Periodic solutions of a neutral differential equation with piecewise constant arguments

J. Math. Anal. Appl. 326 (2007) 736–747 www.elsevier.com/locate/jmaa

Periodic solutions of a neutral differential equation with piecewise constant arguments Gen-Qiang Wang Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, PR China Received 17 October 2005 Available online 18 April 2006 Submitted by William F. Ames

Abstract Existence criteria are proved for the periodic solutions of a neutral differential equation with piecewise constant arguments. © 2006 Published by Elsevier Inc. Keywords: Periodic solution; Neutral differential equation; Piecewise constant argument

1. Introduction Considerable attention has been given to delay differential equations with piecewise constant arguments by several authors including Cooke and Wiener [1], Shah and Wiener [2], Aftabizadeh et al. [3]. This class of differential equations has useful applications in biomedical models of disease that has been developed by Busenberg and Cooke [4]. Studies of such equations were motivated by the fact that they represent a hybrid of discrete and continuous dynamical systems and combine the properties of both differential and differential-difference equations. On the other hand, properties and solutions of delay differential equations with piecewise constant arguments and piecewise constant time delay have received considerable attention by several authors including Wiener [5], Cooke and Wiener [6], Wiener and Cooke [7], Wiener and E-mail address: [email protected]. 0022-247X/$ – see front matter © 2006 Published by Elsevier Inc. doi:10.1016/j.jmaa.2006.02.093

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737

Debnath [8,9], Gopalsamy et al. [10], Lin and Wang [11], Papaschinopoulos and Schinas [12], Huang [13], Shen and Stavroulakis [14] , Wiener and Heller [15], Cooke and Wiener [16], Wang and Cheng [17–19] and Wang and Yan [20]. Carvalho and Wiener [21] considered periodic solutions of the first-order differential equation with piecewise constant argument    x  (t) = ax(t) 1 − x [t] , (1) where a is a constant and [·] is a greatest-integer function. As mentioned by Cooke and Wiener [16], equations such as (1) may be treated as semi-discretization of the ordinary logistic equation and solutions of (1) exhibit a wide variety of properties of interest. It is easy to see that if we let x(t) = ey(t) then (1) can be written as   y  (t) = a 1 − ey([t]) . (2) In this paper, we consider the following equation          x(t) + cx(t + σ ) = f x [t] , x [t − 1] , . . . , x [t − k] ,

(3)

where c is a real number different from −1 and 1, σ is an integer, and f (x0 , . . . , xk ) is a real continuous function defined on R k+1 . The main objective of this paper is to prove several existence criteria for periodic solutions with positive integer period ω of (3) by using Mawhin’s continuous theorem. We know that in case c = 0, (3) has been studied by Wang [22] and we also note that usually, Mawhin’s continuous theorem is directly used to prove the existence of periodic solutions for continuous differential systems with delays or discretion difference systems, but (3) can be a semi-discretion differential system and additionally the neutral term. We introduce a technique to build a relations between (3) and a difference equation. Then, by using Mawhin’s continuous theorem to study ω-periodic solutions of this difference equation, we find the ω-periodic solutions of (3). For the nonlinear difference equations, our results are new and have a strong interest. Also, in case c = 0 in (3) our results are new. By a solution of (3), we mean a function x(t) which is defined on R and which satisfies the following conditions: (i) x(t) is continuous on R; (ii) the derivative (x(t) + cx(t + σ )) exists at each point t ∈ R, with the possible exception of the points [t] ∈ R, where one-side derivatives exist; (iii) Eq. (3) is satisfied on each interval [n, n + 1) ⊂ R with integral endpoints. We also state Mawhin’s continuous theorem (see [23]). Let X and Y be two Banach spaces and L : Dom L ⊂ X → Y is a linear mapping and N : X → Y a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dim Ker L = codim Im L < +∞, and Im L is closed in Y . If L is a Fredholm mapping of index zero, there exist continuous projectors P : X → X and Q : Y → Y such that Im P = Ker L and Im L = Ker Q = Im(I − Q). It follows that L|Dom L∩Ker P : (I − P )X → Im L has an inverse which will be denoted by KP . If Ω is an open and bounded subset of X, the mapping N will be ¯ is bounded and KP (I − Q)N (Ω) ¯ is compact. Since Im Q is called L-compact on Ω¯ if QN (Ω) isomorphic to Ker L there exists an isomorphism J : Im Q → Ker L.

