Applicable Analysis and Discrete Mathematics
available online at http://pefmath.etf.rs
Appl. Anal. Discrete Math. 9 (2015), 271–284.
doi:10.2298/AADM150811016M
EXISTENCE OF NONOSCILLATORY SOLUTIONS FOR
SYSTEM OF NEUTRAL DIFFERENCE EQUATIONS
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
The system of neutral type difference equations with delays

∆ x(n) + p(n) x(n − τ ) = a(n) f (y(n − ℓ))


∆y(n) = b(n) g(z(n − m))


∆z(n) = c(n) h(x(n − k))
is considered. The aim of this paper is to present sufficient conditions for
the existence of nonoscillatory bounded positive solutions of the considered
system with various (p(n)).
1. INTRODUCTION
In this paper we consider a nonlinear difference system of the three equations
of the form


 ∆ x(n) + p(n) x(n − τ ) = a(n) f (y(n − ℓ))
(1)
∆y(n) = b(n) g(z(n − m)),
n ∈ N0


∆z(n) = c(n) h(x(n − k))
where ∆ is the forward difference operator defined by ∆u(n) = u(n + 1) − u(n),
(a(n)), (b(n)), (c(n)) are sequences of real numbers, (p(n)) is a sequence of positive
real numbers, τ, l,m, k are nonnegative integers, and functions f, g, h : R → R. Here
R is a set of real numbers and N = {0, 1, 2, . . . }. By n0 we denote max {τ, l, m, k},
and N0 = {n0 , n0 + 1, . . . }. In the above system (x(n)), (y(n)) and (z(n)) are
real sequences defined for n ∈ N. Throughout this paper, X(n) denotes vector
2010 Mathematics Subject Classification. 39A10, 39A22.
Keywords and Phrases. System of difference equations, three dimensional, neutral type, nonoscillatory solutions, boundedness.
271
272
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
[x(n), y(n), z(n)]T . For the elements of R3 the symbol | · | stands for the maximum
norm. By B we denote the Banach space of all bounded sequences in R3 with the
supremum norm, i.e.,
B = X : N → R3 : kXk = sup |X(n)| < ∞ .
n∈N
A sequence of real numbers is said to be nonoscillatory if it is either eventually
positive or eventually negative. By a solution of system (1) we mean a sequence
(X(n)) which satisfied system (1) for sufficiently large n. A solution X of the system
(1) is called nonoscillatory if all its components, i.e. x, y, z are nonoscillatory.
Note, that system (1) includes different types of third order difference equations. For example, if the sequences a, b are positive, f, g are linear functions and
l = 0, m = 0, the system (1) reduces to the third-order neutral type difference
equation of the form
1
1
∆
∆(x(n) + p(n)x(n − τ ))
= c(n)h(x(n − k)).
∆
b(n)
a(n)
Such equations and their special cases have been studied by many authors, see for
example, [2], [3], [4], [7], [9], [12], [18] and the references cited therein.
