne − ν(n − n0 ) =: c n − n0, (3.2) and henceγ n0≥ c n − n0 Obviouslyc m > 0 for m sufficiently large Thus 0 < c m ≤ γ n0  Now we come to our first main result It characterizes the exponential dichotomy of (1.1) by properties of the operatorT.

Theorem 3.2 Let { A n } n ∈N be a family of bounded linear and uniformly bounded operators

on the Banach space X Then the following assertions are equivalent.

(i) Equation ( 1.1) has an exponential dichotomy.

(ii)T is surjective and X0(0) is complemented in X.

Proof (i) ⇒(ii) Let ( P n)n ≥0be the family of dichotomy projections ThenX0(0)= P0X,

and henceX0(0) is complemented If f ∈ l p, definev = { v n } n ∈Nby

v n =

⎧

⎪

⎪

⎨

⎪

⎪

⎩

n



k =1

U n,k P k f k −1−

∞



k = n+1 U| n,k

Id−P k f k −1 forn ≥1,

−

∞



k =1

U|0,k Id−P k f k −1 forn =0,

(3.3)

thenv n+1 = A n v n+f n Moreover, since







n



k =1

U n,k P k f k −1−

∞



k = n+1 U| n,k

Id−P k f k −1





 ≤ N∞

k =1

e − ν | n − k |f k −1 (3.4) and f ∈ l p, we can easily derive thatv ∈ l p By the definition ofT, we have Tv = f

There-foreT : l p → l pis surjective

(ii)⇒(i) We prove this in several steps.

(A) LetZ ⊆ X be a complement of X0(0) inX, that is, X = X0(0)⊕ Z Set X1(n) =

U n,0 Z Then

U n,s X0(s) ⊆ X0(n), U n,s X1(s) = X1(n), n ≥ s ≥0. (3.5) (B) There are constantsN,ν > 0 such that

U n,0 x  ≥ Ne ν(n − s)U s,0 x forx ∈ X1(0), n ≥ s ≥0. (3.6)

In fact, letY : = {( v n)n ∈N ∈ l p:v0∈ X1(0)}endowed withl p-norm ThenY is a closed

subspace of the Banach spacel p, and henceY is complete ByRemark 1.2, we have kerT : = { v ∈ l p:v n = U n,0 x for some x ∈ X0(0)} SinceX = X0(0)⊕ X1(0) andT is surjective, we

obtain that

is bijective and hence an isomorphism Thus, by Banach isomorphism theorem, there is

a constantη > 0 such that

η  Tv  p ≥  v  p, forv ∈ Y. (3.8)

To prove (3.6), we first prove that there is a positive constantl such that

U n,0 x  ≥ lU s,0 x forx ∈ X1(0),n ≥ s ≥0,n,s ∈ N . (3.9) Indeed, fixx ∈ X1(0),x =0, andn ≥ s ≥0 Ifn =0, there is nothing to do So, assume thatn ≥1 Now taking

v : =v m

withv m:=

⎧

⎨

⎩U m,0 x for 0 ≤ m ≤ n −1,

we have thatv ∈ Y Then, by definition of T, we obtain that

(Tv) m:=

⎧

⎪

⎨

⎪

⎩

0 form > n −1,

− U n,0 x for m = n −1,

0 form < n −1.

(3.11)

Inequality (3.8) yields

ηU n,0 x  ≥⎛⎝n−1

k =0

U k,0 xp

⎞

⎠

1/p

≥U s,0 x  ∀0≤ s ≤ n −1. (3.12) Putting nowl : =min{1/η,1 }, inequality (3.9) follows

We now show that there is a numberK = K(η,l) > 0 such that

U s+n,0 x  ≥2U s,0 x forn ≥ K, s ≥0;x ∈ X1(0). (3.13) Let 0= x ∈ X1(0), setu n:=U n,0 x, n ≥0 ByRemark 1.2we haveu n =0 for alln ≥0 To prove (3.13), leta < b be two natural numbers such that  u b  < 2  u a  From (3.9), we obtain that

2

lu a>u n  ≥ lu a  ∀ a ≤ n ≤ b. (3.14) Take nowv = { v n }, where

v n =

⎧

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎩

− u n

b



k = a+1

1

u k for 0≤ n < a,

− u n b



k = n+1

1

u k for a ≤ n < b,

(3.15)

Then,v ∈ Y By definition of T, we have that

(Tv) n =

⎧

⎪

⎪

⎪

⎪

0 for 0≤ n < a,

u n+1

u n+1 fora ≤ n < b,

0 forn ≥ b.

(3.16)

By inequality (3.8), we obtain

η(b − a)1/p ≥  v  p (3.17) Using H¨older inequality forv and χ[a,b −1], where

χ[a,b −1] n =

⎧

⎨

⎩

1 fora ≤ n ≤ b −1,

we have that

b−1

n = a

v n  ≤(b − a)1−1/p  v  p (3.19) Substituting this into inequality (3.17), we obtain that

η(b − a) ≥ b −

1



n = a

Using now the estimates (3.14), we have

η(b − a) ≥

b−1

n = a

v n  = b−1

n = a

b



k = n+1

u n

u k

≥

b−1

n = a lu a b

k = n+1

l

2u a  = l2(b − a)(b4 − a + 1) > l2(b −4 a)2.

(3.21)

This yieldsb − a < 4η/l2 PuttingK : =4η/l2, the inequality (3.13) follows

We finish this step by proving inequality (3.6) Indeed, ifn ≥ s ∈ N, writing n − s =

n0K + r for 0 ≤ r < K, and n0∈ N, we have

U n,0 x  =  U n − s+s,0 x  =  U n0K+r+s,0 x

by ( 3.13 )

≥ 2n0 U r+s,0 x by ( 3.9 )

≥ l2 n0 U s,0 x  ≥ l

2e((n − s)/K)ln2U s,0 x. (3.22) TakingN : = l/2 and ν : =ln 2/K, inequality (3.6) follows

(C)X = X0(n) ⊕ X1(n), n ∈ N.

LetY ⊂ l pbe as in (B) Then byRemark 1.2, we have thatl0

p ⊂ Y From this fact and

(3.8), we obtain thatη  T0v  l p ≥  v  l p, forv ∈ l0

p Thus,

The relation (3.23) andCorollary 2.2imply thatX0(n) is closed From (3.5), (3.6), and the closedness ofX1(0), we can easily derive thatX1(n) is closed and X1(n) ∩ X0(n) = {0}

forn ≥0

Finally, fixn0> 0, and x ∈ X (note that we already have X = X0(0)⊕ X1(0)) For a natural numbern1> n0+ 1, set

v =v n

withv n =

⎧

⎪

⎨

⎪

⎩

0 for 0≤ n < n0,

n − n0+ 1 U n,n0x for n0≤ n ≤ n1,

f =f n

with f n =

⎧

⎪

⎪

⎪

⎪

⎪

⎪

0 for 0≤ n < n0,

U n+1,n0x forn0≤ n < n1,

− n1− n0+ 1 U n+1,n0x for n = n1,

(3.24)

Thenv, f ∈ l pand satisfy (1.2) for alln ≥ n0> 0 By assumption, there exists w ∈ l psuch thatTw = f By the definition of T, w nis a solution of (1.2) Thus,

v n − w n = U n,n

v n − w n = U n,n

x − w n , n ≥ n0. (3.25)

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v n = < /p>

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