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### DIFFERENCE EQUATIONS

JES ´US RODRIGUEZ AND DEBRA LYNN ETHERIDGE Received 6 August 2004

We establish conditions for the existence of periodic solutions of nonlinear, second-order diﬀerence equations of the formy(t+ 2) +by(t+ 1) +cy(t)= f(y(t)), wherec=0 and f :RRis continuous. In our main result we assume that f exhibits sublinear growth and that there is a constant β >0 such thatu f(u)>0 whenever |u| ≥β. For such an equation we prove that ifNis an odd integer larger than one, then there exists at least one N-periodic solution unless all of the following conditions are simultaneously satisfied:

c=1,|b|<2, andNarccos1(b/2) is an even multiple ofπ.

1. Introduction

In this paper, we study the existence of periodic solutions of nonlinear, second-order, discrete time equations of the form

y(t+ 2) +by(t+ 1) +cy(t)= fy(t), t=0, 1, 2, 3,..., (1.1) where we assume thatbandcare real constants,cis diﬀerent from zero, and f is a real- valued, continuous function defined onR.

In our main result we consider equations where the following hold.

(i) There are constantsa1,a2, ands, with 0s <1 such that

f(u)a1|u|s+a2 uinR. (1.2)

(ii) There is a constantβ >0 such that

u f(u)>0 whenever|u| ≥β. (1.3) We prove that ifNis odd and larger than one, then the diﬀerence equation will have a N-periodic solution unless all of the following conditions are satisfied:c=1,|b|<2, and Narccos1(b/2) is an even multiple ofπ.

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As a consequence of this result we prove that there is a countable subsetSof [2, 2]

such that ifb /S, then

y(t+ 2) +by(t+ 1) +cy(t)= fy(t) (1.4) will have periodic solutions of every odd period larger than one.

The results presented in this paper extend previous ones of Etheridge and Rodriguez  who studied the existence of periodic solutions of diﬀerence equations under signifi- cantly more restrictive conditions on the nonlinearities.

2. Preliminaries and linear theory

We rewrite our problem in system form, letting x1(t)=y(t),

x2(t)=y(t+ 1), (2.1)

wheretis inZ+≡ {0, 1, 2, 3,...}. Then (1.1) becomes x1(t+ 1)

x2(t+ 1)

=

0 1

c b

x1(t) x2(t)

+ 0

fx1(t)

(2.2) fortinZ+. For periodicity of periodN >1, we must require that

x1(0) x2(0)

= x1(N)

x2(N)

. (2.3)

We cast our problem (2.2) and (2.3) as an equation in a sequence space as follows.

LetXN be the vector space consisting of allN-periodic sequencesx:Z+R2, where we use the Euclidean norm| · |onR2. For suchx, ifx =supt∈Z+|x(t)|, then (XN, · ) is a finite-dimensional Banach space.

The “linear part” of (2.2) and (2.3) may be written as a linear operatorL:XNXN, where for eachtZ+,

Lx(t)=

x1(t+ 1) x2(t+ 1)

A x1(t)

x2(t)

, (2.4)

the matrixAbeing

0 1

c b

. (2.5)

The “nonlinear part” of (2.2) and (2.3) may be written as a continuous functionF: XNXN, where fortZ+,

F(x)(t)= 0

fx1(t)

. (2.6)

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We have now expressed (2.2) and (2.3) in an equivalent operator equation form as

Lx=F(x). (2.7)

Following [4,5], we briefly discuss the purely linear problemsLx=0 andLx=h.

Notice thatLx=0 if and only if

x(t+ 1)=Ax(t) tinZ+,

x(0)=x(N), (2.8)

wherex(t) is inR2. But solutions of this system must be in the formx(t)=Atx(0), for t=1, 2, 3,..., where (IAN)x(0)=0. Accordingly, the kernel ofL (henceforth called ker(L)) consists of those sequences inXN for whichx(0)ker(IAN) and otherwise x(t)=Atx(0).

