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Asymptotic behavior of solutions of mixed type impulsive neutral differential equations

  • 1Email author,
  • 2, 3 and
  • 1
Advances in Difference Equations20142014:327

https://doi.org/10.1186/1687-1847-2014-327

  • Received: 29 August 2014
  • Accepted: 9 December 2014
  • Published:

Abstract

This paper investigates the asymptotic behavior of solutions of the mixed type neutral differential equation with impulsive perturbations [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) ] + P ( t ) f ( x ( t δ ) ) + Q ( t ) t x ( β t ) = 0 , 0 < t 0 t , t t k , x ( t k ) = b k x ( t k ) + ( 1 b k ) ( t k δ t k P ( s + δ ) f ( x ( s ) ) d s + β t k t k Q ( s / β ) s x ( s ) d s ) , k = 1 , 2 , 3 ,  . Sufficient conditions are obtained to guarantee that every solution tends to a constant as t . Examples illustrating the abstract results are also presented.

MSC:34K25, 34K45.

Keywords

  • asymptotic behavior
  • nonlinear neutral delay differential equation
  • impulse
  • Lyapunov functional

1 Introduction

The main purpose of this paper is to investigate the asymptotic behavior of solutions of the following mixed type neutral differential equation with impulsive perturbations:
{ [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) ] + P ( t ) f ( x ( t δ ) ) + Q ( t ) t x ( β t ) = 0 , 0 < t 0 t , t t k , x ( t k ) = b k x ( t k ) + ( 1 b k ) ( t k δ t k P ( s + δ ) f ( x ( s ) ) d s + β t k t k Q ( s / β ) s x ( s ) d s ) , k = 1 , 2 , 3 , ,
(1.1)

where τ , δ > 0 , 0 < α , β < 1 , C ( t ) , D ( t ) P C ( [ t 0 , ) , R ) , P ( t ) , Q ( t ) P C ( [ t 0 , ) , R 0 + ) , f C ( R , R ) , 0 < t k < t k + 1 with lim k t k = and b k , k = 1 , 2 , 3 ,  , are given constants. For J R , P C ( J , R ) denotes the set of all functions h : J R such that h is continuous for t k t < t k + 1 and lim t t k h ( t ) = h ( t k ) exists for all k = 1 , 2 ,  .

The theory of impulsive differential equations appears as a natural description of several real processes subject to certain perturbations whose duration is negligible in comparison with the duration of the process. Differential equations involving impulse effects occurs in many applications: physics, population dynamics, ecology, biological systems, biotechnology, industrial robotic, pharmacokinetics, optimal control, etc. The reader may refer, for instance, to the monographs by Bainov and Simeonov [1], Lakshmikantham et al. [2], Samoilenko and Perestyuk [3], and Benchohra et al. [4]. In recent years, there has been increasing interest in the oscillation, asymptotic behavior, and stability theory of impulsive delay differential equations and many results have been obtained (see [520] and the references cited therein).

Let us mention some papers from which are motivation for our work. By the construction of Lyapunov functionals, the authors in [8] studied the asymptotic behavior of solutions of the nonlinear neutral delay differential equation under impulsive perturbations,
{ [ x ( t ) + C ( t ) x ( t τ ) ] + P ( t ) f ( x ( t δ ) ) = 0 , 0 < t 0 t , t t k , x ( t k ) = b k x ( t k ) + ( 1 b k ) t k δ t k P ( s + δ ) f ( x ( s ) ) d s , k = 1 , 2 , 3 , .
(1.2)
A similar method was used in [21] by considering an impulsive Euler type neutral delay differential equation with similar impulsive perturbations
{ [ x ( t ) D ( t ) x ( α t ) ] + Q ( t ) t x ( β t ) = 0 , 0 < t 0 t , t t k , x ( t k ) = b k x ( t k ) + ( 1 b k ) β t k t k Q ( s / β ) s x ( s ) d s , k = 1 , 2 , 3 , .
(1.3)

In this paper we combine the two papers [8, 21] and we study the mixed type impulsive neutral differential equation problem (1.1). By using a similar method with the help of four Lyapunov functionals, sufficient conditions are obtained to guarantee that every solution of (1.1) tends to a constant as t . We note that problems (1.2) and (1.3) can be derived from the problem (1.1) as special cases, i.e., if D ( t ) 0 and Q ( t ) 0 , then (1.1) reduces to (1.2) while if C ( t ) 0 and P ( t ) 0 , then (1.1) reduces to (1.3). Therefore, the mixed type of nonlinear delay with an Euler form of impulsive neutral differential equations gives more general results than the previous one.

Setting η 1 = max { τ , δ } , η 2 = min { α , β } , and η = min { t 0 η 1 , η 2 t 0 } , we define an initial function as
x ( t ) = φ ( t ) , t [ η , t 0 ] ,
(1.4)

where φ P C ( [ η , t 0 ] , R ) = { φ : [ η , t 0 ] R | φ is continuous everywhere except at a finite number of point s, and φ ( s ) and φ ( s + ) = lim s s + φ ( s ) exist with φ ( s + ) = φ ( s ) }.