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Theorem A (Mawhin’s continuation theorem [23]). Let L be a Fredholm mapping of index zero, ¯ Suppose and let N be L-compact on Ω. (i) for each λ ∈ (0, 1) and x ∈ ∂Ω, Lx = λN x; and (ii) for each x ∈ ∂Ω ∩ Ker L, QN x = 0 and deg(J QN, Ω ∩ Ker L, 0) = 0. Then the equation Lx = nx has at least one solution in Ω¯ ∩ dom L. 2. A necessary and sufficient condition Consider (3) and the following difference equation: xn + cxn+σ = f (xn , xn−1 , . . . , xn−k ).

(4)

Let Z be the set of integers. By a solution of (4) we mean a sequence {xn }n∈Z , which satisfies (4). A sequence {xn }n∈Z is said to be with period ω, if xn+ω = xn for n ∈ Z. Theorem 2.1. If |c| = 1, then (3) has an ω-periodic solution if and only if (4) has an ω-periodic solution. Lemma 2.1. If |c| = 1, then for each ω-periodic continuous function u(t) which is defined on R, there is a unique ω-periodic continuous function x(t) such that u(t) = x(t) + cx(t + σ ),

t ∈ R.

(5)

Proof. For each ω-periodic continuous function u(t) which is defined on R, in case |c| < 1, it is i i easy to see that ∞ i=0 (−1) c u(t + iσ ) is uniformly convergent on compact intervals of R. If we define x(t) by x(t) =

∞  (−1)i ci u(t + iσ ),

t ∈ R,

(6)

i=0

then x(t) is an ω-periodic continuous function. Furthermore, it is not difficult to check that x(t) satisfies (5). Similarly, in case |c| > 1, if we define x(t) as  i+1 ∞    i 1 x(t) = (−1) u t − (i + 1)σ , t ∈ R, (7) c i=0

then x(t) is an ω-periodic continuous function such that (5) holds. To show uniqueness, let y(t) be an ω-periodic continuous function which is defined on R and which also satisfies u(t) = y(t) + cy(t + σ ),

t ∈ R.

From (5) and (8), we see that for any t ∈ R,     x(t) − y(t) = |c|x(t + σ ) − y(t + σ ). By (9) and the fact that x(t) and y(t) are ω-periodic, we have

(8)

(9)

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    max x(t) − y(t) = supx(t) − y(t) 0tω t∈R   = |c| supx(t + σ ) − y(t + σ ) t∈R   = |c| max x(t) − y(t).

(10)

0tω

Since |c| = 1, (10) implies x(t) = y(t) for t ∈ R. The proof of Lemma 2.1 is complete.

2

Similarly, we know that the following lemma is true. Lemma 2.2. If |c| = 1, then for each ω-periodic sequence {un }n∈Z , there is a unique ω-periodic sequence {xn }n∈Z such that un = xn + cxn+σ ,

n ∈ Z.

(11)

Proof of Theorem 2.1. Let x(t) be an ω-periodic solution of (3). It is easy to see that for any n ∈ Z,          x(t) + cx(t + σ ) = f x [t] , x [t − 1] , . . . , x [t − k] , n  t < n + 1. Integrating (3) from n to t, we have x(t) + cx(t + σ ) − x(n) − cx(n + σ )   = (t − n)f x(n), x(n − 1), . . . , x(n − k) ,

n  t < n + 1.

