Applied Mathematics E-Notes, 11(2011), 208–215 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Positive Periodic Solutions For A Class Of Higher
Order Functional Di¤erence Equations
Xinhong Chen and Weibing Wangy
Received 6 September 2010
Abstract
In this paper, we apply a …xed point theorem to obtain su¢ cient conditions
for the existence, multiplicity and nonexistence of positive !-periodic solutions
for a class of higher-order functional di¤erence equations.
1
Introduction
In this paper, we investigate the existence, multiplicity and nonexistence of positive
!-periodic solutions for the periodic equation.
x(n + m) = g(x(n))x(n)
b(n)f (x(n
(n));
(1)
where > 0 is a positive parameter and we make the assumptions:
(H1 ) b; : Z ! Z are !-periodic sequences, b(n) > 0, !; m 2 N; (m; !) = 1, here
(m; !) is the greatest common divisor of m and !.
(H2 ) f; g : [0; +1) ! [0; +1) are continuous. 1 < l < g(u) < L < +1 for
u 0; f (u) > 0 for u > 0.
The existence of positive periodic solutions of discrete mathematical models has
been studied extensively in recent years, see [1, 2, 5, 6, 7, 9, 10, 11] and the references
therein. For example, Ra¤oul [3] considered the existence of positive periodic solutions
for functional di¤erence equations with parameter
x(n + 1) = a(n)x(n) + h(n)g(x(n
(n))):
(2)
Jiang [4] obtained the optimal existence theorem for single and multiple positive periodic solutions to general functional di¤erence equations
x(n) =
a(n)x(n) + g(n; x(n
(n))):
(3)
However, relatively few paper has discussed existence of positive periodic solutions for
higher-order functional di¤erence equations. In this paper, we apply a …xed point theorem to discuss the existence, multiple and nonexistence of positive !-periodic solutions
of (1).
Mathematics Subject Classi…cations: 38A12
of Mathematics, Hunan University of Science and Technology, Xiangtan, Hunan
411201, P.R. China
y Department
208
X. H. Chen and W. B. Wang
2
209
Preliminaries
Let X = fx : Z ! R; x(n + !) = x(n)g: When endowed with the maximum norm
kxk = maxn2[0;!] jx(n)j, X is a Banach space. From (1), we have that for any x 2 X,
1
x(n + m)
g(x(n))
b(n)
f (x(n
g(x(n))
x(n) =
(n)));
1
1
x(n + 2m)
x(n + m)
g(x(n))g(x(n + m))
g(x(n))
b(n + m)
f (x(n + m
(n + m)));
g(x(n))g(x(n + m))
=
::::::
!
Y1
(
i=0
=
!
Y2
1
)x(n + !m)
g(x(n + im))
!
Y1
(
i=0
(
i=0
1
)b(n + (!
g(x(n + im))
1
)x(n + (!
g(x(n + im))
1)m)f (x(n + (!
1)m
1)m)
(n + (!
1)m))):
By summing the above equations and using periodicity of x, we obtain
x(n) =
!
X1
i=0
1
Qi
1
j=0 g(x(n+jm))
1
1
t=0 g(x(n+tm))
Q!
b(n + im)f (x(n + im
(n + im))):
De…ne the map T : X ! X and a cone P in X by
!
X1
T x(n) =
G(n; i)b(n + im)f (x(n + im
(n + im)));
i=0
P = fx 2 X : x(n)
kxk; n 2 [0; !]g;
respectively, where
0
and
1
1
A 1
G(n; i) = @
g(x(n
+
jm))
j=0
i
Y
=
Clearly,
l!
L!
!
Y1
t=0
1
g(x(n + tm))
1
:
1
2 (0; 1) and
1
L!
1
G(n; i)
1
l!
1
;0
i
!
1:
!
1
;
(4)
210
Periodic Solutions of Di¤erence Equations
Further, one can easily show that the …xed point of T in P is the positive periodic
solution of (1). The following well-known result of the …xed point theorem is crucial
in our arguments.
LEMMA 2.1 ([8]). Let E be a Banach space and P be a cone in E. Suppose
and 2 are open subsets of E such that 0 2 1
1
2 and suppose that
T :P \(
2n 1)
1
!P
is a completely continuous operator. If one of the following conditions is satis…ed,
(i) kT xk
kxk for x 2 P \ @
1 ; kT
xk
kxk for x 2 P \ @
2;
(ii) kT xk
kxk for x 2 P \ @
1 ; kT
xk
kxk for x 2 P \ @
2:
Then T has a …xed point in P \ (
2 n 1 ).
LEMMA 2.2. Assume (H1 )-(H2 ) hold. Then T (P )
completely continuous.