A function x ( t ) is said to be a solution of (1.1) satisfying the initial condition (1.4) if
  1. (i)

    x ( t ) = φ ( t ) for η t t 0 , x ( t ) is continuous for t t 0 and t t k , k = 1 , 2 , 3 ,  ;

     
  2. (ii)

    x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) is continuously differentiable for t > t 0 , t t k , k = 1 , 2 , 3 ,  , and satisfies the first equation of system (1.1);

     
  3. (iii)

    x ( t k + ) and x ( t k ) exist with x ( t k + ) = x ( t k ) and satisfy the second equation of system (1.1).

     

A solution of (1.1) is said to be nonoscillatory if it is eventually positive or eventually negative. Otherwise, it is said to be oscillatory.

2 Main results

We are now in a position to establish our main results.

Theorem 2.1 Assume that:

(H1) There exists a constant M > 0 such that
| x | | f ( x ) | M | x | , x R , x f ( x ) > 0 ,  for  x 0 .
(2.1)
(H2) The functions C, D satisfy
lim t | C ( t ) | = μ < 1 , lim t | D ( t ) | = γ < 1 with  μ + γ < 1 ,
(2.2)
and
C ( t k ) = b k C ( t k ) , D ( t k ) = b k D ( t k ) .
(2.3)

(H3) t k τ and α t k are not impulsive points, 0 < b k 1 , k = 1 , 2 ,  , and k = 1 ( 1 b k ) < .

(H4) The functions P, Q satisfy
{ lim sup t [ t δ t + δ P ( s + δ ) d s + β t t + δ Q ( s / β ) s d s + μ ( 1 + P ( t + τ + δ ) P ( t + δ ) ) + γ ( 1 + P ( ( t / α ) + δ ) α P ( t + δ ) ) ] < 2 M
(2.4)
and
{ lim sup t [ t δ t / β P ( s + δ ) d s + β t t / β Q ( s / β ) s d s + μ ( 1 + t Q ( ( t + τ ) / β ) ( t + τ ) Q ( t / β ) ) + γ ( 1 + Q ( t / ( α β ) ) Q ( t / β ) ) ] < 2 .
(2.5)

Then every solution of (1.1) tends to a constant as t .

Proof Let x ( t ) be any solution of system (1.1). We will prove that the lim t x ( t ) exists and is finite. Indeed, the system (1.1) can be written as
[ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ] + P ( t + δ ) f ( x ( t ) ) + Q ( t / β ) t x ( t ) = 0 , t t 0 , t t k ,
(2.6)
x ( t k ) = b k x ( t k ) + ( 1 b k ) ( t k δ t k P ( s + δ ) f ( x ( s ) ) d s + β t k t k Q ( s / β ) s x ( s ) d s ) , k = 1 , 2 , .
(2.7)
From (H2) and (H4), we choose constants ε , λ , υ , ρ > 0 sufficiently small such that μ + ε < 1 and γ + λ < 1 and T > t 0 sufficiently large, for t T ,
[ t δ t + δ P ( s + δ ) d s + β t t + δ Q ( s / β ) s d s + ( μ + ε ) ( 1 + P ( t + τ + δ ) P ( t + δ ) ) + ( γ + λ ) ( 1 + P ( ( t / α ) + δ ) α P ( t + δ ) ) ] 2 M υ ,
(2.8)
[ t δ t / β P ( s + δ ) d s + β t t / β Q ( s / β ) s d s + ( μ + ε ) ( 1 + t Q ( ( t + τ ) / β ) ( t + τ ) Q ( t / β ) ) + ( γ + λ ) ( 1 + Q ( t / ( α β ) ) Q ( t / β ) ) ] 2 ρ ,
(2.9)
and, for t T ,
| C ( t ) | μ + ε , | D ( t ) | γ + λ .
(2.10)
From (2.1), (2.10), we have
| C ( t ) | μ + ε 1 f 2 ( x ( t τ ) ) x 2 ( t τ ) , | D ( t ) | γ + λ 1 f 2 ( x ( α t ) ) x 2 ( α t ) , t T ,
which lead to
| C ( t ) | x 2 ( t τ ) ( μ + ε ) f 2 ( x ( t τ ) ) , | D ( t ) | x 2 ( α t ) ( γ + λ ) f 2 ( x ( α t ) ) , t T .
(2.11)

In the following, for convenience, the expressions of functional equalities and inequalities will be written without its domain. This means that the relations hold for all sufficiently large t.