Since limt→n+1 −x(t) = x(n + 1), we see further that   x(n) + cx(n + σ ) = f x(n), x(n − 1), . . . , x(n − k) ,

(12)

n ∈ Z.

If we now let xn = x(n) for n ∈ Z, then {xn }n∈Z is an ω-periodic solution of (4). Conversely, let {xn }n∈Z be an ω-periodic solution of (4). Set x(n) = xn , for n ∈ Z, and let the function y(t) on each interval [n, n + 1) be defined by   u(t) = x(n) + cx(n + σ ) + (t − n)f x(n), x(n − 1), . . . , x(n − k) , n  t < n + 1.

(13)

Then it is not difficult to check that this u(t) is an ω-periodic continuous function. Furthermore, by Lemma 2.1, there is a unique ω-periodic continuous function y(t) such that u(t) = y(t) + cy(t + σ ),

t ∈ R.

(14)

From (13) and (14), we see that u(n) = x(n) + cx(n + σ ) = y(n) + cy(n + σ ),

n ∈ Z.

(15)

Note that {u(n)}n∈Z , {y(n)}n∈Z and {x(n)}n∈Z are ω-periodic sequences. In view of Lemma 2.2, we have y(n) = x(n),

n ∈ Z.

(16)

Thus, we can let x(t) = y(t), t ∈ R. From (13) and (14), it yields (12). We know that x(t) is an ω-periodic solution of (3). The proof of Theorem 2.1 is complete. 2

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3. Main results We use the following conditions, where D is a positive constant, α and β are nonnegative constants: (a) (b1 ) (b2 ) (c1 ) (c2 )

|f (x0 , . . . , xk )|  β max0ik |xi | + α, for (x0 , . . . , xk ) ∈ R k+1 ; |f (x0 , . . . , xk )| > 0 for xi  D (i = 0, . . . , k); |f (x0 , . . . , xk )| < 0 for xi  D (i = 0, . . . , k); |f (x0 , . . . , xk )| < 0 for xi  −D (i = 0, . . . , k); |f (x0 , . . . , xk )| > 0 for xi  −D (i = 0, . . . , k).

Our main results are the following two theorems: Theorem 3.1. If |c| = 1, the conditions (a) , (b1 ) and (c1 ) are satisfied, then for (3) has an ω-periodic solution.

1 2|1−|c|| ωβ

< 1,

Theorem 3.2. If |c| = 1, the conditions (a), (b2 ) and (c2 ) are satisfied, then for (3) has an ω-periodic solution.

1 2|1−|c|| ωβ

< 1,

We only give the proof of Theorem 3.1, as Theorem 3.2 can be proved similarly. We first need some basic tools. First of all, for any real sequence {un }n∈Z , we define a nonstandard “summation” operation ⎧ β β α  β, ⎨ n=α un , (17) un =  0, β = α − 1, ⎩ α−1 n=α u , β < α − 1. n n=β+1 It is then easy to see that {xn }n∈Z is an ω-periodic solution of (4) if and only if {xn }n∈Z is an ω-periodic solution of the following equation: (xn + cxn+σ ) − (x0 + cxσ ) =

n−1

f (xi , xi−1 , . . . , xi−k ),

n ∈ Z.

(18)

i=0

Let Xω be the Banach space of all real ω-periodic sequences of the form x = {xn }n∈Z which is defined on R and endowed with the usual linear structure as well as the norm x 1 = max0iω−1 |xi |. Let Yω be the Banach space of all real sequences of the form y = {yn }n∈Z = {nα + hn }n∈Z such that y0 = 0, where α ∈ R and {hn }n∈Z ∈ Xω and endowed with the usual linear structure as well as the norm y 2 = |α| + h . Let the zero elements of Xω and Yω be denoted by θ1 and θ2 , respectively. Define the mappings L : Xω → Yω and N : Xω → Yω respectively by (Lx)n = (xn + cxn+σ ) − (x0 + cxσ ),

n ∈ Z,

(19)

and (N x)n =

n−1

f (xi , xi−1 , . . . , xi−k ),

n ∈ Z.

