Advances in Pure Mathematics, 2011, 1, 118-127
doi:10.4236/ apm.2011.14024 Published Online July 2011 (http://www.SciRP.org/journal/apm)
Copyright © 2011 SciRes. APM
Discrete Pseudo Almost Periodic Solutions for Some
Difference Equations
Elhadi Ait Dads*, Khalil Ezzinbi, Lahcen Lhachimi
University Cadi Ayyad, Faculty of Sciences Semlalia, Department of Mathematics, Marrakesh, Morocco
E-mail: *eaitdads@gmail.com, ezzinbi@ucam.ac.ma, lllahcen@gmail.com
Received February 23, 2011; revised April 26, 2011; accepted May 10, 2011
Abstract
In this work, we study the existence and uniqueness of pseudo almost periodic solutions for some difference
equations. Firstly, we investigate the spectrum of the shift operator on the space of pseudo almost periodic
sequences to show the main results of this work. For the illustration, some applications are provided for a
second order differential equation with piecewise constant arguments.
Keywords: Difference Equations, Pseudo Almost Periodic Sequences, Schift Operator
1. Introduction
Difference equations have many applications in popula-
tions dynamics, they are used to describe the evolution of
many phenomena over the course of time. For example,
if a certain population has discrete generations, the size
of the th generation is a function of the
th generation
(1)n(1)xn
n()
x
n. This relation expresses itself in
the following difference equatio n
 
1= ,xnf xnn. (1)
The discrete processes occur in the investigation of
many phenomena, mainly in the case of use of computers.
One of the most widely adopted definition of a discrete
process can be formulated as follows: a discrete process
is a map from the additive group of the integers , into
a complete metric space
(,)
X
d, such as or
with the distance function induced by the vector norm.
m
 m
We use two different notations to designate a discrete
process, namely, if :
f
X


n
fn
is a discrete process, we
shall write instead
or

, dropping
nn
f
usually the subscript “ n”, since no confusion can
occur (indeed, we are not going to consider in this work
discrete processes defined o n a group, othe r t han ).
Of course, one of the most common sources for the
discrete processes is the theory of difference equations,
such as
1=,
nnn
xAxbn
, (2)
where n
x
stands for the unknown process, with values
in or
m
 m
A
is a square matrix of order with
real or complex entries and n
b stands for a given
discrete process, with values in the same space as
m
n
x
.
In practice, we deal with solutions of (2) which are only
defined on subsets of , and therefore, they might be
regarded as restrictions of a “complete” process to a
subset of its domain of definition.
1
m

Difference equations and discrete dynamic systems
represent two sides of the same coin. For instance, when
mathematicians talk about difference equations, they
usually refer to the analytic theory of the subject, and
when they talk about discrete dynamic systems, they
generally refer to its geometrical and topological aspects.
More sophisticated equations (or systems) than (2) are
those described by the following discrete eq uation

=,n
m

,
nn
xfxn (3)
where (or ) is a given map, in
general nonlinear in both arguments.
:fm

Another example, let
n be the size of a
population at time . If n
is the rate of growth of the
population from one generation to another, then we may
consider a mathematical model in the form

yn
1=, >0yn
. (4)
If the initial population is given by

0= 0,
y
y
>1, then
the solutions are given by . If

0
n
yn y
=
then
yn increases infinitely, and If

= .lim
n yn =1,
then
0
=
y
ny for all , which means that the
size of the population is constan t for the indefinite future.
>0n
E. A. DADS ET AL.
119
However, for <1,
we have and the

lim=0,
nyn

population eventually becomes extinct.
Since our main objective is to provide a criteria to get
the existence of a pseudo almost periodic solution for
equations of the form (2) or (3), we shall first review the
basic properties of pseudo almost periodic discrete
processes.
This work is motivated by the results obtained in [1,2],
and the main results would be some extension for some
well-established results in the literature, more details can
be found in [3].
This work is organized as follows. In Section , we
consider geometrical properties of the shift operator in
general case and, we deal with the properties of shift
operator the spaces of almost periodic and on ergodic
sequences. In Section 3, we a consider the existence and
uniqueness solutions of some difference equations using
polynomial functions. In the last section, we deal with
the application of the previous results to some second
order differential equation with a piecewise constant
argument.
2
2. Shift Operator Acting on the Space of
Pseudo Almost Periodic Sequences
In this section, we give some properties on pseudo
almost periodic sequences that will be used in this work.
For more details in this connexion, the reader will see
[4-12].
Definition 2.1: A sequence
nn
x with values in
is called almost periodic if for all
m
>0,
the set