Let V ( t ) = V 1 ( t ) + V 2 ( t ) + V 3 ( t ) + V 4 ( t ) , where
V 1 ( t ) = [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ] 2 , V 2 ( t ) = t δ t P ( s + 2 δ ) s t P ( u + δ ) f 2 ( x ( u ) ) d u d s + β t t P ( ( s + β δ ) / β ) β s t Q ( u / β ) u x 2 ( u ) d u d s , V 3 ( t ) = t δ t Q ( ( s + δ ) / β ) s + δ s t P ( u + δ ) f 2 ( x ( u ) ) d u d s + β t t Q ( s / β 2 ) s s t Q ( u / β ) u x 2 ( u ) d u d s ,
and
V 4 ( t ) = ( μ + ε ) t τ t P ( s + τ + δ ) f 2 ( x ( s ) ) d s + ( μ + ε ) t τ t Q ( ( s + τ ) / β ) s + τ x 2 ( s ) d s + ( γ + λ ) α t t Q ( s / ( α β ) ) s x 2 ( s ) d s + γ + λ α α t t P ( ( s / α ) + δ ) f 2 ( x ( s ) ) d s .
Computing d V 1 / d t along the solution of (1.1) and using the inequality 2 a b a 2 + b 2 , we have
d V 1 d t = 2 [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ] × ( P ( t + δ ) f ( x ( t ) ) + Q ( t / β ) t x ( t ) ) P ( t + δ ) [ 2 x ( t ) f ( x ( t ) ) | C ( t ) | x 2 ( t τ ) | C ( t ) | f 2 ( x ( t ) ) | D ( t ) | x 2 ( α t ) | D ( t ) | f 2 ( x ( t ) ) t δ t P ( s + δ ) f 2 ( x ( s ) ) d s f 2 ( x ( t ) ) t δ t P ( s + δ ) d s β t t Q ( s / β ) s x 2 ( s ) d s f 2 ( x ( t ) ) β t t Q ( s / β ) s d s ] Q ( t / β ) t [ 2 x 2 ( t ) | C ( t ) | x 2 ( t ) | C ( t ) | x 2 ( t τ ) | D ( t ) | x 2 ( t ) | D ( t ) | x 2 ( α t ) t δ t P ( s + δ ) f 2 ( x ( s ) ) d s x 2 ( t ) t δ t P ( s + δ ) d s β t t Q ( s / β ) s x 2 ( s ) d s x 2 ( t ) β t t Q ( s / β ) s d s ] .
Calculating directly for d V i / d t , i = 2 , 3 , 4 , t t k , we have
d V 2 d t = P ( t + δ ) f 2 ( x ( t ) ) t δ t P ( s + 2 δ ) d s P ( t + δ ) t δ t P ( s + δ ) f 2 ( x ( s ) ) d s + Q ( t / β ) β t x 2 ( t ) β t t P ( ( s + β δ ) / β ) d s P ( t + δ ) β t t Q ( s / β ) s x 2 ( s ) d s , d V 3 d t = P ( t + δ ) f 2 ( x ( t ) ) t δ t Q ( ( s + δ ) / β ) s + δ d s Q ( t / β ) t t δ t P ( s + δ ) f 2 ( x ( s ) ) d s + Q ( t / β ) t x 2 ( t ) β t t Q ( s / β 2 ) s d s Q ( t / β ) t β t t Q ( s / β ) s x 2 ( s ) d s ,
and
d V 4 d t = ( μ + ε ) P ( t + τ + δ ) f 2 ( x ( t ) ) ( μ + ε ) P ( t + δ ) f 2 ( x ( t τ ) ) + ( μ + ε ) ( t + τ ) Q ( ( t + τ ) / β ) x 2 ( t ) ( μ + ε ) t Q ( t / β ) x 2 ( t τ ) + ( γ + λ ) Q ( t / ( α β ) ) t x 2 ( t ) ( γ + λ ) Q ( t / β ) t x 2 ( α t ) + ( γ + λ ) α P ( ( t / α ) + δ ) f 2 ( x ( t ) ) ( γ + λ ) P ( t + δ ) f 2 ( x ( α t ) ) .
Summing for d V i / d t , i = 1 , 2 , 3 , we obtain
d V 1 d t + d V 2 d t + d V 3 d t P ( t + δ ) [ 2 x ( t ) f ( x ( t ) ) | C ( t ) | x 2 ( t τ ) | C ( t ) | f 2 ( x ( t ) ) | D ( t ) | x 2 ( α t ) | D ( t ) | f 2 ( x ( t ) ) f 2 ( x ( t ) ) t δ t P ( s + δ ) d s f 2 ( x ( t ) ) β t t Q ( s / β ) s d s f 2 ( x ( t ) ) t δ t P ( s + 2 δ ) d s f 2 ( x ( t ) ) t δ t Q ( ( s + δ ) / β ) s + δ d s ] Q ( t / β ) t [ 2 x 2 ( t ) | C ( t ) | x 2 ( t ) | C ( t ) | x 2 ( t τ ) | D ( t ) | x 2 ( t ) | D ( t ) | x 2 ( α t ) x 2 ( t ) t δ t P ( s + δ ) d s x 2 ( t ) β t t Q ( s / β ) s d s x 2 ( t ) β β t t P ( ( s + β δ ) / β ) d s x 2 ( t ) β t t Q ( s / β 2 ) s d s ] .