To characterize the image ofL(henceforth called Im(L)), we observe that ifhis an element ofXN, andx(t) is inR2for alltinZ+, thenhis an element of Im(L) if and only if x(t+ 1)=Ax(t) +h(t) tinZ+, (2.9)

x(0)=x(N). (2.10)

It is well known [1,6,7] that solutions of (2.9) are of the form x(t)=Atx(0) +Att

1 l=0

Al+11h(l) (2.11)

fort=1, 2, 3,.... For such a solution also to satisfy theN-periodicity condition (2.10), it follows thatx(0) must satisfy

IANx(0)=ANN 1 l=0

Al+11h(l), (2.12)

which is to say thatANNl=01(Al+1)1h(l) must lie in Im(IAN). Because Im(IAN)= [ker(IAN)T], it follows that if we construct matrixWby letting its columns be a basis for ker(I AN)T, then for h in XN, h is an element of Im(L) if and only if WTANNl=01(Al+1)1h(l)=0. See .

Following , we let

Ψ(0)=

ANTW, Ψ(l+ 1)=

Al+1TANTW forlinZ+. (2.13) Thenhis in Im(L) if and only ifNl=01ΨT(l+ 1)h(l)=0.

As will become apparent inSection 3, in which we construct the projectionsU and IEfor specific cases, it is useful to know that the columns ofΨ(·) span the solution space of the homogeneous “adjoint” problem

Lx=0, (2.14)

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whereL=XNXNis given by

Lx(t) =x(t+ 1)ATx(t) fortinZ+. (2.15) Further, this solution space and ker(L) are of the same dimension. See [4,5,9].

The proof appears in Etheridge and Rodriguez . One observes that x(t+ 1)= (AT)x(t) if and only ifx(t)=(AT)tx(0) and next, by direct calculation, that

Ψ(t+ 1)=

ATΨ(t). (2.16) Furthermore,

IATNΨ(0)=0 (2.17)

so thatΨ(0)=Ψ(N), whence the columns ofΨ(·) lie inXN. One then observes that, just as the dimension of ker(L) is equal to that of ker(IAN), the dimension of ker(L) is equal to that of ker(I(AT)N). The two matrices have kernels of the same dimension.

Our eventual aim is to analyze (2.7) using the alternative method [2,3,8,9,10,11,12]

and degree-theoretic arguments [3,12,13]. To begin, we will “split”XNusing projections U:XNker(L) andE:XNIm(L). The projections are those of Rodriguez . See also [4,5]. A sketch of their construction is given here.

Just as we let the columns ofWbe a basis for ker((IAN)T), we let the columns of the matrixVbe a basis for ker(IAN). Note that the dimensions of these two spaces are the same. LetCUbe the invertible matrixNl=01(AlV)T(AlV) andCIEthe invertible matrix N1

l=0 ΨT(l+ 1)Ψ(l+ 1). ForxinXN, define Ux(t)=AtVCU1N1

l=0

AlVTx(l), (2.18)

(IE)x(t)=Ψ(t+ 1)CI1E N1

l=0

ΨT(l+ 1)x(l) (2.19)

for eachtinZ+. Rodriguez  shows that these are projections which splitXN, so that XN=ker(L)Im(IU),

XN=Im(L)Im(IE), (2.20)

where

Im(E)=Im(L),

Im(U)=ker(L), (2.21)

the spaces Im(IE) and Im(U) having the same dimension.

Note that if we let ˜L be the restriction to Im(IU) of L, then ˜L is an invertible, bounded linear map from Im(IU) onto Im(E). If we denote byM the inverse of ˜L, then it follows that

(i)LMh=hfor allhin Im(L), (ii)MLx=(IU)xfor allxinXN, (iii)UM=0,EL=L, and (IE)L=0.

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3. Main results

We haveXN=ker(L)Im(IU). Letting the norms on ker(L) and Im(IU) be the norms inherited from XN, we let the product space ker(L)×Im(IU) have the max norm, that is,(u,v) =max(u,v).

Proposition3.1. The operator equationLx=F(x)is equivalent to vMEF(u+v)=0,

Q(IE)Fu+MEF(u+v)=0, (3.1)

whereuis inker(L)=Im(U),vIm(IU), andQmapsIm(IE)linearly and invertibly ontoker(L).

Proof.