(20)

i=0

Let h¯ n =

n−1 i=0

n f (xi , xi−1 , . . . , xi−k ), ω ω−1

f (xi , xi−1 , . . . , xi−k ) −

i=0

n ∈ Z.

(21)

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Since {h¯ n }n∈Z ∈ Xω and h¯ 0 = 0, N is a well-defined operator from Xω into Yω . Let us define P : Xω → Xω and Q : Yω → Yω respectively by (P x)n = x0 ,

n ∈ Z,

(22)

for x = {xn }n∈Z ∈ Xω , and (Qy)n = nα,

n ∈ Z,

(23)

for y = {nα + hn }n∈Z ∈ Yω . Lemma 3.1. If |c| = 1 and let the mapping L be defined by (19). Then we have Ker L = {x ∈ Xω | xn = x0 , n ∈ Z, x0 ∈ R}.

(24)

Proof. It suffices to show that for each x = {xn }n∈Z ∈ Ker L, then xn = x0 ∈ R, for any n ∈ Z. Indeed, let un = xn + cxn+σ , for n ∈ Z, from (19), we see that Lx = θ2 if and only if un = u0 , for any n ∈ Z. Thus u = {un }n∈Z is a constant sequence. Let vn = u0 /(1 + c), for n ∈ Z, we furthermore see that un = vn + cvn+σ = xn + cxn+σ . By using the uniqueness of Lemma 2.2, we know that xn = vn = u0 /(1 + c) ∈ R, for any n ∈ Z. The proof of Lemma 3.1 is complete. 2 Lemma 3.2. If |c| = 1 and let the mapping L be defined by (19). Then we have Im L = {y ∈ Xω | y0 = 0} ⊂ Yω .

(25)

Proof. It suffices to show that for each y = {yn }n∈Z ∈ Xω that satisfies y0 = 0, there is x = {xn }n∈Z ∈ Xω such that yn = (xn + cxn+σ ) − (x0 + cxσ ),

n ∈ Z.

(26)

Indeed, from Lemma 2.2, there is x¯ = {x¯n }n∈Z ∈ Xω , such that yn = x¯n + cx¯n+σ ,

n ∈ Z,

(27)

and note that y0 = x¯0 + cx¯0+σ = 0, if we let x = x, ¯ then (26) is satisfied. The proof of Lemma 3.2 is complete.

(28) 2

In view of (22), (23), Lemmas 3.1 and 3.2, we know that the operators P and Q are projections and Xω = Ker P ⊕ Ker L, Yω = Im L ⊕ Im Q. It is easy to see that dim Ker L = codim Im L = 1, and that Im L = {y ∈ Xω | y0 = 0} is closed in Yω . Thus the following lemma is true. Lemma 3.3. If |c| = 1, and let the mapping L defined by (19). Then L is a Fredholm mapping of index zero. Indeed, from Lemmas 3.1 and 3.2 and the definition of Yω , dim Ker L = codim Im L = 1 < +∞. From (25), we see that Im L is closed in Yω . Hence L is a Fredholm mapping of index zero. Lemma 3.4. If |c| = 1, and let L and N be defined by (19) and (20), respectively. Suppose Ω is ¯ an open and bounded subset of Xω . Then N is L-compact on Ω.

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¯ Proof. It is easy to see that for any x = {xn }n∈Z ∈ Ω, n f (xi , xi−1 , . . . , xi−k ), ω ω−1

(QN x)n =

n ∈ Z.