Since
t δ t P ( s + 2 δ ) d s = t t + δ P ( s + δ ) d s , β t t Q ( s / β 2 ) s d s = t t / β Q ( s / β ) s d s , t δ t Q ( ( s + δ ) / β ) s + δ d s = t t + δ Q ( s / β ) s d s , 1 β β t t P ( ( s + β δ ) / β ) d s = t t / β P ( s + δ ) d s ,
it follows that
d V 1 d t + d V 2 d t + d V 3 d t P ( t + δ ) [ 2 x ( t ) f ( x ( t ) ) | C ( t ) | x 2 ( t τ ) | C ( t ) | f 2 ( x ( t ) ) | D ( t ) | x 2 ( α t ) | D ( t ) | f 2 ( x ( t ) ) f 2 ( x ( t ) ) t δ t + δ P ( s + δ ) d s f 2 ( x ( t ) ) β t t + δ Q ( s / β ) s d s ] Q ( t / β ) t [ 2 x 2 ( t ) | C ( t ) | x 2 ( t ) | C ( t ) | x 2 ( t τ ) | D ( t ) | x 2 ( t ) | D ( t ) | x 2 ( α t ) x 2 ( t ) t δ t / β P ( s + δ ) d s x 2 ( t ) β t t / β Q ( s / β ) s d s ] .
Adding the above inequality with d V 4 / d t and using condition (2.11), we have
d V 1 d t + d V 2 d t + d V 3 d t + d V 4 d t P ( t + δ ) [ 2 x ( t ) f ( x ( t ) ) | C ( t ) | f 2 ( x ( t ) ) | D ( t ) | f 2 ( x ( t ) ) f 2 ( x ( t ) ) t δ t + δ P ( s + δ ) d s f 2 ( x ( t ) ) β t t + δ Q ( s / β ) s d s ] Q ( t / β ) t [ 2 x 2 ( t ) | C ( t ) | x 2 ( t ) | D ( t ) | x 2 ( t ) x 2 ( t ) t δ t / β P ( s + δ ) d s x 2 ( t ) β t t / β Q ( s / β ) s d s ] + ( μ + ε ) P ( t + τ + δ ) f 2 ( x ( t ) ) + ( μ + ε ) t + τ Q ( ( t + τ ) / β ) x 2 ( t ) + ( γ + λ ) Q ( t / ( α β ) ) t x 2 ( t ) + ( γ + λ ) α P ( ( t / α ) + δ ) f 2 ( x ( t ) ) .
Applying (2.8), (2.9), and (2.10), it follows that
d V d t = d V 1 d t + d V 2 d t + d V 3 d t + d V 4 d t P ( t + δ ) f 2 ( x ( t ) ) [ 2 x ( t ) f ( x ( t ) ) | C ( t ) | | D ( t ) | t δ t + δ P ( s + δ ) d s β t t + δ Q ( s / β ) s d s ( μ + ε ) P ( t + τ + δ ) P ( t + δ ) ( γ + λ ) α P ( ( t / α ) + δ ) P ( t + δ ) ] Q ( t / β ) t x 2 ( t ) [ 2 | C ( t ) | | D ( t ) | t δ t / β P ( s + δ ) d s β t t / β Q ( s / β ) s d s ( μ + ε ) t t + τ Q ( ( t + τ ) / β ) Q ( t / β ) ( γ + λ ) Q ( t / ( α β ) ) Q ( t / β ) ] P ( t + δ ) f 2 ( x ( t ) ) [ 2 M t δ t + δ P ( s + δ ) d s β t t + δ Q ( s / β ) s d s ( μ + ε ) ( 1 + P ( t + τ + δ ) P ( t + δ ) ) ( γ + λ ) ( 1 + P ( ( t / α ) + δ ) α P ( t + δ ) ) ] Q ( t / β ) t x 2 ( t ) [ 2 t δ t / β P ( s + δ ) d s β t t / β Q ( s / β ) s d s ( μ + ε ) ( 1 + t t + τ Q ( ( t + τ ) / β ) Q ( t / β ) ) ( γ + λ ) ( 1 + Q ( t / ( α β ) ) Q ( t / β ) ) ] P ( t + δ ) f 2 ( x ( t ) ) υ Q ( t / β ) t x 2 ( t ) ρ .
(2.12)
For t = t k , we have
V 1 ( t k ) = [ x ( t k ) + C ( t k ) x ( t k τ ) D ( t k ) x ( α t k ) t k δ t k P ( s + δ ) f ( x ( s ) ) d s β t k t k Q ( s / β ) s x ( s ) d s ] 2 = [ b k x ( t k ) + b k C ( t k ) x ( t k τ ) b k D ( t k ) x ( α t k ) b k ( t k δ t k P ( s + δ ) f ( x ( s ) ) d s + β t k t k Q ( s / β ) s x ( s ) d s ) ] 2 = b k 2 V 1 ( t k ) .