Lx=F(x) (3.2)

⇐⇒

EL(x)F(x)=0,

(IE)LxF(x)=0 (3.3)

⇐⇒

L(x)EF(x)=0,

(IE)F(x)=0 (3.4)

⇐⇒

MLxMEF(x)=0,

Q(IE)F(x)=0 (3.5)

⇐⇒

x=Ux+MEF(x),

Q(IE)F(x)=0 (3.6)

⇐⇒

(IU)xMEF(x)=0,

Q(IE)FUx+MEF(x)=0. (3.7) Now, eachxinXNmay be uniquely decomposed asx=u+v, whereu=Uxker(L) andv=(IU)x. So (3.7) is equivalent to

vMEF(u+v)=0, (3.8)

Q(IE)Fu+MEF(u+v)=0. (3.9) By means of (2.18) and (2.19), we have split our operator equation (2.7);vMEF(u+ v) is in Im(IU), whileQ(IE)F(u+MEF(u+v)) is in Im(U).

Proposition 3.2. IfN is odd, c=0, and Narccos(b/2) is not an even multiple ofπ whenc=1and|b|<2, then eitherker(L)is trivial or bothker(L)andIm(IE)are one- dimensional. In the latter case, the projectionsUandIEand the bounded linear mapping

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Q(IE)may be realized as follows. ForxinXN, and for alltZ+ Ux(t)=

1 1

1

2N

N1

l=0

x1(l)

+

N1

l=0

x2(l)

, (3.10)

(IE)x(t)= c

1

1

Nc2+ 1

(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

. (3.11) Proof. The “homogeneous linear part” of our scalar problem (corresponding toLx=0) is

y(t+ 2) +by(t+ 1) +cy(t)=0, (3.12) where

y(0)=y(N), y(1)=y(N+ 1). (3.13) Calculations, detailed in the appendix of this paper, show that under the hypotheses of Proposition 3.2, the homogeneous linear part of our scalar problem has either only the trivial solutiony(t)=0 for alltinZ+or the constant solutiony(t)=1 for alltinZ+.

In the latter case, the constant function 1

1

(3.14) spans ker(L), so that for everytZ+,AtVof (2.18) may be taken to be

1 1

. (3.15)

Then

CU1= N

l=0

AlVTAlV 1

=

N1

l=0

[1, 1]

1 1

1

=(2N)1,

N1 l=0

AlVTx(l)=

N1 l=0

[1, 1]

x1(l) x2(l)

=

N1

l=0

x1(l)

+

N1

l=0

x2(l)

(3.16)

wheneverxis inXN. Therefore Ux(t)=

1 1

1

2N

N1

l=0

x1(l)

+

N1

l=0

x2(l)

(3.17)

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fortZ+, a constant multiple of

1 1

. (3.18)

In the appendix, we also show that under the hypotheses ofProposition 3.2, the homo- geneous adjoint problemLx=0 has either only the trivial solution or a one-dimensional solution space spanned by the constant function

c 1

. (3.19)

Therefore in (2.19), we may take

Ψ(t)= c

1

(3.20) for alltinZ+, so that

CIE1

=

N1

l=0

[c 1]

c 1

1

=

Nc2+ 11,

N1 l=0

ΨT(l+ 1)x(l)=

N1 l=0

[c 1]

x1(l) x2(l)

=(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

.

(3.21)

Therefore forxinXN, for alltZ+ (IE)x(t)=

c 1

1 Nc2+ 1

(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

, (3.22) a constant multiple of

c 1

. (3.23)

Furthermore, since Q must map Im(IE) linearly and invertibly onto ker(L)= Im(U), our simplest choice forQis as follows.

Each element of Im(IE) is of the form (3.11) for somexinXN. Now let Q(IE)x(t)=

1 1

1 Nc2+ 1

(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

, (3.24) fortinZ+. We notice thatQis clearly linear, bounded, and maps onto Im(U), and that

Q((IE)x)=0 if and only if (IE)x=0.

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Remark 3.3. In the case for which ker(L)= {0}, each ofUandIEis the zero projection onXN,Eis the identity onXN, andM isL1. Equation (3.9) then becomes trivial and (3.8) becomesL1F(v)=v, obviously equivalent to (2.7).