The existence of a bounded nonoscillatory solution of nonlinear neutral type
difference system of two second-order equations has been studied in [17].
Some oscillation results, classification of nonoscillatory solutions or boundedness criteria for system (1) have been presented in [13], [14], [15] and [16] under
the assumption
∞
∞
X
X
a(n) =
b(n) = ∞.
n=1
n=1
In this paper we consider the case when
∞
X
n=1
|a(n)| < ∞,
∞
X
n=1
|b(n)| < ∞,
∞
X
|c(n)| < ∞.
n=1
We establish sufficient conditions for the existence of nonoscillatory bounded positive solutions of the considered system with various (p(n)). The results are illustrated by examples.
The following definition and theorems will be used in the sequel.
Definition 1 (Uniformly Cauchy subset, [6]). A subset Ω of the Banach space B
is said to be uniformly Cauchy if for every ε > 0 there exists a positive integer N
such that |X(i) − X(j)| < ε whenever i, j > N for any X ∈ Ω.
Theorem 1 (Arzelá–Ascoli’s Theorem, [1]). A bounded and uniformly Cauchy
subset of B is relatively compact.
Theorem 2 (Krasnoselskii’s Fixed Point Theorem, [8]). Let B be a Banach space,
let Ω be a bounded closed convex subset of B and let F, T be maps of Ω into B such
that F x + T y ∈ Ω for every pair x, y ∈ Ω. If F is a contraction and T is completely
continuous, then the equation F x + T x = x has a solution in Ω.
273
Existence of nonoscillatory solutions . . .
Theorem 3 (Schauder’s Fixed Point Theorem, [5]). Let M be a nonempty, compact
and convex subset of a Banach space and let T : M → M be a continuous. Then T
has a fixed point in M.
2. MAIN RESULTS
In this section, using the Krasnoselskii’s fixed point theorem and Schauder
fixed point theorem, we establish sufficient conditions for the existence of nonoscillatory bounded solutions of system (1).
Theorem 4. Assume that
∞
X
(2)
|a(n)| < ∞,
∞
X
n=1
(3)
∞
X
|b(n)| < ∞,
n=1
|c(n)| < ∞,
n=1
f, g, h : R → R are continuous functions.
If there exists a real number cp such that
(4)
0 < p(n) ≤ cp < 1,
n ∈ N0 ,
then system (1) has a bounded nonoscillatory solution.
Proof. For a fixed positive real number r we define the set
Ω1 = {X ∈ B : x(n), y(n), z(n) ∈ I1 , n ∈ N} ,
i
1
where I1 = (1 − cp )r, r . Ω1 is bounded closed convex subset of the Banach space
h
3
B. Since condition (3) is satisfied, we can set
Mf = max {|f (t)| : t ∈ I1 },
Mg = max {|g(t)| : t ∈ I1 },
Mh = max {|h(t)| : t ∈ I1 }.
From (2), there exists n1 ∈ N0 such that
∞
X
n=n1
(1 − cp )r
|a(n)| ≤
,
3Mf
∞
X
n=n1
(1 − cp )r
|b(n)| ≤
,
3Mg
∞
X
n=n1
|c(n)| ≤
(1 − cp )r
.
3Mh
Next, we define the maps F, T : Ω1 → B where