P and T
: P ! P is
PROOF. In view of the de…nition of P , for x 2 P , we have
T x(n + !)
=
!
X1
G(n + !; i)b(n + ! + im)f (x(n + ! + im
(n + ! + im)))
i=0
=
!
X1
G(n; i)b(n + im)f (x(n + im
(n + im))) = T x(n):
i=0
On the other hand,
1
T x(n)
L!
=
=
1
1
L!
1
1
L!
1
!
X1
b(n + im)f (x(n + im
(n + im)))
i=0
!
X1
b(jm)f (x(jm
(jm)))
j=0
!
X1
b(j)f (x(j
(j)));
j=0
and
T x(n)
Hence,
T x(n)
1
l!
1
l!
L!
!
X1
b(j)f (x(j
(j))):
j=0
1
kT xk = kT xk:
1
Thus T (P )
P and according to Arzela-Ascoli’s Theorem, it is easy to show that
T : P ! P is completely continuous. The proof is complete.
X. H. Chen and W. B. Wang
211
LEMMA 2.3. Assume (H1 )-(H2 ) hold and let " > 0. If f (u)
then for any x 2 P ,
! 1
l! 1 X
b(j)kxk:
kT xk
" !
(L
1)2 j=0
PROOF. Since x 2 P and f (u)
1
T x(n)
L!
1
1
L!
1
1
L!
=
1
u" for any u > 0,
u", we have
!
X1
b(n + im)f (x(n + im
(n + im)))
i=0
!
X1
b(n + im)x(n + im
(n + im))"
i=0
!
X1
b(j)
j=0
l!
L!
1
kxk"
1
!
X1
1
b(j)kxk":
1)2 j=0
l!
(L!
Thus
kT xk
"
! 1
1 X
b(j)kxk:
1)2 j=0
l!
(L!
LEMMA 2.4. Assume (H1 )-(H2 ) hold . If there exists an
for any u > 0 , then for x 2 P
kT xk
1
l!
1
!
X1
> 0 such that f (u)
u
b(j)kxk:
j=0
This Lemma can be shown in a similar manner as in Lemma 2.3.
3
Main Results
Let
r
= fx 2 P : kxk < rg: Then @
f0 = lim+
u!0
r
= fx 2 P : kxk = rg: Put
f (u)
f (u)
; f1 = lim
;
u!1 u
u
(5)
I0 = number of zeros in the setff0 ; f1 g; I1 = number of in…nitions in the setff0 ; f1 g;
m(r) = minff (x) : x 2 [ r; r]; r > 0g; M (r) = maxff (x) : x 2 [ r; r]; r > 0g:
At …rst, we discuss the existence and multiplicity of positive periodic solutions for
(1).
212
Periodic Solutions of Di¤erence Equations
THEOREM 3.1. Assume (H1 )-(H2 ) hold. If I0 = 1 or 2, then (1) has at least one
I0 positive !-periodic solution for >
where
r
L! 1
inf
:
= P! 1
r>0
m(r)
j=0 b(j)
PROOF. Choose r1 > 0 such that
P!
1
j=0
L!
r
r1
inf
<
r>0 m(r)
m(r1 )
Noting that f (u)
b(j)
1
m(r1 ) for u 2 @ r1 , we can easily get
P! 1
j=0 b(j)m(r1 )
kT xk
;x 2 @
L! 1
:
r1 :
Hence,
kT xk > kxk; for x 2 @
r1
and
>
:
If f0 = 0, we choose 0 < r2 < r1 such that f (x)
x for 0
satis…es
! 1
1 X
b(j) < 1:
l! 1 j=0
According to Lemma 2.4, we have for x 2 @
l!
1
r2 , where
>0
r2 ,
!
X1
1
kT xk
x
b(j)kxk
j=0
kxk:
Then T has a …xed point in P \ ( r1 n r2 ), which is a positive !-periodic solution of
(1) for > :
If f1 = 0, there is a K > 0 such that f (x)
x for x K, where > 0 satis…es
1
l!
1
!
X1
b(j) < 1:
j=0
Let r3 = maxf2r1 ; K g, then x(n)
kxk
K for x 2 @
f (x)
x for x 2 @ r3 . In view of Lemma 2.4, we have
kT xk
1
l!
1
!
X1
j=0
b(j)kxk
r3
kxk; for x 2 @
and n 2 [0; !]: Thus
r3 :
Then T has a …xed point in P \( r3 n r1 ), and (1) has at least one positive !-periodic
solution for > .
If f0 = f1 = 0, it is easy to see from the above proof that T has a …xed point x1
in r1 n r2 and a …xed point x2 in r3 n r1 such that
r2 < kx1 k < r1 < kx2 k < r3 :
X. H. Chen and W. B. Wang
213
Consequently, (1) has at least two positive !-periodic solutions for > : The proof
is complete.