,: :,<
nn
Txfor all nxx
  
is rela-
tively dense.
The space of almost periodic sequences is denoted by
. If we use the notation Let
(, )
m
AP  =1,m().AP
(, )B
denote the space of bounded complex sequen-
ces provide with the supremum norm. 0 denote
the space of bounded complex sequences
()PAP
nn
x
satisfying the ergodicity condition
=
1
lim= 0
2
N
n
NnN
x
N

.
Remark 2.2: =
1
lim= 0
2
N
n
NnN
x
N

, doesnt imply that
is bounded. In fact, let us consider the sequence
defined by

nn
x
3
if =
=0otherwise.
n
pnp
x
Let be such that . Then
p
3
3<1pN p

3
==1
1
11
=0
22
p
N
np
nN k
pp
xk
NNp

 .
For a function we define
:
m
f,(,)Tf
by
 
,=:for all ,<.Tft ftft


Definition 2.3: A bounded continuous function
x
is
said to be almost periodic if the set
,Tf
is rela-
tively dense for all >0
.
For the next denotes the space of all
almost periodic functions from to .
(, )
m
AP 
m
Proposition 2.4: Let and m

=nn
xx
be a
sequence with values in . Let define the function
m
x
:
by
m

=forall
n
xn xn,
and
x
is affine in [, 1].nn
Then the following re-
sults are true.
1)
sup= sup,
n
tn
x
tx

 
,,Tx Tx
and
(, )
m
xAP if and only if , (, )
m
xAP
2) if and only if
0(, )
m
xPAP(, )
m
xPAP.
Proof. 1) is a consequence of results taken from [1].
For the proof of 2), by taking the components, real
part and imaginary part, we can consider the case where
0(, )xPAP
[,tnn
. Then, one has
For 1]
one has

11
==
nn nnn
1
x
txxtnx x tnxnt

 
Two cases to be considered:
a) If
10
nn
xx

11
|||
d= 2
nnn
n
|
x
x
xt t
.
b) If
1<0
nn
xx


22 1
11
1
11
|||| =d
22||||
||||3||
|| .
22
n
nn nn
n
nn
nnn n
n
xx xx
x
tt
xx
xxx x
x





The result is a consequence of the fact that
x
if and only if
0(,)PAP 

1
0
d(,
)
n
nn
xt tPAP
.
Definition 2.5: We define the space of pseudo almost
periodic sequences by
0
() ()()PAP AP PAP
 .
Proposition 2.6: [2] Let be such that ()x PAP
Copyright © 2011 SciRes. APM
120 E. A. DADS ET AL.
=,
x
yz for some and Then ()yAP0()zPAP
y
x

.
Difference calculus is the discrete analogue of the
familiar differential and integral calculus. In this section,
we introduce some basic properties of two the following
operators that are essential in difference equations
 
=1
 
x
nxn

xn
and the shift operator
=1Ex nx n

.
Then
n k=
k
Exn x.
Let
I
be the identity operator. Then and
. The following formula are true =E

 
1
kk
k
I
=EI
.
i

 

ki
i
 
=0
=0
==
=1
k
i
k
i
x
nEI
k
xn Exn
i
x
nki




i





i
xn

=0
=k
i
k
ni



(5)
Similarly we have
kk
Ex
.
We should point out here that the operator
is the
counterpart of the derivative operator in calculus.
Both operators and share one of the helpful fea-
tures of the derivative operator , namely, the property
of linearity. Another interesting difference, parallel to
differential calculus, is the discrete analogue of the
fundamental theorem of calculus.
D
n
a
1.