(29)

i=0

Thus

 ω−1  1    QNx 2 =  f (xi , xi−1 , . . . , xi−k ), ω 

(30)

i=0

¯ is bounded and so that QN(Ω) n−1 ω−1   n (I − Q)N x n = f (xi , xi−1 , . . . , xi−k ) − f (xi , xi−1 , . . . , xi−k ), ω i=0

n ∈ Z. (31)

i=0

By direct calculations, we know that in the case |c| < 1,

n+j σ ∞    j j (−1) c f (xi , xi−1 , . . . , xi−k ) KP (I − Q)N x n = j =0

i=0

n + jσ − ω

ω−1



f (xi , xi−1 , . . . , xi−k ) ,

n ∈ Z,

(32)

i=0

and in the case |c| > 1,

 i+1 n−(j +1)σ ∞    j 1 (−1) f (xi , xi−1 , . . . , xi−k ) KP (I − Q)N x n = c j =0 i=0  ω−1 n − (j + 1)σ f (xi , xi−1 , . . . , xi−k ) , n ∈ Z. − ω

(33)

i=0

It follows from (32) and (33) that   KP (I − Q)N x   1

ω−1   2 f (xi , xi−1 , . . . , xi−k ). |1 − |c||

(34)

i=0

¯ is bounded and dim Xω = ω < ∞, we know that KP (I − Q)N (Ω) ¯ is Thus KP (I − Q)N (Ω) ¯ The proof of Lemma 3.4 is complete. 2 relatively compact. Thus N is L-compact on Ω. Lemma 3.5. If |c| = 1, then for any x = {xn }n∈Z ∈ Xω we have 1 |xk + cxk+σ |. 2|1 − |c|| ω−1

max

0s,iω−1

|xs − xi | 

(35)

k=0

Proof. Let x = {xn }n∈Z ∈ Xω and s, i ∈ {0, 1, . . . , ω − 1}. Without loss of generality, let s ∈ {i + 1, . . . , z + ω − 1}, we have xs + cxs+σ = xi + cxi+σ +

s−1 (xk + cxk+σ ) k=i

(36)

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743

and xi + cxi+σ = xi+ω + cxi+ω+σ = xs + cxs+σ +

i+ω−1

(xk + cxk+σ ).

(37)

k=s

From (36) and (37), we see that 2(xs + cxs+σ ) = 2(xi + cxi+σ ) +

s−1 i+ω−1 (xk + cxk+σ ) − (xk + cxk+σ ), k=i

that is xs + cxs+σ = xi + cxi+σ

(38)

k=s



s−1 i+ω−1 1 + (xk + cxk+σ ) − (xk + cxk+σ ) . 2 k=i

(39)

k=s

Thus for any s, i ∈ {0, 1, . . . , ω − 1}, i+ω−1     (xk + cxk+σ ) xs − xi + c(xs+σ − xi+σ )  1 2 k=i

ω−1  1  (xk + cxk+σ ), = 2

(40)

k=0

so that ω−1  1      (xk + cxk+σ ). max xs − xi + c(xs+σ − xi+σ )  2 0s,iω−1

(41)

k=0

Since in the case |c| < 1,   1 − |c| max |xs − xi | = 0s,iω−1

 and in the case |c| > 1,   |c| − 1 max

0s,iω−1

max

max

0s,iω−1

|xs − xi | = |c| 

|xs − xi | − |c| max |xs+σ − xi+σ | 0s,iω−1   xs − xi + c(xs+σ − xi+σ ),

0s,iω−1

max

(42)

|xs+σ − xi+σ | − max |xs − xi | 0s,iω−1   xs − xi + c(xs+σ − xi+σ ),

0s,iω−1

max

0s,iω−1

so that if |c| = 1, then we have max

0s,iω−1

|xs − xi | 

  1 max xs − xi + c(xs+σ − xi+σ ). |1 − |c|| 0s,iω−1

It follows from (41) and (43) that (35) holds. The proof of Lemma 3.5 is complete.

(43) 2

Now, we consider the following auxiliary equation (xn + cxn+σ ) − (x0 + cxσ ) = λ

n−1 i=0

where λ ∈ (0, 1).

f (xi , xi−1 , . . . , xi−k ),

n ∈ Z,

(44)

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Lemma 3.6. If the conditions of Theorem 3.1 are satisfied, then there is a positive constant D1 such that for any ω-periodic solution x = {xn }n∈Z of (44), x 1  D1 .