T1
F1
T =  T2  ,
F =  F2  ,
T3
F3
as follows



(F X)(n) = 


−p(n)x(n − τ ) +
2(1 − cp )r
3
2(1 − cp )r
3
(2 + cp )r
3






for n ≥ n1 ,
274
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
and
(5)
(F X)(n) = (F X)(n1 ) for 0 ≤ n < n1 ;


∞
X
a(s) f (y(s − ℓ)) 
 −

 s=n
∞

 X
 −
b(s) g(z(s − m)) 
(T X)(n) = 


 s=n

 X
∞


c(s) h(x(s − k))
−
for
n ≥ n1 ,
s=n
and
(T X)(n) = (T X)(n1 ) for 0 ≤ n < n1 .
We will show that F and T satisfy the conditions of Theorem 2. First we show
that if X, X̄ ∈ Ω1 , then F X + T X̄ ∈ Ω1 . For n ≥ n1 we have
∞
(F1 X)(n) + (T1 X̄)(n) = −p(n) x(n − τ ) +
X
(2 + cp )r
−
a(s) f (ȳ(s − ℓ))
3
s=n
∞
X
(2 + cp )r
a(s) f (ȳ(s − ℓ))
+
≤
3
s=n
2
1
(1 − cp )r
≤ r + cp r + M f ·
= r,
3
3
3Mf
∞
(F1 X)(n) + (T1 X̄)(n) = −p(n) x(n − τ ) +
X
(2 + cp )r
−
a(s) f (ȳ(s − ℓ))
3
s=n
∞
X
(2 + cp )r
a(s) f (ȳ(s − ℓ)) − p(n) x(n − τ )
−
≥
3
2
≥ r+
3
2
= r+
3
s=n
1
(1 − cp )r
cp r − M f ·
− cp r
3
3Mf
1
1
1
1
cp r − r + cp r − cp r = (1 − cp )r.
3
3
3
3
Below we present reasoning for F2 ; the same conclusions can be drawn for F3 . For
n ≥ n1
∞
(F2 X)(n) + (T2 X̄)(n) =
≤
X
2(1 − cp )r
−
b(s) g(z̄(s − m))
3
2(1 − cp )r
+
3
s=n
∞
X
s=n
b(s) g(z̄(s − m))
2
(1 − cp )r
2
= (1 − cp )r ≤ r,
≤ r − cp r + M g ·
3
3
3Mg
∞
(F2 X)(n) + (T2 X̄)(n) =
X
2(1 − cp )r
−
b(s) g(z̄(s − m))
3
s=n
Existence of nonoscillatory solutions . . .
275
∞
X
2(1 − cp )r
b(s) g(z̄(s − m))
−
≥
3
s=n
2
1
2
(1 − cp )r
≥ r − cp r − M g ·
= (1 − cp )r.
3
3
3Mg
3
The task is now to prove that F is a contraction mapping. It is easy to see
(F1 X)(n) − (F1 X̄)(n) ≤ p(n) |x(n − τ ) − x̄(n − τ )|
≤ cp |x(n − τ ) − x̄(n − τ )|,
(F2 X)(n) − (F2 X̄)(n) = 0,
(F3 X)(n) − (F3 X̄)(n) = 0,
for X, X̄ ∈ Ω1 and n ≥ n1 . Hence
kF X − F X̄k ≤ cp kX − X̄k,
where, by (4), there is 0 < cp < 1.
The next step is to show continuity of T. Let Xj = [xj , yj , zj ]T ∈ Ω1 for any
j ∈ N, and let (Xj (n)) be such that xj (n) → x(n), yj (n) → y(n), zj (n) → z(n)
as j → ∞. Since Xj ∈ Ω1 and Ω1 is closed, we have X = [x, y, z]T ∈ Ω1 , so
x(n), y(n), z(n) ∈ I1 for n ∈ N. By (5), (2) and (3) we obtain
|(T1 Xj )(n) − (T1 X)(n)| ≤
∞
X
a(s) f (yj (s − ℓ)) − f (y(s − ℓ)) → 0 if j → ∞,
s=n
for any n ∈ N. Analogously
|(T2 Xj )(n) − (T2 X)(n)| → 0 and |(T3 Xj )(n) − (T3 X)(n)| → 0 if j → ∞.
Therefore
k(T Xj ) − (T X)k → 0 if j → ∞.
We get that T is a continuous mapping.
Next we will demonstrate that T Ω1 is uniformly Cauchy. As before we only