E
exp

dsgs s

.k
 
:
nn
xx



D
Remark 2.7: Exponential in differential equa-
tions corresponds to the exponential and the
at
integral corresponds to the sum-
mation:
0exp
tat
11
=0
nnk
k
ag

Let us consider the linear map defined by
nn
T

Let
F
be a subspace of that is invariant by ,
for example B T
F
could be one of the following spaces
0 Let (, ,AP ) (,),PAPPAP (,).
F
T be the
linear map induced by on
T
F
and take
y
F
and
where
[],PX[]
X
is the space of polynomial
functions over . Next, we study the existence of
solutions in
F
for a given
y
F for the following
algebraic equation
=
F
PT xy.
This equation has solutions if

Im ,
F
y
PT but we
have to compute
Im .
F
PTker
The uniqueness problem is
equivalent to determine The following re-
sult is well-established.
PT
.
F
Lemma 2.8: Let []PX
be non constant. Then
 
=1
ker=: withdeg() <
Nn
iiii
in
PTQ nQm




where the i
s
are non zero roots of with res-
pective multiplicities P
P
.
i
m
Remark 2.9 : If is the unique root of , then 0
ker=0.PT
Lemma 2.10: Let 1|
=
B
TT (the restriction of to
B). Then T

1
ker=,=1,=1,,.
n
ii
n
P Tvectir


Proof. Let
be a complex number such that =1.
Then


1
ker =,
n
n
TI cc


.
Let
2
1
keryTI

and

=.
B
x
TIy
=
Then
11
and
nn nn
=0 n
x
xyyx

,
which implies that 10
=
n
nn ,
y
yx

and
10
1=
nn
nn
yyx

.
Since n
n
y
is also bounded, because =1,
then
and
0=0x1=
n
yy,
n
also

2
11
ker=ker.TI TI

By simple recurrence on we deduce that
,m
for all
1,m



11
ker= ker=:
mn
n
TITI cc


.
Consequently:




11
1
ker= ker
=:=1,=1,,
mi
i
ir
n
ii
n
PTT I
vecti r




.
Lemma 2.11: Let ,
x
. Consider y
x
y
the
sequence defined by

1
01
1
1
if> 0
=0 if=
if< 0.
kn k
kn
n
kn k
nk
xy n
xy n
xyn




0.
Copyright © 2011 SciRes. APM
E. A. DADS ET AL.
121
Then

,
x
yxy,
is bilinear and symmetric. More-
over, if we denote by =TI
 then


0
=
x
yxy xy

 
Remark 2.12: If then ,x
x
can be extended
to a function
x
which is of stepping type on in the
following manner:
 
x
tx
Et where
Et de-
notes the greatest integer function of then one has
for ,t
,,xy

0d
n
x
yn xtyntt 
 .
Proof. Using the above remark, one can see that the
following map

x
yxy
is bilinear and symme-
tric. On the other ha nd, one has
  
 
  

 
 





1
00
1
0
1
10
00
0
00
0
1d d
1d
1dd
1d
d
1d
nn
nn
n
n
n
n
n
n
x
ynxtynt txtyn t t
xt yntt
x
tynttxtyn t t
xyxuyn u u
xt yntt
xyxtxtyn t t
xyxy n
 


 

 



 

 



p
In the sequel, we denote by


,=(), suchthat
anddeg .
n
pn n
nn
FbQn
bF Q

We define the following polynomials
 
011
=1and =if
!
p
XX
XXXp
CCp
p
  .
Lemma 2.13: Let and y
be a complex
number such that =1
, Then for , the
following are true p
1) where


1
,=
p
p
cy

y

,=pnp
pn
n
cC
.
2) such that

1
Im= :
p
F
TI yF
,,pp
cyF
 .
Proof. 1) For we claim that
>0,p

,1,
=.
pp
cyc
 
 y
.
y
y
It follows that
From lemma (2.11) one has :
 

,, ,
0
,1,
=
==
pp p
pp
cycyc
cyc
 
 
 
 
for all0p
 
10,
0, 0,
0
=
==
py cy
cycy



 
,p
c


.y
2) One has

1
Im
p
F
yT
 if and only if there
exists
x
F
such that

1

=
.
p
F
e has:

=n
Txy Frommma
(2.8) n
le
and 1) o

,
np
n
x
Qn
cy
is in
F
if and only if ,p
cy
is in ,.
p
F
Proposition 2.14: complex
number such that
Let y, λ be a
=1
nomial of
degree .p Thfollowin ue
,be a poly
en the
.
In particular
and Q
g are tr




