PERIODIC SOLUTIONS OF NONLINEAR SECOND-ORDER
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 form y(t+2)+by(t+1)+cy(t)=f(y(t)), where c= 0and
f:R→Ris continuous. In our main result we assume that fexhibits sublinear growth
and that there is a constant β>0suchthatuf(u)>0whenever|u|≥β.Forsuchan
equation we prove that if Nis 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, and Narccos−1(−b/2) is an even multiple of π.
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)
whereweassumethatband care real constants, cis different from zero, and fisareal-
valued, continuous function defined on R.
In our main result we consider equations where the following hold.
(i) There are constants a1,a2,ands,with0≤s<1suchthat
f(u)
≤a1|u|s+a2∀uin R.(1.2)
(ii) There is a constant β>0suchthat
uf(u)>0whenever|u|≥β. (1.3)
We prove th at if Nis 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
Narccos−1(−b/2) is an even multiple of π.
Copyright ©2005 Hindawi Publishing Corporation
Advances in Difference Equations 2005:2 (2005) 173–192
DOI: 10.1155/ADE.2005.173
174 Periodic solutions of nonlinear second-order difference equations
As a consequence of this result we prove that there is a countable subset Sof [−2,2]
such that if b/∈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
[4] 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)
where tis in Z+≡{0,1,2,3, ...}.Then(
1.1)becomes
x1(t+1)
x2(t+1)
=01
−c−bx1(t)
x2(t)+0
fx1(t)(2.2)
for tin Z+. For periodicity of period N>1, we must require that
x1(0)
x2(0)=x1(N)
x2(N).(2.3)
We cast ou r problem ( 2.2)and(2.3) as an equation in a sequence space as follows.
Let XNbe the vector space consisting of all N-periodic sequences x:Z+→R2,where
we use the Euclidean norm |·|on R2.Forsuchx,ifx=supt∈Z+|x(t)|,then(XN,·)
is a finite-dimensional Banach space.
The “linear part” of (2.2)and(2.3)maybewrittenasalinearoperatorL:XN→XN,
where for each t∈Z+,
Lx(t)=x1(t+1)
x2(t+1)
−Ax1(t)
x2(t), (2.4)
the matrix Abeing
01
−c−b.(2.5)
The “nonlinear part” of (2.2)and(2.3) may be written as a continuous function F:
XN→XN,wherefort∈Z+,
F(x)(t)=0
fx1(t).(2.6)
J. Rodriguez and D. L. Etheridge 175
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 problems Lx =0andLx =h.
Notice that Lx =0ifandonlyif
x(t+1)=Ax(t)∀tin Z+,
x(0) =x(N), (2.8)
where x(t)isinR2. But solutions of this system must be in the form x(t)=Atx(0), for
t=1,2,3,...,where(I−AN)x(0) =0. Accordingly, the kernel of L(henceforth called
ker(L)) consists of those sequences in XNfor which x(0)∈ker(I−AN) and otherwise
x(t)=Atx(0).
To ch a r ac te r iz e t h e im a g e of L(henceforth called Im(L)), we observe that if his an
element of XN,andx(t)isinR2for all tin Z+,thenhis an element of Im(L)ifandonlyif
x(t+1)=Ax(t)+h(t)∀tin Z+, (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) + At
t−1
l=0Al+1−1h(l) (2.11)
for t=1,2,3,.... For such a solution also to satisfy the N-periodicity condition (2.10), it
follows that x(0) must satisfy
I−ANx(0) =AN
N−1
l=0Al+1−1h(l), (2.12)
which is to say that ANN−1
l=0(Al+1)−1h(l) must lie in Im(I−AN). Because Im(I−AN)=
[ker(I−AN)T]⊥, it follows that if we construct matrix Wby letting its columns be a basis
for ker(I−AN)T,thenforhin XN,his an element of Im(L)ifandonlyif
WTANN−1
l=0(Al+1)−1h(l)=0. See [4].
Following [4], we let
Ψ(0) =ANTW,
Ψ(l+1)=Al+1−TANTWfor lin Z+.(2.13)
Then his in Im(L)ifandonlyifN−1
l=0ΨT(l+1)h(l)=0.