It is easy to see that V 2 ( t k ) = V 2 ( t k ) , V 3 ( t k ) = V 3 ( t k ) , and V 4 ( t k ) = V 4 ( t k ) .

Therefore,
V ( t k ) = V 1 ( t k ) + V 2 ( t k ) + V 3 ( t k ) + V 4 ( t k ) = b k 2 V 1 ( t k ) + V 2 ( t k ) + V 3 ( t k ) + V 4 ( t k ) V 1 ( t k ) + V 2 ( t k ) + V 3 ( t k ) + V 4 ( t k ) = V ( t k ) .
(2.13)

From (2.12) and (2.13), we conclude that V ( t ) is decreasing. In view of the fact that V ( t ) 0 , we have lim t V ( t ) = ψ exist and ψ 0 .

By using (2.8), (2.9), (2.12), and (2.13), we have
υ T P ( t + δ ) f 2 ( x ( t ) ) d t + ρ T Q ( t / β ) t x 2 ( t ) d t V ( T ) ,
which yields
P ( t + δ ) f 2 ( x ( t ) ) , Q ( t / β ) t x 2 ( t ) L 1 ( t 0 , ) .
Hence, for any ϕ > 0 and ξ ( 0 , 1 ) , we get
lim t t ϕ t P ( s + δ ) f 2 ( x ( s ) ) d s = 0 , lim t ξ t t Q ( s / β ) s x 2 ( s ) d s = 0 .
Thus, it follows from (2.4) and (2.5) that
t δ t P ( s + 2 δ ) s t P ( u + δ ) f 2 ( x ( u ) ) d u d s + β t t P ( ( s + β δ ) / β ) β s t Q ( u / β ) u x 2 ( u ) d u d s t δ t + δ P ( s + δ ) d s t δ t P ( u + δ ) f 2 ( x ( u ) ) d u + t δ t / β P ( s + δ ) d s β t t Q ( u / β ) u x 2 ( u ) d u 2 M t δ t P ( u + δ ) f 2 ( x ( u ) ) d u d s + 2 M β t t Q ( u / β ) u x 2 ( u ) d u 0 , as  t , t δ t Q ( ( s + δ ) / β ) s + δ s t P ( u + δ ) f 2 ( x ( u ) ) d u d s + β t t Q ( s / β 2 ) s s t Q ( u / β ) u x 2 ( u ) d u d s β t t + δ Q ( s / β ) s d s t δ t P ( u + δ ) f 2 ( x ( u ) ) d u + β t t / β Q ( s / β ) s d s β t t Q ( u / β ) u x 2 ( u ) d u 2 M t δ t P ( u + δ ) f 2 ( x ( u ) ) d u + 2 β t t Q ( u / β ) u x 2 ( u ) d u 0 , as  t ,
and
( μ + ε ) t τ t P ( s + τ + δ ) f 2 ( x ( s ) ) d s + ( μ + ε ) t τ t Q ( ( s + τ ) / β ) s + τ x 2 ( s ) d s + ( γ + λ ) α t t Q ( s / ( α β ) ) s x 2 ( s ) d s + γ + λ α α t t P ( ( s / α ) + δ ) f 2 ( x ( s ) ) d s = ( μ + ε ) t τ t P ( s + τ + δ ) P ( s + δ ) P ( s + δ ) f 2 ( x ( s ) ) d s + ( μ + ε ) t τ t s Q ( ( s + τ ) / β ) Q ( s / β ) ( s + τ ) Q ( s / β ) s x 2 ( s ) d s + ( γ + λ ) α t t Q ( s / ( α β ) ) Q ( s / β ) Q ( s / β ) s x 2 ( s ) d s + γ + λ α α t t P ( ( s / α ) + δ ) P ( s + δ ) P ( s + δ ) f 2 ( x ( s ) ) d s 2 M t τ t P ( s + δ ) f 2 ( x ( s ) ) d s + 2 t τ t Q ( s / β ) s x 2 ( s ) d s + 2 α t t Q ( s / β ) s x 2 ( s ) d s + 2 M α t t P ( s + δ ) f 2 ( x ( s ) ) d s 0 , as  t .

Therefore, from the above estimations, we have lim t V 2 ( t ) = 0 , lim t V 3 ( t ) = 0 , and lim t V 4 ( t ) = 0 , respectively.

Thus, lim t V 1 ( t ) = lim t V ( t ) = ψ , that is,
lim t [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ] 2 = ψ .
(2.14)
Now, we will prove that the limit
lim t [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ]
(2.15)
exists and is finite. Setting
y ( t ) = x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ,
(2.16)
and using (1.1) and condition (H3), we have
y ( t k ) = x ( t k ) + C ( t k ) x ( t k τ ) D ( t k ) x ( α t k ) t k δ t k P ( s + δ ) f ( x ( s ) ) d s β t k t k Q ( s / β ) s x ( s ) d s = b k [ x ( t k ) + C ( t k ) x ( t k τ ) D ( t k ) x ( α t k ) t k δ t k P ( s + δ ) f ( x ( s ) ) d s β t k t k Q ( s / β ) s x ( s ) d s ] = b k y ( t k ) .
(2.17)
In view of (2.14), it follows that
lim t y 2 ( t ) = ψ .
In addition, from (2.16) and (2.17), system (2.6)-(2.7) can be written as
{ y ( t ) + P ( t + δ ) f ( x ( t ) ) + Q ( t / β ) t x ( t ) = 0 , 0 < t 0 t , t t k , y ( t k ) = b k y ( t k ) , k = 1 , 2 , 3 , .
(2.18)
If ψ = 0 , then lim t y ( t ) = 0 . If ψ > 0 , then there exists a sufficiently large T such that y ( t ) 0 for any t > T . Otherwise, there is a sequence { a k } with lim k a k = such that y ( a k ) = 0 , and so y 2 ( a k ) 0 as k . This contradicts ψ > 0 . Therefore, for any t k > T , and t [ t k , t k + 1 ) , we have y ( t ) > 0 or y ( t ) < 0 from the continuity of y on [ t k , t k + 1 ) . Without loss of generality, we assume that y ( t ) > 0 on [ t k , t k + 1 ) . It follows from (H3) that y ( t k + 1 ) = b k y ( t k + 1 ) > 0 , and thus y ( t ) > 0 on [ t k + 1 , t k + 2 ) . By using mathematical induction, we deduce that y ( t ) > 0 on [ t k , ) . Therefore, from (2.14), we have
lim t y ( t ) = lim t [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) t δ t P ( s + δ ) f ( x ( s ) ) d s β t t Q ( s / β ) s x ( s ) d s ] = κ ,
(2.19)
where κ = ψ and is finite. In view of (2.18), for sufficient large t, we have
β t δ t P ( s + δ ) f ( x ( s ) ) d s + β t δ t Q ( s / β ) s x ( s ) d s = y ( β t δ ) y ( t ) β t δ < t k < t [ y ( t k ) y ( t k ) ] = y ( β t δ ) y ( t ) β t δ < t k < t ( 1 b k ) y ( t k ) .
Taking t and using (H3), we have
lim t [ β t δ t P ( s + δ ) f ( x ( s ) ) d s + β t δ t Q ( s / β ) s x ( s ) d s ] = 0 ,
which leads to
lim t t δ t P ( s + δ ) f ( x ( s ) ) d s = 0 and lim t β t t Q ( s / β ) s x ( s ) d s = 0 .
This implies that
lim t [ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) ] = κ .
(2.20)
Next, we shall prove that
lim t x ( t )  exists and is finite .
(2.21)
Further, we first show that | x ( t ) | is bounded. Actually, if | x ( t ) | is unbounded, then there exists a sequence { z n } such that z n , | x ( z n ) | , as n and
| x ( z n ) | = sup t 0 t z n | x ( t ) | ,
(2.22)
where, if z n is not an impulsive point, then x ( z n ) = x ( z n ) . Thus, we have
| x ( z n ) + C ( z n ) x ( z n τ ) D ( z n ) x ( α z n ) | | x ( z n ) | | C ( z n ) | | x ( z n τ ) | | D ( z n ) | | x ( α z n ) | | x ( z n ) | [ 1 μ ε γ λ ] ,