1
Im
=, suchthat
p
F
n
TI
F Qn

,p
n
yyF

1
,
Im=, suchthat .
ppn
Fp
n
TI yFnyF

 
and
 
11,
Im=:suchthatfor all1,,
F
mn
i
im
ii
n
PTy Fir
nyF


Proof. LetThen for all

1
Im .
p
F
yTI
 ,qp

1q
I
Im F
yT
, by lemma (2.13) we have
For all ,qp
()
nn
Cy
qnq is in ,q
F
,
to which is equivalent
For all (, )
qn
nn
qpC y
is in ,q
F
,
and as 0
()
q
Xqp
C
is a [],
p
X
basis of then
,
n isin
p
n
Qny F
.
Conversely, assume that


is
n
n
Qn y
in ,
p
F
.
One has from lemma (2.11)
,
andareinat
p
xyyF which impliesth
is ,
in
p
x
yF
. 
But


1
1
=
nn
nn
QnQ n



,
then
1,
isin
n
p
n
Qny F

,
by iteration, we see that

1,
for all[0,],in
qnis
p
n
qp QnyF

,
Copyright © 2011 SciRes. APM
122
is a basis of
E. A. DADS ET AL.
and since


1
qQX

0qp [],
p
X
then
for


,
all [],isin
n
p
p
n
RXRn yF

,
In particular
F

,
is in
pnp
np
n
Cy
,
and consequently

1
p
F
yImT I

the fact that . The end of the
proof results from

i
Let

1
Im= Immi
FF
ir
PTT I

.
be the linear map induced by T on
if (,)AP 
It is clear that1,
then

I
of is invertible.
On the other hand if the roots
P
are of modulus
ditfferen from 1, then

P
is invertible, in this case
 
r=0P
and we have theness of the
solutions. Let 1
()
iir
ke e uniqu
 be the rts of P with
modulus 1. Th


oo
en

is a polynomial whose roots are of modulus
differerom
Proposition 2.15:
1ir
P

=mi
i
XX QX
where Q
nt f1.



ker=:=1, =1,,.
n
ii
P vectir


n
(2.10), one has Proof. From lemma
 


=,=1,for =1
n
ii
vect i

ker=,,=1, =1,
,...,,
n
ii
n
n
P APvectir
r

 

(since is periodic).
It is well known that if ] and
then
,

n
in
12
,[QQ X
12

=1,QQ

1
Im= ImImQQQ Q
12 2

,
and


1
Im= Immi
i
ir
P
I


.
3. Existence of Pseudo Almost Periodic
Sequences
It is known that if
,
n
AM
,m
bAP
and
x
is a bounded solution of
1
ten
=,
nnn
xAxbn

h
x
is almost periodic.
If
,yAP and [],PX
=xy , PP
to a system, we deduce the following lemma.
Lemma 3.1: For and ,

yAP,[]PX
is almost every bounded solution
mma (3.1) anroposition
lt.
ion 3.2: Let
of

PT
=xy
periodic.
Consequently from led the p
(2.14), we get the following resu
Proposit
be a complex number such
that 0
,
p
and Q be a polynomial with degree
p. Then

1
Im
=
p
I
yA
 

,
,
, suchthat n
p
n
PQn yB


.
In particular


1
,
Im
,, suchthat
p
pn
p
n
I
n yB
=yA
P

.
and
 

11,
Im=,, suchthatforall [1,],
.
n
i
im
ii
n
PyAPi
nyB



m
r
In the next, we are concerned with the solutions in
,AP of the following equation

1=,for
p
Ixyp
. (6)
tion 3.3: Let .y
Then we Proposi have
1)

1=
p
p
TIycy
, (where =(
p
c) ),
p
nn
C
2)

1
Im
p
yI
 bB
if and only if there exists
by transforming
the scalar equation
such that =
p
yc bQ
w
the solutions of
ith Q[].
pX
3)

1=
p
I
xy
in
,AP
are =
nn
x
bc where d
c is a c onstant an


0
=pk
kp
byc c


1
2
12
1
=l
im
()
=lim
k
pn
pk
kn pn
kn
yc n
n
yc nCC
n




Proof. Since
.
0
k
Xkp
C
is a basis of [],
p
X
then
there exit scalars
0
k
kp
such that kk
Qc
0
=
kp
in this case, one h
0
as
=
p
kk
kp
c