Similar to that of the Theorem 3.1, we have
THEOREM 3.2. Assume (H1 )-(H2 ) hold. If I1 = 1 or 2, then (1) has at least one
!
I1 positive !-periodic solution for 0 < < P!l 1 1b(j) supr>0 Mr(r) :
j=0
Next, we consider the nonexistence of positive !-periodic solutions for (1).
THEOREM 3.3. Assume (H1 )-(H2 ) hold. If I0 = 0 (or I1 = 0), then (1) has no
positive !-periodic solutions for su¢ ciently large > 0 (or su¢ ciently small > 0).
PROOF. Since I0 = 0, we have f0 > 0 and f1 > 0, there exist "1 > 0; "2 > 0 and
>
2
1 > 0 such that
f (x)
"1 x for x 2 [0;
1 ];
f (x)
"2 x for x 2 [
Let
c1 = min "1 ; "2 ; min
1
x
2
f
2 ; 1):
f (x)
g :
x
Then f (x) c1 x for x 2 [0; 1): Assume y is a positive !-periodic solution of (1). We
show that this leads to a contradiction for > ; where
=
(L!
(l!
1)c1
1)2
P! 1
j=0
b(j)
Since T y = y; it follows from Lemma 2.3 that for
kyk = kT yk
:
> ;
! 1
X
1
c1
b(j)kyk > kyk;
2
1)
j=0
l!
(L!
which is a contradiction.
If I1 = 0; then f0 < 1 andf1 < 1: There exists 1 > 0; 2 > 0; 2 >
that
f (x)
f (x)
1 x for x 2 [0; 1 ];
2 x for x 2 [ 2 ; 1):
Let c2 = maxf 1 ;
f (x)
2 ; maxf x gg;
f (x)
1
> 0 such
we have
c2 x for x 2 [0; 1):
Assume y is a positive !-periodic solution of (1). We show that this leads to a contradiction for 0 < < ; where
l! 1
= P! 1
:
c2 j=0 b(j)
Since T y = y, it follows from Lemma 2.4 that for 0 <
kyk = kT yk
1
l!
1
c2
!
X1
j=0
< ;
b(j)kyk < kyk;
214
Periodic Solutions of Di¤erence Equations
which is a contradiction. The proof is complete.
COROLLARY 3.1. Assume that (H1 )-(H2 ) hold. If there is a M1 > 0 such that
!
2
f (x) M1 x for x 2 [0; 1), then there exists a = (l! 1)(LP! 1)1 b(j)M such that for all
1
j=0
> , (1) has no positive !-periodic solution.
COROLLARY 3.2. Assume that (H1 )-(H2 ) hold. If there is a M2 > 0 such that
!
1
f (x)
M2 x for x 2 [0; 1), then there exists a
= P! l 1 b(j)M
such that for all
2
j=0
0<
< , (1) has no positive !-periodic solutions.
4
Examples
In this section, we illustrate our main results obtained in the previous sections with
several examples.
EXAMPLE 4.1. Consider the di¤erence equation
x(n + 3) = (3 + sin x(n))x(n)
b(n)x3 (n
9);
(6)
here b(n) > 0 is a 4-periodic sequences,
In fact 2 3 + sin(x(n)) 4 for n 2 [0; 4]. f (u) = u3 and
f0 = lim+
u!0
f (u)
f (u)
= 0; f1 = lim
= 1:
u!1
u
u
By Theorems 3.1 and 3.2, (6) has at least one positive !-periodic solution for su¢ ciently
large > 0 or su¢ ciently small > 0.
EXAMPLE 4.2. Consider the di¤erence equation
x(n + 1) = 3x(n)
sin x(n)
b(n)x(n
(n));
(7)
here b(n) > 0 and : Z ! Z are !-periodic sequences.
Clearly, the positive periodic solutions of (7) are the positive periodic solutions of
the following di¤erence equation
x(n + 1) = g(x(n))x(n)
where
g(u) =
3
2;
sin u
u ;
b(n)x(n
(n));
(8)
if u > 0;
if u = 0:
Note that
f (u)
f (u)
= 1; f1 = lim
= 1:
u!1 u
u
By Theorem 3.3, (8) has no positive !-periodic solutions for su¢ ciently large or small
> 0. Hence, (7) has no positive !-periodic solutions for su¢ ciently large or small
> 0.
f0 = lim+
u!0
Acknowledgement. Supported by the Scienti…c Research Fund of Hunan Provincial Education Department (09B033) and by Hunan Provincial Natural Science Foundation of China (09JJ3010).
X. H. Chen and W. B. Wang
215
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