as n , which contradicts (2.20). Therefore, | x ( t ) | is bounded.

If μ = 0 and γ = 0 , then lim t x ( t ) = κ , which implies that (2.21) holds. If 0 < μ < 1 and 0 < γ < 1 , then we deduce that C ( t ) and D ( t ) are eventually positive or eventually negative. Otherwise, there are two sequences { w k } and { w j } with lim k w k = and lim j w j = such that C ( w k ) = 0 and D ( w j ) = 0 . Therefore, C ( w k ) 0 and D ( w j ) 0 as k , j . It is a contradiction to μ > 0 and γ > 0 .

Now, we will show that (2.21) holds. By condition (H2), we can find a sufficiently large T 1 such that for t > T 1 , | C ( t ) | + | D ( t ) | < 1 . Set
ω = lim inf t x ( t ) , θ = lim sup t x ( t ) .
Then we can choose two sequences { u n } and { v n } such that u n , v n as n , and
lim n x ( u n ) = ω , lim n x ( v n ) = θ .

For t > T 1 , we consider the following eight possible cases.

Case 1. When lim t C ( t ) = 0 and 1 < D ( t ) < 0 for t > T 1 , we have
κ = lim n [ x ( u n ) D ( u n ) x ( α u n ) ] ω + γ θ ,
and
κ = lim n [ x ( v n ) D ( v n ) x ( α v n ) ] θ + γ ω .
Thus, we obtain
ω + γ θ θ + γ ω ,
that is,
ω ( 1 γ ) θ ( 1 γ ) .
Since 0 < γ < 1 and θ ω , it follows that θ = ω . By (2.20), we obtain
θ = ω = κ 1 γ ,

which shows that (2.21) holds.

Case 2. When lim t D ( t ) = 0 and 1 < C ( t ) < 0 for t > T 1 , we get
κ = lim n [ x ( u n ) + C ( u n ) x ( u n τ ) ] ω μ ω
and
κ = lim n [ x ( v n ) + C ( v n ) x ( v n τ ) ] θ μ θ ,
which leads to
ω ( 1 μ ) θ ( 1 μ ) .
Since 0 < μ < 1 and θ ω , we conclude that
θ = ω = κ 1 μ ,

which implies that (2.21) holds.

Case 3. lim t C ( t ) = 0 , 0 < D ( t ) < 1 for t > T 1 . The method of proof is similar to the above two cases. Therefore, we omit it.

Case 4. lim t D ( t ) = 0 , 0 < C ( t ) < 1 for t > T 1 . The method of proof is similar to the above two first cases. Therefore, we omit it.

Case 5. When 1 < D ( t ) < 0 and 0 < C ( t ) < 1 for t > T 1 , we have
κ = lim n [ x ( u n ) + C ( u n ) x ( u n τ ) D ( u n ) x ( α u n ) ] ω + μ θ + γ θ
and
κ = lim n [ x ( v n ) + C ( v n ) x ( v n τ ) D ( v n ) x ( α v n ) ] θ + μ ω + γ ω ,
which yields
ω ( 1 μ γ ) θ ( 1 μ γ ) .
Since 0 < μ + γ < 1 and θ ω , we have θ = ω . Thus
θ = ω = κ 1 μ γ ,

and so (2.21) holds.

Using similar arguments, we can prove that (2.21) also holds for the following cases:

Case 6. 1 < C ( t ) < 0 , 0 < D ( t ) < 1 .

Case 7. 1 < C ( t ) < 0 , 1 < D ( t ) < 0 .

Case 8. 0 < C ( t ) < 1 , 0 < D ( t ) < 1 .

Summarizing the above investigation, we conclude that (2.21) holds and so the proof is completed. □

Theorem 2.2 Let conditions (H1)-(H4) of Theorem 2.1 hold. Then every oscillatory solution of (1.1) tends to zero as t .