. ) yc b (7
Copyright © 2011 SciRes. APM
E. A. DADS ET AL.
123
Then, by applying

1
p
TI
to the equation (7), we
obtain

1
11
=p
1
p
p
ycT Ibc
 ,
then

1
=lim
pn
yc n
n

. (8)
ose that
Supp 12
,,,
pp k


 are known and let us
compute 1:
k
e has by applying

k
TI to the equation
(7), on

2 2
=

11
k
p
kkk ppk
cT Icc
 

 
,
so
k
yb c





2
2
1=lim
p
k
pkknp n
kn
yc nCC
n



 . (9
We conclude that for all the equation
)

,,yAP

1=
p
I
xy
admits the solutions in
if
d (9)
,AP
(8) anand only if, the limits given by equations
exist and the sequence 0
p
kk
c
kp
yc

is bounded, in
this case the solutions are given by
0
=
p
kk
kp
x
yc c


,
with 0
is any constant and the (1
)
kkp
 are given by
(8) a).
Remark 3.4: By a change of variables, the equation
nd (9

1=
p
I
xy

when =1
becpre-
vious form (6).
Indeed, let uso
omes in the
Since
cnsider the following operator
 
,APP

:,
n
M A
.
nn
xx
=,MM


then

==
I
MM MM
 
 
 . I
So
1
=
I
MIM
 
,
an
1
d
 
11
1
=
pp
p
I
MIM




.
Then eq uation

1=
p
I
xy

, (10)
be
comes
 
1
11
p
p
=
M
IMx



y,
 
1
11
1
1
=
p
p
I
Mx M


, y
ing
differently
then by putt

1
=
X
Mx
and 1
1
1
=,
p
YM
y
and we are comhe followi equation ing down to t
=.
ng

1p
I
XY
Theorem 3.5 Fol equat
=,Pxy
r the generaion
the solution is of the form: 01
=,
x
xx such that
r
r
1=1
=i
i
x
x
, (with is the number of different roots with
modulus equal to 1, i
x
is the solution of the equ

1=
pi
ation
iii
I
xy

) and 0
x
af n element o
ker .P
Proof.
w
We writeP under the form

1
=pi
X Q
1i
i
r
P
ith i
s
are the roots of P which are of modulus
equal and to 1 ,
i
p
e may c o
r
Evef it means to replace
by wmew n t o Indeed
ttin
n i
do
1
y
let

1
Qy, =1.Q
us pug

=1,
=
p
k
k
=1 =1.
ii
AU Then
kk
i
PX
, then using the
Bezout idntity, we get that there exist polynomials i
U
r
e
such that
i

=1
=ii
i
Ay
with
r
y
=
ii
y
Uy
and equation
=Px
y(11)
becomes
t a solution of Equation (12), it suffices to
determine a solution
 
=
=r
i
Px A

1ii
y (12)
then to buil
i
x
of the following equation
i

1=
pi
ii
I
xy

,
the solution is easily determined and after we take
r
=1
=i
i
x
x
. To obtain all solutions , we add elements of


ker =,1
n
in
Pvecti r


.
ple 3.6: For all polynomial Q with all rootExam s
are of modulus different from in
1, []
X
one has the
following decomposition

,
1=m
riij
=1 =1
j
ij i
a
QX
the i
 where
s
are two by two dtinct of modulus isand
different from 1, then we have
Copyright © 2011 SciRes. APM
124 E. A. DADS ET AL.


1
,
==1 =1
m
ri
j
iji
QaI

 g down
to the

=m
QX
ij
e cas
so we are comin
with m
and 1
.
First case: >1:
 
=0k
I=
mm
I=
mk
m
k
m
C



so for xA has
 
 



,P one
=
m

I
xy
where
ase:


=km
x


.
=0
k
nm
k
yC

nk
Second c<1



1
m
II C
 

 
,
=0
==
mk
m
k

as
mk
km
hence for one h

,xAP

=
m
I
xy

with k
Let be bounded. Thk for a bounded
r the followingtion :
n

=.
k
nm
x

=0
k
nm
k
yC

en we loo
qua
2
28
n

nn
h
solution fo difference e
43 1
213 8=
nn nn
30
x
xx x

 . x
 
8 =2
h

1 2Qxxxxx x
3
43 2
= 2133028
 
32
12
1
392722721
22
=
Qx
xx
gg


where
1112141 8
:= 
x
181
:= 27 21
g
x
and
 
232
11 2141
:= 3927
22
gx
xx

 .
2
For 1<||< 2
2x, we have
1=1
81
=,:=
27 2
n
n
nn
n
a
ga
x

2
211
:=:= 48 22
n
gb
=0
171 371
,144216 2
n
nnnn
n
xbn n
 
.
the solution n
x
is given by:
kn
=1 =0
:=
knknn
nn
x
ah bh



=1
2
11 17
n
n
=0
81
:= 27 2
1 371
48144 216
222
kkn
n
n
kn
nn
n
xh
n h



Let

0,yPAP
nce and uniqueness
and the
existe of[].PX
solutions in We study
0,PAP of
the following equ ation
0=Px
, y
where 0
is the linear map induced by on T
PAP
00 ch that . Let be su

,x PAP,
0=0.Px
From lemma (2.10), we have



 
00
0
, =1,,,
,,=0
nir PAP
AP PAP
 


 
ker =n
i
P vect

.
Thenhave the uniqueness of the solution.
3.7: Unlike to the almost periodic ca
we
Remark se,
x
bounded and
0=Pxy
is not enough to get that
In fact, we
0,xPAP
example: have the following counter
1=2 ,,
n
nn
xx n
the solutions are given by
1
0=0
2,if >0
k
xn
.
0
1
0=
=,
2,
if<0
nk
n
k
kn
xx
xn
unded, on the other part one
has:
Then all solutions are bo
0
lim =2
n
nxx

and 0
lim =1
.
n
nxx
 . If 0
x
PAP
we will have 00
2= 1=0xx
e solution is not in
which is absurd, con-
sequently, th
As a consequence of proposition (2.14), we get the
following result.
0.PAP
Proposition 3.8: Let
be a complex number such
p
,that 0
, and Qal wit a polynomih degree
p. Then





,uchthat
.Qny PAP


1
00
Im=, s
p
IyPAP


0,p
n
In paular
n
rtic




1
00
0,
Im=,, suchthat
.
p
pn
p
n
IyPAP
nyPAP




Copyright © 2011 SciRes. APM
E. A. DADS ET AL.
125
In ergodic case, for the calculation of
the solutions, the method is similar to the one given in
the almost periodic case, firstly we begin with solution of
the following equation



00
101,
Im=,,such that
[1, ],().
mn
i
in mii
PyPAP
irn yPAP
 

Remark 3.9:


1
00
=, =
p
p
kk
kp
I
xyxyc c



with 1
()
kkp

his time th
are determined by equations (8 and (9)
but te
)
0
is not arbitrarily, but 0
is the
mean value of 1
p
kk
c
kp
solutions needs more 1
yc, then the etence of xis
p
kk
kp
ycc


to have a mean
value 0
then p
y c 0
0kp
c

kk
PAP.
Example 3.10: Let be an absolutely convergent
,
0k
k
a
series ]X, 0
()
kk
z a family of complex numbers
with modulus equa1, such that
[P
l 0
inf( )0
k
kPz
and
=0
n
k
>
=.
nk
k
y
az Th

has ast per
en the following equation
y
lmoiodic solutions. In fact, if we put

=Px
,

=0
=n
k
nk
a
kk
x
z
Pz
, one n
 has
x
is well defined and



=0
forall 0,==
ini
k
ni
nk
a
ixx

k
k
z
Pz
it results that



=0 =0
==
nn
k
kkkk n
nkk
k
a
Px Pz=azy
Pz
 

the equation admits solutions in
z
,AP The hypo-
thesis

0
inf>0 is necessary, as we rema
k
kPz
rk it
through the fo llowing counter example :
122
=0
1
=exp
nn
k
in
xx kk




.
If the solution exists, then
22
11
,exp1=
i
ax 2
k




kk



and
2
1,1ax

k
k


which contradicts the Parsevall’s iden tity, we d educ e that
thlution.
4. Application
More details and the motivation on this applications can
d in [4,14-20] and the references cited therein.
To apply the previous results, we consider the
following system
e equation does not have a so
be foun


21 2122
11 =1
2
nnn
q
px pxx


 


22
2
13
2
12 = 2
nn
n
qpx px a
xpxpqxpxb

21 21222nn nnn
  



where
,,ab AP ,pq with 0.q
nn
Remark 4.1: The last system comes from the re-
search of solutions of the following second order
differential equation with piecewise constant argument:
 