show transformations for T1 , since similar arguments apply to T2 and T3 . Let X ∈
Ω1 . We conclude from the assumptions (2) and (3) that for any given ε > 0 there
exists an integer n2 > n1 such that for n ≥ n2
∞
X
a(s) f (y(s − ℓ)) < ε .
2
s=n
Hence for n4 > n3 ≥ n2 we obtain
X
∞
X
∞
a(s) f (y(s − ℓ)) < ε.
a(s) f (y(s − ℓ)) −
|(T1 X)(n4 ) − (T1 X)(n3 )| = s=n4
s=n3
276
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
Therefore T Ω1 is uniformly Cauchy.
By Theorem 2, there exists (X(n)) such that (F X)(n)+(T X)(n) = X(n). Finally, we verify that X(n) satisfies system (1) for n ≥ n1 . As (F1 X)(n)+(T1 X)(n) =
x(n) we have
∞
−p(n) x(n − τ ) +
X
(2 + cp )r
−
a(s)f (y(s − ℓ)) = x(n).
3
s=n
Next we move the term −p(n) x(n−τ ) to the right hand side of the equation. Using
the forward difference operator to the obtained equation we get
∆ (x(n) + p(n) x(n − τ )) = −∆
∞
X
a(s)f (y(s − ℓ)).
s=n
Hence
∆ (x(n) + p(n) x(n − τ )) = a(n)f (y(n − ℓ)).
Similarly, if (F2 X)(n) + (T2 X)(n) = y(n), then
∞
X
2(1 − cp )r
−
b(s) g(z(s − m)) = y(n).
3
s=n
Using the forward difference operator we get
∆y(n) = −
∞
X
b(s) g(z(s − m)) +
∞
X
b(s) g(z(s − m)),
s=n
s=n+1
and hence
∆y(n) = b(n) g(z(n − m)).
In the same manner we verify that (F3 X)(n) + (T3 X)(n) = z(n) implies the third
equation of (1). The proof is complete.
Example 1. Consider the difference system
−3n2 + n
1
x(n − 1) =
y(n)
∆ x(n) +
2n
4n4 − 2n2 − 2
−2n2 + n + 1
z(n − 1)
+ n4 − 4n3 − 3n2
n−1
∆z(n) = 3
x(n − 1).
n + n2
∆y(n) =
2n5
All assumptions of Theorem 4 are satisfied. One of the bounded solutions of the above
h
1
1
1 iT
system is X(n) = 1 + , 2 + 2 , 2 −
.
n
n
n
Theorem 5. Assume that conditions (2) and (3) are satisfied. If there exists a
real number c˜p such that
(6)
1 < c˜p ≤ p(n), n ∈ N0 ,
then system (1) has a bounded nonoscillatory solution.
277
Existence of nonoscillatory solutions . . .
Proof. We define subset Ω2 of B in the following way
Ω2 = {X ∈ B : x(n), y(n), z(n) ∈ I2 , n ∈ N0 } ,
i
1
where I2 = (c˜p − 1)r, c˜p r , r is a fixed positive real number. Obviously Ω2 is a
3
bounded, closed and convex subset of B. Let us set
h
M̃f = max {|f (t)| : t ∈ I2 },
M̃g = max {|g(t)| : t ∈ I2 },
M̃h = max {|h(t)| : t ∈ I2 }.
From the assumption (2), we conclude that there exists n1 ∈ N0 such that
∞
X
|a(n)| ≤
n=n1
∞
∞
(c˜p − 1)r X
(c˜ − 1)r
(c˜ − 1)r X
|c(n)| ≤ p
|b(n)| ≤ p
,
,
.
3M̃f
3
M̃
3M̃h
g
n=n
n=n
1
1
We define the maps F, T : Ω2 → B in the following way