As will become apparent in Section 3, in which we construct the projections Uand
I−Efor specific cases, it is useful to know that the columns of Ψ(·) span the solution
space of the homogeneous “adjoint” problem
Lx=0, (2.14)
176 Periodic solutions of nonlinear second-order difference equations
where
L=XN→XNis given by
Lx(t)=
x(t+1)−A−Tx(t)fortin Z+.(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 [4]. One observes that x(t+1)=
(A−T)x(t)ifandonlyifx(t)=(A−T)tx(0) and next, by direct calculation, that
Ψ(t+1)=A−TΨ(t).(2.16)
Furthermore,
I−A−TNΨ(0) =0 (2.17)
so that Ψ(0) =Ψ(N), whence the columns of Ψ(·) lie in XN. One then observes that, just
as the dimension of ker(L) is equal to that of ker(I−AN), the dimension of ker(
L)isequal
to that of ker(I−(A−T)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:XN→ker(L)andE:XN→Im(L). The projections are those of Rodriguez [9]. See also
[4,5]. A sketch of their construction is given here.
Just as we let the columns of Wbe a basis for ker((I−AN)T), we let the columns of the
matrix Vbe a basis for ker(I−AN). Note that the dimensions of these two spaces are the
same. Let CUbe the invertible matrix N−1
l=0(AlV)T(AlV)andCI−Ethe invertible matrix
N−1
l=0ΨT(l+1)Ψ(l+1).Forxin XN,define
Ux(t)=AtVC−1
U
N−1
l=0AlVTx(l), (2.18)
(I−E)x(t)=Ψ(t+1)C−1
I−E
N−1
l=0
ΨT(l+1)x(l) (2.19)
for each tin Z+.Rodriguez[
9] shows that these are projections which split XN,sothat
XN=ker(L)⊕Im(I−U),
XN=Im(L)⊕Im(I−E), (2.20)
where
Im(E)=Im(L),
Im(U)=ker(L), (2.21)
the spaces Im(I−E)andIm(U) having the same dimension.
Note that if we let ˜
Lbe the restriction to Im(I−U)ofL,then˜
Lis an invertible,
bounded linear map from Im(I−U)ontoIm(E). If we denote by Mthe inverse of ˜
L,
then it follows that
(i) LMh =hfor all hin Im(L),
(ii) MLx =(I−U)xfor all xin XN,
(iii) UM =0, EL =L,and(I−E)L=0.
J. Rodriguez and D. L. Etheridge 177
3. Main results
We have XN=ker(L)⊕Im(I−U). Letting the norms on ker(L)andIm(I−U)bethe
norms inherited from XN,welettheproductspaceker(L)×Im(I−U)havethemax
norm, that is, (u,v)=max(u,v).
Proposition 3.1. The operator equation Lx =F(x)is equivalent to
v−MEF(u+v)=0,
Q(I−E)Fu+MEF(u+v)=0, (3.1)
where uis in ker(L)=Im(U),v∈Im(I−U),andQmaps Im(I−E)linearly and invertibly
onto ker(L).
Proof.
Lx =F(x) (3.2)
⇐⇒
EL(x)−F(x)=0,
(I−E)Lx −F(x)=0(3.3)
⇐⇒
L(x)−EF(x)=0,
(I−E)F(x)=0(3.4)
⇐⇒
MLx −MEF(x)=0,
Q(I−E)F(x)=0(3.5)
⇐⇒
x=Ux+MEF(x),
Q(I−E)F(x)=0(3.6)
⇐⇒
(I−U)x−MEF(x)=0,
Q(I−E)FUx+MEF(x)=0.(3.7)
Now, each xin XNmaybeuniquelydecomposedasx=u+v,whereu=Ux ∈ker(L)
and v=(I−U)x.So(
3.7)isequivalentto
v−MEF(u+v)=0, (3.8)
Q(I−E)Fu+MEF(u+v)=0.(3.9)
By means of (2.18)and(2.19), we have split our operator equation (2.7); v−MEF(u+
v)isinIm(I−U), while Q(I−E)F(u+MEF(u+v)) is in Im(U).
Proposition 3.2. If Nis odd, c= 0,andNarccos(−b/2) is not an even multiple of π
when c=1and |b|<2, then either ker(L)is trivial or both ker(L)and Im(I−E)are one-
dimensional. In the latter case, the projections Uand I−Eand the bounded linear mapping