Corollary 2.1 Assume that (H3) holds and
lim sup t [ t δ t + δ P ( s + δ ) d s + β t t + δ Q ( s / β ) s d s ] < 2
(2.23)
and
lim sup t [ t δ t / β P ( s + δ ) d s + β t t / β Q ( s / β ) s d s ] < 2 .
(2.24)
Then every solution of the equation
{ x ( t ) + P ( t ) x ( t δ ) + Q ( t ) t x ( β t ) = 0 , 0 < t 0 t , t t k , x ( t k ) = b k x ( t k ) + ( 1 b k ) ( t k δ t k P ( s + δ ) x ( s ) d s x ( t k ) = + β t k t k Q ( s / β ) s x ( s ) d s ) , k = 1 , 2 , 3 , ,
(2.25)

tends to a constant as t .

Corollary 2.2 The conditions (2.23) and (2.24) imply that every solution of the equation
x ( t ) + P ( t ) x ( t δ ) + Q ( t ) t x ( β t ) = 0 , 0 < t 0 t ,
(2.26)

tends to a constant as t .

Theorem 2.3 The conditions (H1)-(H4) of Theorem 2.1 together with
t 0 P ( s + δ ) d s = , t 0 Q ( s / β ) s d s = ,
(2.27)

imply that every solution of (1.1) tends to zero as t .

Proof From Theorem 2.2, we only have to prove that every nonoscillatory solution of (1.1) tends to zero as t . Without loss of generality, we assume that x ( t ) is an eventually positive solution of (1.1). As in the proof of Theorem 2.1, (1.1) can be written as in the form (2.18). Integrating from t 0 to t both sides of the first equation of (2.18), one has
t 0 t P ( s + δ ) f ( x ( s ) ) d s + t 0 t Q ( s / β ) s x ( s ) d s = y ( t 0 ) y ( t ) t 0 < t k < t ( 1 b k ) y ( t k ) .
Applying (2.19) and (H3), we have
t 0 P ( s + δ ) f ( x ( s ) ) d s < and t 0 Q ( s / β ) s x ( s ) d s < .

This, together with (2.27), implies that lim inf t f ( x ( t ) ) = 0 and lim inf t x ( t ) = 0 . By Theorem 2.1, lim t x ( t ) = 0 . This completes the proof. □

Corollary 2.3 Assume that (2.1), (2.2), (2.4), (2.5), and (2.27) hold. Then every solution of the equation
[ x ( t ) + C ( t ) x ( t τ ) D ( t ) x ( α t ) ] + P ( t ) f ( x ( t δ ) ) + Q ( t ) t x ( β t ) = 0 ,
(2.28)

0 < t 0 t , tends to zero as t .

3 Examples

In this section, we present two examples to illustrate our results.

Example 3.1 Consider the following mixed type neutral differential equation with impulsive perturbations:
{ [ x ( t ) + ( k + 3 ) t 3 k 2 + 3 k 6 x ( t 1 2 ) ( 3 k + 9 ) t 8 k 2 + 8 k 16 x ( t e ) ] + ( 2 + t 5 t 2 ) ( 1 + 1 4 cos 2 x ( t π ) ) x ( t π ) + 1 t ( ln t + 2 ) x ( t e 2 ) = 0 , t 1 , x ( k ) = k 2 + 6 k + 8 ( k + 3 ) 2 x ( k ) + ( 1 k 2 + 6 k + 8 ( k + 3 ) 2 ) ( t k π t k 2 + s + π 5 ( s + π ) 2 x ( k ) = × ( 1 + 1 4 cos 2 x ( s ) ) x ( s ) d s + t k e 2 t k 1 s ( ln ( e 2 s ) + 2 ) x ( s ) d s ) , k = 2 , 3 , 4 , .
(3.1)
Here C ( t ) = ( ( k + 3 ) t ) / ( 3 k 2 + 3 k 6 ) , D ( t ) = ( ( 3 k + 9 ) t ) / ( 8 k 2 + 8 k 16 ) , P ( t ) = ( 2 + t ) / ( 5 t 2 ) , Q ( t ) = 1 / ( ln t + 2 ) , t [ k 1 , k ) , b k = ( k 2 + 6 k + 8 ) / ( ( k + 3 ) 2 ) , t 0 = 1 , k = 2 , 3 , 4 ,  , f ( x ) = x ( 1 + ( ( 1 / 4 ) ( cos 2 x ) ) ) , τ = 1 / 2 , δ = π , α = 1 / e , and β = 1 / e 2 . We can find that
  1. (i)

    | x | | ( 1 + 1 4 cos 2 x ) x | 5 4 | x | , x R , ( 1 + 1 4 cos 2 x ) x 2 > 0 for x 0 ;

     
  2. (ii)

    lim t | C ( t ) | = 1 3 = μ < 1 , lim t | D ( t ) | = 3 8 = γ < 1 with μ + γ = 17 24 < 1 , and C ( k ) = k 2 + 6 k + 8 ( k + 3 ) 2 C ( k ) , D ( k ) = k 2 + 6 k + 8 ( k + 3 ) 2 D ( k ) ;

     
  3. (iii)
    t k ( 1 / 2 ) and ( 1 / e ) t k are not impulsive points, 0 < ( k 2 + 6 k + 8 ) / ( ( k + 3 ) 2 ) 1 for k = 1 , 2 ,  , and
    k = 1 ( 1 k 2 + 6 k + 8 ( k + 3 ) 2 ) = k = 1 1 ( k + 3 ) 2 < ;
     
  4. (iv)
    { lim sup t [ t δ t + δ P ( s + δ ) d s + β t t + δ Q ( s / β ) s d s + μ ( 1 + P ( t + τ + δ ) P ( t + δ ) ) + γ ( 1 + P ( ( t / α ) + δ ) α P ( t + δ ) ) ] = 17 12 < 8 5
     
and
{ lim sup t [ t δ t / β P ( s + δ ) d s + β t t / β Q ( s / β ) s d s + μ ( 1 + t Q ( ( t + τ ) / β ) ( t + τ ) Q ( t / β ) ) + γ ( 1 + Q ( t / ( α β ) ) Q ( t / β ) ) ] = 109 60 < 2 .