(2c˜p + 1)r
x(n + τ )
+
−

 p(n + τ )
3


(2
c
˜
−
1)r
p
 for n ≥ n1

(F X)(n) = 

3


(2c˜p − 1)r
3
and
(F X)(n) = (F X)(n1 ) for 0 ≤ n < n1 ;

∞
X
1
a(s)f (y(s − ℓ))
 − p(n + τ )

s=n+τ
∞

X

b(s) g(z(s − m))
−
(T X)(n) = 

s=n

∞
X

c(s) h(x(s − k))
−
s=n









for n ≥ n1
and
(T X)(n) = (T X)(n1 ) for 0 ≤ n < n1 .
Let X, X̄ ∈ Ω2 , n ≥ n1 . To prove F X+T X̄ ∈ Ω2 , we will present all transformations
only for the first and the second components of F and T. For n ≥ n1 we have
(F1 X)(n) + (T1 X̄)(n) = −
(2c˜p + 1)r
1
x(n + τ ) +
p(n + τ )
3
−
∞
X
1
a(s)f (ȳ(s − ℓ))
p(n + τ )
s=n+τ
278
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
∞
X
(2c˜p + 1)r
1
≤
+
|a(s)| |f (ȳ(s − l))|
3
p(n + τ )
s=n+τ
2
3
1
3
≤ c˜p r + r + M̃f ·
(c˜p − 1)r
= c˜p r,
3M̃f
and
(F1 X)(n) + (T1 X̄)(n) = −
1
(2c˜p + 1)r
x(n + τ ) +
p(n + τ )
3
−
∞
X
1
a(s)f (ȳ(s − ℓ))
p(n + τ )
s=n+τ
∞
X
(2c˜p + 1)r
a(s) f (ȳ(s − ℓ)) −
≥
−
3
s=n
1
x(n + τ )
p(n + τ )
2
1
(c˜ − 1)r
1
≥ c˜p r + r − M̃f · p
− r = (c˜p − 1)r.
3
3
3
3M̃f
Similarly,
∞
(F2 X)(n) + (T2 X̄)(n) =
≤
X
(2c˜p − 1)r
−
b(s)g(z̄(s − m))
3
(2c˜p − 1)r
+
3
s=n
∞
X
s=n
b(s) g(z̄(s − m))
2
2
1
(c˜ − 1)r
≤ c˜p r − r + M̃g · p
r ≤ c˜p r,
= c˜p −
3
3
3
3M̃g
and
∞
(F2 X)(n) + (T2 X̄)(n) =
X
(2c˜p − 1)r
−
b(s)g(z̄(s − m))
3
s=n
∞
X
(2c˜p − 1)r
b(s) g(z̄(s − m))
−
≥
3
s=n
2
1
(c˜ − 1)r
1
1
≥ c˜p r − r − M̃g · p
= c˜p r ≥ (c˜p − 1)r.
3
3
3
3
3M̃g
To see that F is a contraction mapping, let us observe that
1
x(n + τ ) − x̄(n + τ )
p(n + τ )
1 x(n + τ ) − x̄(n + τ )
≤
c˜p
(F1 X)(n) − (F1 X̄)(n) ≤
and
(F2 X)(n) − (F2 X̄)(n) = 0,
(F3 X)(n) − (F3 X̄)(n) = 0.
279
Existence of nonoscillatory solutions . . .
Hence
kF X − F X̄k ≤
but
1
kX − X̄k,
c˜p
1
< 1 by (6).
c˜p
The proof of the continuity of the mapping T goes exactly in the same way
as previously.
By virtue of Theorem 2, there exists (X(n)) such that (F X)(n) + (T X)(n) =
X(n). Finally, we show that such (X(n)) satisfy the system (1) for n ≥ n1 . Let
(F1 X)(n) + (T1 X)(n) = x(n). Thus
−
∞
X
x(n + τ )
(2c˜p + 1)r
1
+
−
a(s)f (y(s − ℓ)) = x(n).
p(n + τ )
3
p(n + τ )
s=n+τ
Therefore
x(n + τ )
= −∆
∆ x(n) +
p(n + τ )
∞
X
1
a(s)f (y(s − ℓ)) .
p(n + τ )
s=n+τ
Then, by obtaining a common denominator, we have
1
∆ x(n + τ ) + p(n + τ )x(n)
p(n + τ + 1)
(7)
+ ∆
1
p(n + τ )
x(n + τ ) + p(n + τ )x(n)
X
∞
1
=−
∆
p(n + τ + 1)
− ∆
Since
s=n+τ
1
p(n + τ )
a(s)f (y(s − ℓ))
X
∞
s=n+τ
a(s)f (y(s − ℓ)) .
X
∞
−∆
a(s)f (y(s − ℓ)) = a(n + τ )f (y(n + τ − ℓ)),
s=n+τ
from (7) we get
∆ x(n + τ ) + p(n + τ )x(n) = a(n + τ )f (y(n + τ − ℓ)).