2
2
d1
1= 22
d
t
x
tpxtqxft
t


 






.
. where denotes the greatest integer function.
In the case where 1,p
the system has a unique
solution
nn
x
in
AP ,
=1,the heintend to study
sysomes
re we
the situtem bec ation where p

222 22
0=12 3
22
=1 ,
nnnn
qq
21 212 22nn nnn
x
xx a
xx qxxb

 

 
 
 

or more
22
=2
nn
Px a

21 22,
nn
n
=1
I
xqIxb




(13)
where
 
2
=24 32PX qXqX
 .
P
th m
We know that is invertible if and only if the
roots of are wiodulus different from
tion 4.2: 1) Let . The equation
has roo if and only
P 1.
Proposi
2
21
ax ax012
,,aaa
ts of modulus
0
=0a
if
1
201
=aaa or
20
10
=
<2.
aa
aa
3,a
the following equation 2) For
Copyright © 2011 SciRes. APM
126
ha d only if
E. A. DADS ET AL.
3
x a
2
3210
=0axax a , (14)
s roots of modulus 1 if an
20 1
=aaaa
3
or
22
020313
=aaa aa

Proof. It is clear that are roots of (
if
20 3
<2 .aa a
a
1 14) if and only
201
=aaa If the eq
of modulus if
and only if their pro, which is equint
to the abeomes then
2<4
is e
e cond
12
0
two conjugate compleroots which are
,aaa
x
ductqual
ovition
uation admits
1
vale
1b
20
=,aa c1
a<
0
2a For the secondmits as roots if
and only if eqtion , it adua1
20 1
=aaa, we can assume that
if not we divide the etion by, it is a
prove that
3
a
3=1,a
matter toqua a3
2
020 1
20
=1aaa a
aa

since the equation admits always a real root ,r it will
haval roots with modulus equal to 1 if and
ill be factorized as follows
<2
,
e non re only
if it w

32 2
210 1with<2,xaxacxc
whch implies that =r
=
xa xrx
is a root, in the sequel
.
If we obtain admits com-
plex th modulen from the
0
a

0
1=a
2
0201 0aaaa 
2
0=0,a
roots wi 21
=0xaxa
us equal to 1 th
previous result, we d educe that
1=1a
2<2
.a
1
, the equ as If 02
0
=1aaa ation can be written
follows
2a

3232 2
220
=1
 

1
020 0
2
020
=1
x
axax axaxaaa x a

xa xa ax
We will have no real roots with modulus equal to
if and only if 1
20
<2.aa
4,Corollary 4.3: 1) Ifthe roots of are
modulus different from
q Pof
1 (we assume that 0q).
2) If
 
=2131.X X
=4,q
PX
Proof. It suffices to apply the previou s proposition.
Proposition 4.4: If the system (13) admits
solutions if and only if
4,q

2Im
nn
ab I
 .
If =4,q
the system (13) admits solutions if an
only if d

Im
n
aI
ab

2I
m
nn I

Proof. First case: 4:q
the system becomes



1
21
() =21
nn
1
22
=2( )
nn
xPa
.
n
I
xqIPa

b
 
Th admits so lutions if and only if is system


1
21 Im
nn
qIPabI

,
or yet

21Im=Im ,
nn
qIaPb PII

 
since
P
respis invertible. Make the Euclidean dision
of ectively iv
P
11qX
by 1
X
, we s
th condt toee that
e previousition is equivalen

24Im
nn
qa qbI
,
identically

2Im
nn
ab I
.
Second case: =4:q
The system becomes
 
22
21 22
23 =2
=3 ,
nn
nnn
IIxa
I
xIx



 
b
equivalently



22
1
22 21
3=
=3 .
nn
nn
IIxa
n
x
IIxb



 

Let us consider the following system
(( )II



21
1
22 21
) =
=3 ,
nnn
nn
xb a
n
x
IIxb


 

 
which has solutions if and only if

 
Im
=ImIm,
nn
I
baII
II


 
 
which is equivalent to
Im
n
aI


2I
m
nn
ab I

.
5. Acknowledgements
The authors would like to thank the referees for their
careful reading of the paper.
Copyright © 2011 SciRes. APM
E. A. DADS ET AL.
Copyright © 2011 SciRes. APM
127
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