Theorem3.4. Suppose thatN3is odd,c=0, and f :RRis continuous. Assume also that

(i)there are nonnegative constants,a,˜ b, and˜ swiths <1such that|f(z)| ≤a˜|z|s+ ˜b for allzR,

(ii)there is a positive numberβsuch that for allz > β,f(z)>0andf(z)<0, (iii)whenc=1and|b|<2, thenNarccos(b/2)is not an even multiple ofπ.

Then there is at least oneN-periodic solution ofy(t+ 2) +by(t+ 1) +cy(t)=f(y(t)).

Proof. We have already seen that this scalar problem may be written equivalently as equa- tions of the form

0=Q(IE)Fu+MEF(u+v),

0=vMEF(u+v). (3.25)

Recall that our norm on ker(L)×Im(IU) is(u,v) =max{u,v}, whereuand vare, respectively, the norms onuandvas elements ofXN.

We defineH: ker(L)×Im(IU)ker(L)×Im(IU) by H(u,v)=

Q(IE)Fu+MEF(u+v) vMEF(u+v)

. (3.26)

We know that solving our scalar problem is equivalent to finding a zero of the continuous mapH.

We have shown that under the hypotheses of this theorem, ker(L) is either trivial or one-dimensional, and that when ker(L) is one-dimensional, it consists of the span of the constant function

1 1

. (3.27)

We will establish the existence of a zero ofH by constructing a bounded open subset, Ω, of ker(L)×Im(IU) and showing that the topological degree ofHwith respect toΩ and zero is diﬀerent from zero. We will do this using a homotopy argument.

The reader may consult Rouche and Mawhin  and the references therein as a source of ideas and techniques in the application of degree-theoretic methods in the study of nonlinear diﬀerential equations.

We writeH(u,v)=(IG)(u,v), whereIis the identity and G(u,v)=

uQ(IE)Fu+MEF(u+v) MEF(u+v)

. (3.28)

It is obvious that ifΩcontains (0, 0), then the topological degree ofIwith respect toΩ and zero is one.

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For 0τ1,

τH+ (1τ)I=τ(IG) + (1τ)I=IτG. (3.29) Therefore, if we can show that(IτG)(u,v)>0 for all (u,v) in the boundary ofΩ, then by the homotopy invariance of the Brouwer degree, it follows that the degree ofH with respect toΩand zero will be one, and consequentlyH(u,v)=(0, 0) for some (u,v) inΩ. Since, for 0τ1,

(IτG)(u,v)>(u,v)τG(u,v), (3.30) it suﬃces to show thatG(u,v)<(u,v)for all (u,v) in the boundary ofΩ.

We will let Ωbe the open ball in ker(L)×Im(IU) with center at the origin and radiusr, whereris chosen such thatr/2> β+ (2 ˜ars+ ˜b)(1 +ME). Observe that since 0< s <1, such a choice is always possible.

We will show that the second component function ofGmaps each boundary point of ΩintoΩitself and then, by breaking up the boundary ofΩinto separate pieces, consider the eﬀect of the first component function ofGon those pieces.

The pieces will be, respectively, those boundary elements (u,v) for whichu[r,r]

and those for whichu[0,r), where r=

2(β+ME(2 ˜ars+ ˜b))< r.

Observation 3.5. For (u,v)Ω, (i)F(u+v)2 ˜ars+ ˜b,

(ii)MEF(u+v)ME(2 ˜ars+ ˜b).

Proof. For (u,v)Ω,

F(u+v)=sup

t∈Z+

fu1(t) +v1(t)

sup

t∈Z+

a˜u1(t) +v1(t)s+ ˜b

2sa˜(u,v)s+ ˜b2 ˜ars+ ˜b.

(3.31)

This establishes (i), from which (ii) follows immediately.

Observation 3.6. If (u,v) is in the boundary ofΩ, thenMEF(u+v)< r.

Proof.