Now we can transform the last equation into
∆ x(n) + p(n)x(n − τ ) = a(n)f (y(n − ℓ)).
Assume that (F2 X)(n) + (T2 X)(n) = y(n). It is easier than above to see that X
satisfies (1). In fact, then we have
∞
X
(2c˜p − 1)r
−
b(s) g(z(s − m)) = y(n).
3
s=n
280
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
Then acting with the forward difference operator and simplifying, we arrive to
∆y(n) = −
∞
X
b(s) g(z(s − m)) +
∞
X
b(s) g(z(s − m)),
s=n
s=n+1
and hence
∆y(n) = b(n) g(z(n − m)).
Using the third equation (F3 X)(n) + (T3 X)(n) = z(n) we conclude in exactly the
same manner. The proof is now complete.
Note, that the assumptions of Theorem 2 and Theorem 3 ensure, that system (1) has not only one bounded nonoscillatory solution, but uncountably many
bounded nonoscillatory solutions.
Example 2. Now, let us consider the difference system
1 7 · 2n + 6
1 y(n − 1)
∆ x(n) + 2 + n x(n − 2) = · n
2
2 4 + 2n+1
4n
1
∆y(n) = −
z 3 (n − 1)
n−1
16 (3 · 2
+ 1)3
2n
1
x2 (n − 2).
∆z(n) = −
n−1
32 (2
− 1)2
It is easy to see that all assumptions of Theorem 5 are satisfied. Hence the above system
has bounded solutions. One of such solutions is
T
1
1
1
X(n) = 2 − n , 1 + n , 3 + n
.
2
2
2
In the next theorem, for the special case p(n) ≡ 1, we prove an even better
result. We give sufficient conditions under which for any real constants d1 , d2 , d3
there exists a solution of system (1) convergent to [d1 , d2 , d3 ]T .
Theorem 6. Assume that conditions (2) and (3) are satisfied. If p(n) ≡ 1, then
for any real constants d1 , d2 , d3 there exists a solution (X(n)) of the system (1)
such that lim X(n) = [d1 , d2 , d3 ]T .
n→∞
Proof. Let d1 , d2 , d3 ∈ R and let us choose a real number e such that e > 0. There
exist constants Mi > 1, i = 1, 2, 3 such that
|f (t)| ≤ M1 for every t ∈ [d1 − e, d1 + e],
|g(t)| ≤ M2 for every t ∈ [d2 − e, d2 + e],
|h(t)| ≤ M3 for every t ∈ [d3 − e, d3 + e].
Let us denote M = max {M1 , M2 , M3 } and
Sa (n) =
∞
X
j=n
|a(j)|, Sb (n) =
∞
X
j=n
|b(j)|, Sc (n) =
∞
X
j=n
|c(n)|.
281
Existence of nonoscillatory solutions . . .
By (2), there exists an index n1 ≥ n0 such that for n ≥ n1 we have
e
,
M
Sa (n) ≤
Sb (n) ≤
e
,
M
Sc (n) ≤
e
.
M
We define a subset Ω3 of B by
Ω3 = {X ∈ B : X(0) = · · · = X(n1 −1) = D and |X(n)−D|≤ M |S(n)| for n ≥ n1 },
where D = [d1 , d2 , d3 ]T and S = [Sa , Sb , Sc ]T . It is easy to check, that Ω3 is a
convex subset of B. It can be also shown that Ω3 is compact (see, for example, the
proofs of Theorem 1 in [10] or Lemma 4.7 in [11]). Now, for n ≥ 0, we define a
map