Hence, by (i)-(iv) all assumptions of Theorem 2.1 are satisfied. Therefore, we conclude that every solution of (3.1) tends to a constant as t .

Example 3.2 Consider the following mixed type neutral differential equation with impulsive perturbations
{ [ x ( t ) + ( 12 k + 16 ) t 54 k 2 + 27 k 27 x ( t 2 3 ) ( 12 k + 16 ) t 42 k 2 + 21 k 21 x ( t 2 e 3 ) ] + ( 2 t + 1 ( 4 t + 3 ) 2 ) ( 1 + 2 5 sin 2 x ( t π 2 ) ) x ( t π 2 ) + 4 t ( 2 ln t + 3 ) x ( t 3 e ) = 0 , t 1 , x ( k ) = 6 k 2 + 17 k + 7 6 k 2 + 17 k + 12 x ( k ) + ( 1 6 k 2 + 17 k + 7 6 k 2 + 17 k + 12 ) ( t k π 2 t k 2 s + 1 + π ( 4 s + 3 + 2 π ) 2 x ( k ) = × ( 1 + 2 5 sin 2 x ( s ) ) x ( s ) d s + t k 3 e t k 4 s ( 2 ln ( 3 e s ) + 3 ) x ( s ) d s ) , k = 2 , 3 , 4 , .
(3.2)
Here C ( t ) = ( ( 12 k + 16 ) t ) / ( 54 k 2 + 27 k 27 ) , D ( t ) = ( ( 12 k + 16 ) t ) / ( 42 k 2 + 21 k 21 ) , P ( t ) = ( 2 t + 1 ) / ( ( 4 t + 3 ) 2 ) , Q ( t ) = 4 / ( 2 ln t + 3 ) , t [ k 1 , k ) , b k = ( 6 k 2 + 17 k + 7 ) / ( 6 k 2 + 17 k + 12 ) , t 0 = 1 , k = 2 , 3 , 4 ,  , f ( x ) = x ( 1 + ( ( 2 / 5 ) sin 2 x ) ) , τ = 2 / 3 , δ = π / 2 , α = 1 / ( 2 e 3 ) , and β = 1 / ( 3 e ) . We can show that
  1. (i)

    | x | | ( 1 + 2 5 sin 2 x ) x | 7 5 | x | , x R , ( 1 + 2 5 sin 2 x ) x 2 > 0 for x 0 ;

     
  2. (ii)

    lim t | C ( t ) | = 2 9 = μ < 1 , lim t | D ( t ) | = 2 7 = γ < 1 with μ + γ = 32 63 < 1 , and C ( k ) = 6 k 2 + 17 k + 7 6 k 2 + 17 k + 12 C ( k ) , D ( k ) = 6 k 2 + 17 k + 7 6 k 2 + 17 k + 12 D ( k ) ;

     
  3. (iii)
    t k ( 2 / 3 ) and ( 1 / ( 2 e 3 ) ) t k are not impulsive points, 0 < ( 6 k 2 + 17 k + 7 ) / ( 6 k 2 + 17 k + 12 ) 1 for k = 1 , 2 ,  , and
    k = 1 ( 1 6 k 2 + 17 k + 7 6 k 2 + 17 k + 12 ) = k = 1 5 6 k 2 + 17 k + 12 < ;
     
  4. (iv)
    { lim sup t [ t δ t + δ P ( s + δ ) d s + β t t + δ Q ( s / β ) s d s + μ ( 1 + P ( t + τ + δ ) P ( t + δ ) ) + γ ( 1 + P ( ( t / α ) + δ ) α P ( t + δ ) ) ] = 64 63 < 10 7
     
and
{ lim sup t [ t δ t / β P ( s + δ ) d s + β t t / β Q ( s / β ) s d s + μ ( 1 + t Q ( ( t + τ ) / β ) ( t + τ ) Q ( t / β ) ) + γ ( 1 + Q ( t / ( α β ) ) Q ( t / β ) ) ] = 1.2781996 < 2 ;
  1. (v)
    1 P ( s + δ ) d s = 1 2 s + 1 + π ( 4 s + 3 + 2 π ) 2 d s =
     
and
1 Q ( s / β ) s d s = 1 4 s ( 2 ln ( 3 e s ) + 3 ) d s = .

Hence, all assumptions of Theorem 2.3 are satisfied and therefore every solution of (3.2) tends to zero as t .

Declarations

Acknowledgements

We would like to thank the reviewers for their valuable comments and suggestions on the manuscript. This research was funded by King Mongkut’s University of Technology North Bangkok. Contract no. KMUTNB-GOV-57-08.

Authors’ Affiliations

(1)
Nonlinear Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok, 10800, Thailand
(2)
Department of Mathematics, University of Ioannina, Ioannina, 451 10, Greece
(3)
Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

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