MEF(u+v)ME

2 ˜ars+ ˜b<1 +ME

2 ˜ars+ ˜b+β <r

2< r. (3.32) For convenience’s sake, we will letg(u,v)(t)=[MEF(u+v)]1(t) for eachtZ+. The functiong maps ker(L)×Im(IU) continuously intoR. Keep in mind that for eachu in ker(L), there is a uniquely determinedαfor whichuis the constant function

α 1

1

. (3.33)

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Observation 3.7. For (u,v) in the boundary of Ω, and for every lZ+, if α > β+ ME(2 ˜ars+ ˜b), then f(α+g(u,v)(l))>0, while if α <(β+ME(2 ˜ars+ ˜b)), then

f(α+g(u,v)(l))<0.

Proof. When (u,v) lies in the boundary ofΩandαβ+ME(2 ˜ars+ ˜b), we have for eachlZ+,

0< β=

β+ME

2 ˜ars+ ˜bME

2 ˜ars+ ˜b

αME

2 ˜ars+ ˜bαMEF(u+v)

αMEF(u+v)(l)αMEF(u+v)1(l)

=αg(u,v)(l)α+g(u,v)(l)

(3.34)

so that for eachl,f(α+g(u,v)(l))>0.

Similarly, if (u,v) lies in the boundary ofΩandα≤ −βME(2 ˜ars+ ˜b), then for eachlinZ+,

0>β= −

β+ME

2 ˜ars+ ˜b+ME

2 ˜ars+ ˜b

α+ME

2 ˜ars+ ˜bα+MEF(u+v)

α+MEF(u+v)(l)α+MEF(u+v)1(l)

=α+g(u,v)(l)α+g(u,v)(l)

(3.35)

so that for eachl,f(α+g(u,v)(l))<0.

Observation 3.8. If (u,v) is in the boundary ofΩ, uQ(IE)Fu+MEF(u+v)=

2α 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l), (3.36) where

u=α 1

1

. (3.37)

Proof. Since for alltinZ+,

F(x)(t)= 0

fx1(t)

, Fu+MEF(u+v)(t)=

0

fu1(t) +MEF(u+v)1(t)

=

0 fα+g(u,v)(t)

(3.38)

so that

u(t)Q(IE)F(u+v)(t)

=

α 1

Nc2+ 1

(c)(0) +

N1

l=0

fα+g(u,v)(l)

1

1

, (3.39)

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a constant function oft, hence

uQ(IE)Fu+MEF(u+v)=

2α 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l). (3.40) Observation 3.9. For (u,v) in the boundary ofΩ,uQ(IE)F(u+MEF(u+v))< r.

Proof. For (u,v) in the boundary ofΩ,(u,v) =max{u,v} =r. We consider first those elements of the boundary ofΩfor whichu[r,r], and then those for which u[0,r).

For (u,v) in the boundary ofΩ, u is in [r,r]=

2β+ME

2 ˜ars+ ˜b,r (3.41) if and only if

|α| is in β+ME

2 ˜ars+ ˜b,r 2

!

= r 2,r

2

!

. (3.42)

We consider the subcases (i)α >0 and (ii)α <0.

(i)

α is in β+ME

2ars+ ˜b,r 2

!

= r 2,r

2

!

. (3.43)

Then byObservation 3.7, we have f(α+g(u,v)(l))>0 for eachlinZ+, so that α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)< αr

2. (3.44)

To show that

α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)<r

2, (3.45)

it remains to show that 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l)< α+r

2. (3.46)

Now sinceβ+ME(2 ˜ars+ ˜b)α, and each f(α+g(u,v)(l)) in the sum just above is positive, it suﬃces to show that

1 c2+ 1

2 ˜ars+ ˜b<r

2+β+ME

2 ˜ars+ ˜b (3.47)

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or equivalently, that

2 ˜ars+ ˜b 1

c2+ 1ME

β < r

2. (3.48)

This follows, of course, from our having chosenrso that

r

2>2 ˜ars+ ˜b1 +ME

+β. (3.49)

(ii)

α is in r

2,βME

2 ˜ars+ ˜b!= r 2,r

2

!