T1
T =  T 2  : Ω3 → B
T3
as follows
(T1 X)(n) =


d1

∞


d − X
for n < n1
n+2jτ −1
X
a(s)f (y(s − ℓ)) for n ≥ n1
1
j=1 s=n+(2j−1)τ


∞


1X

a(s)f (y(s − ℓ))
for n ≥ n1
d1 −
2 s=n
(T2 X)(n) =
and
(T3 X)(n) =


d2

d2 −


d3

d3 −
and τ > 0,
and τ = 0,
for n < n1
∞
X
b(s) g(z(s − m))
for n ≥ n1
s=n
for n < n1
∞
X
c(s) h(x(s − k)) for n ≥ n1 .
s=n
We will show that T (Ω3 ) ⊆ Ω3 . It is easy to check, that
(8)
∞
X
n+2jτ
X−1
j=1 s=n+(2j−1)τ
|a(s)| ≤
∞
X
|a(s)| .
s=n
Moreover, if X ∈ Ω3 , then by the definition of Ω3 we get |y(n) − d2 | ≤ e for all
n ∈ N. Hence |f (y(n))| ≤ M1 for every (X(n)) ∈ Ω3 , n ∈ N. Therefore by (8), for
n ≥ n1 and τ > 0, we get
(9)
X
∞
|(T1 X)(n) − d1 | = n+2jτ
X−1
j=1 s=n+(2j−1)τ
a(s)f (y(s − ℓ)) 282
Malgorzata Migda, Ewa Schmeidel, Malgorzata Zdanowicz
≤ M1
∞
X
|a(s)| ≤ M Sa (n).
s=n
For n ≥ n1 and τ = 0, we have
X
1 ∞
|(T1 X)(n) − d1 | = a(s)f (y(s − ℓ)) ≤ M Sa (n).
2
s=n
Similarly, we get
(10)
|(T2 X)(n) − d2 | ≤ M Sb (n)
and
|(T3 X)(n) − d3 | ≤ M Sc (n).
So, T (X) ∈ Ω3 for every X ∈ Ω3 and T (Ω3 ) ⊆ Ω3 . Similarly as in the proof
of Theorem 4, it can be shown that T is continuous. By Schauder’s fixed point
theorem there exists X ∈ Ω3 such that T (X) = X, which is a solution of system
(1). In fact, for n ≥ n1 and τ > 0, we have
x(n) = d1 −
n+2jτ
X−1
∞
X
a(s)f (y(s − ℓ)) ,
j=1 s=n+(2j−1)τ
x(n − τ ) = d1 −
∞ n+(2j−1)τ
X −1
X
a(s)f (y(s − ℓ)) .
j=1 s=n+(2j−2)τ
Hence
x(n)+x(n−τ ) = 2d1 −
∞
X
n+2jτ
X−1
a(s)f (y(s − ℓ)) = 2d1 −
∞
X
a(s)f (y(s − ℓ)) .
s=n
j=1 s=n+2(j−1)τ
Therefore
∆ (x(n) + x(n − τ )) = −
∞
X
a(s)f (y(s − ℓ)) +
∞
X
a(s)f (y(s − ℓ)) ,
s=n
s=n+1
and hence
∆ (x(n) + x(n − τ )) = a(n)f (y(n − ℓ)) .
In the case τ = 0, we obtain
∞
1X
∆ (x(n) + x(n)) = 2∆x(n) = 2∆ d1 −
a(s)f (y(s − ℓ)) = a(n)f (y(n − ℓ)) .
2
s=n
Similarly, we get
∆y(n) = b(n) g(z(n − m)),
∆z(n) = c(n) h(x(n − k)).
This means that the sequence (X(n)) fulfills system (1) for n ≥ n1 . By (2), the
sequences Sa , Sb , and Sc tend to zero. Hence and from (9), (10) we get lim X(n) =
n→∞
[d1 , d2 , d3 ]T . This completes the proof.
Existence of nonoscillatory solutions . . .
283
Example 3. Let us consider the following system
∆ (x(n) + x(n − 2)) = −
1
20
·
y(n − 1)
9 2 · 3n−1 + 1
2
3n
· n
z 2 (n − 1)
27 (3 − 1)2
1
2
x(n − 2).
∆z(n) = ·
3 4 · 3n + 9
∆y(n) = −
All assumptions of Theorem 6 are satisfied. It is easy to check that
T
1
1
1
X(n) = 4 + n , 2 + n , 3 − n
3
3
3
is the solution of the above system having the property lim X(n) = [4, 2, 3]T .
n→∞
We remark that the results obtained for system (1) can be extended analogically for a system of the form


 ∆ x(n) + p1 (n) x(n − τ1 ) = a(n) f (y(n − ℓ))
∆ y(n) + p2 (n) y(n − τ2 ) = b(n) g(z(n − m)), n ∈ N0 .


∆ z(n) + p3 (n) z(n − τ3 ) = c(n) h(x(n − k))
Acknowledgments. The first author was supported by the project PB-43-081/14DS
of Ministry of Science and Higher Education of Poland.
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Poznan University of Technology
Poland
E-mail: [email protected]
University of Bialystok
Poland
E-mails: [email protected]
[email protected]
(Received January 6, 2015)
(Revised August 11, 2015)