. (3.50)

Then byObservation 3.7, we have f(α+g(u,v)(l))<0 for eachlZ+, so that α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)> α≥ −r

2. (3.51)

To show that

α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)<r

2, (3.52)

it remains to show that α

1 Nc2+ 1

N1 l=0

fα+g(u,v)(l)<r

2, (3.53)

or equivalently, to show that

1

Nc2+ 1 N1

l=0

fα+g(u,v)(l)<r

2α. (3.54)

Now sinceβME(2 ˜ars+ ˜b)α, so thatβ+ME(2 ˜ars+ ˜b)≤ −α, it suﬃces to show that

1

Nc2+ 1 N1

l=0

fα+g(u,v)(l)< β+ME

2 ˜ars+ ˜b+r

2. (3.55)

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Further, for eachlin the sum just above, f(α+g(u,v)(l)) is negative and

fα+g(u,v)(l)a˜α+g(u,v)(l)s+ ˜bar˜ s+ ˜b (3.56) so that

N1 l=0

fα+g(u,v)(l)=

N1 l=0

fα+g(u,v)(l)

N1 l=0

2 ˜ars+ ˜b, (3.57) hence,

1

Nc2+ 1 N1

l=0

fα+g(u,v)(l)≤ − 1

Nc2+ 1 N1

l=0

2 ˜ars+ ˜b

= 1

c2+ 1

2 ˜ars+ ˜b,

(3.58)

so that it suﬃces to show that 1

c2+ 1

2 ˜ars+ ˜b< β+ME

2 ˜ars+ ˜b+r

2. (3.59)

This, as we have seen in the proof of (i), follows from our choice ofr.

Finally, we consider those elements (u,v) of the boundary ofΩfor which

|α|< β+ME

2 ˜ars+ ˜b=r. (3.60) Clearly

α 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l)≤ |α|+ N Nc2+ 1

2 ˜ars+ ˜b

β+1 +ME

2 ˜ars+ ˜b

<r 2

(3.61)

so thatuQ(IE)F(u+MEF(u+v))< r.

For (u,v) in the boundary ofΩ,(u,v) =r. For each such (u,v), we have shown by means ofObservation 3.6, thatMEF(u+v)< r, and, by means ofObservation 3.9, that uQ(IE)F(u+MEF(u+v))< r. Therefore for such (u,v),G(u,v)<(u,v), so that no element of the boundary ofΩis a zero ofH; hence the degree ofHwith respect toΩand zero is 1, so that at least one solution of (2.7) exists insideΩ. Remark 3.10. If inTheorem 3.4, we change (ii) so that we require f(z)<0 andf(z)>0 for allz > β, the conclusions of the theorem still hold.

We let

=

"

2kπ

j :kandjare integers, 02k < jand jis odd

#

. (3.62)

(14)

It is easy to see that if arccos(b/2)/∆, then for any odd integerN,Narccos(b/2) cannot be an even multiple ofπ. It is also obvious thatS≡ {b: arccos(b/2)}is a countable subset of [2, 2]. The following result is now evident.

Corollary 3.11. Suppose f :RR is continuous,c=0, and the following conditions hold:

(i)there are constantsa,˜ b, and˜ s, with0s <1such that

f(u)a˜|u|s+ ˜b uinR, (3.63)

(ii)there is a constantβ >0, such thatu f(u)>0whenever|u| ≥β.

Then, if eitherbR\S, orc=1, then

y(t+ 2) +by(t+ 1) +cy(t)= fy(t) (3.64) will haveN-periodic solutions for every odd periodN >1.

Appendix

We demonstrate here that, providedNis odd,c=0, andNarccos(b/2) is not an even multiple ofπwhenc=1 and|b|<2, the kernel ofLis either trivial or is a one-dimensional space spanned by the constant function

1 1

. (A.1)

We show also that the solution space of the homogeneous adjoint problem (2.14) is, in the latter case, the span of the constant function

c 1

. (A.2)

Of course, if ker(L) is trivial, so is ker(L).

As we have seen,Lis invertible if and only if the matrixIANis invertible. That matrix is invertible if and only if no eigenvalue ofAis anNth root of unity. Those eigenvalues may be complex conjugates, real and repeated, or real and distinct. We will consider each of those three cases after we examine the kernels ofLand ofLin more detail than before.

(i) The kernel ofLconsists of all functionsxinXNfor whichx(t)=Atx(0) fortinZ+, where

A=

0 1

c b

(A.3)

of ISE

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