(45)

Proof. Let x = {xn }n∈Z be any ω-periodic solution of (44). Furthermore, we set xα = max0iω−1 xi and xβ = min0iω−1 xi , where 0  α, β  ω − 1. By Lemma 3.5 and (44), we have xα − x β =

max

0s,iω−1

|xs − xi |

1 |xk + cxk+σ | 2|1 − |c|| ω−1



k=0



1 2|1 − |c||

ω−1

  f (xi , xi−1 , . . . , xi−k ).

(46)

k=0

If there is some xl , 0  l  ω − 1, such that |xl |  D, then in view of (46), for any n ∈ {0, 1, . . . , ω − 1}, |xn | = |xl | + |xn − xl | D+

max

0s,iω−1

D+

|xs − xi |

ω−1   1 f (xi , xi−1 , . . . , xi−k ). 2|1 − |c||

(47)

k=0

Otherwise, from (44) we see that ω−1

f (xi , xi−1 , . . . , xi−k ) = 0.

(48)

i=0

By (b1 ), (c1 ) and (48), we know that xα  D and xβ  −D. Thus, from (46) we have ω−1   1 f (xi , xi−1 , . . . , xi−k ) xα  xβ + 2|1 − |c|| k=0

 −D +

ω−1   1 f (xi , xi−1 , . . . , xi−k ) 2|1 − |c||

(49)

k=0

and ω−1   1 f (xi , xi−1 , . . . , xi−k ) xβ  x α − 2|1 − |c|| k=0

D−

1 2|1 − |c||

ω−1   f (xi , xi−1 , . . . , xi−k ). k=0

It follows that for n ∈ {0, 1, . . . , ω − 1},

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D−

745

ω−1   1 f (xi , xi−1 , . . . , xi−k ) 2|1 − |c|| k=0

 xn  −D +

ω−1   1 f (xi , xi−1 , . . . , xi−k ), 2|1 − |c||

(51)

k=0

or |xn |  D +

ω−1   1 f (xi , xi−1 , . . . , xi−k ). 2|1 − |c||

(52)

k=0

It is easy to see from condition (a) that ω−1 ω−1   f (xi , xi−1 , . . . , xi−k )  β max |xi−j | + αω. i=0

i=0

0j k

(53)

In view of (52) and (53), we have x 1  D +

ω−1   1 f (xi , xi−1 , . . . , xi−k ) 2|1 − |c|| k=0

 1 αω β max |xi−j | + 2|1 − |c|| 2|1 − |c|| 0j k ω−1 k

D+

k=0 j =0

 σ1 x 1 + C1 , where C1 = D + x 1 

αω 2|1−|c||

and σ1 =

(54) 1 2|1−|c|| ωβ.

It follows that

C1 . 1 − σ1

The proof of Lemma 3.6 is complete.

2

We now turn to the proof of Theorem 3.1. Let L, N , P and Q be defined by (19), (20), (22) and (23), respectively. By Lemma 3.6, there is a positive constant D1 such that any ω-periodic solution x = {xn }n∈Z of (44) satisfies (45). Set   Ω = x ∈ Xω | x 1 < D¯ , where D¯ is a fixed number which satisfies D¯ > D + D1 , Ω is an open and bounded subset of Xω . Furthermore, in view of Lemmas 3.3 and 3.4, L is a Fredholm mapping of index zero and N is L¯ Noting that D¯ > D1 , by Lemma 3.6, for each λ ∈ (0, 1) and x = {xn }n∈Z ∈ ∂Ω, compact on Ω. ¯ LH x = λNx. Next note that a function x ∈ ∂Ω ∩ Ker L must be constant: xn ≡ D¯ or xn = −D, for n ∈ Z. Hence by (20) and (23), n f (xi , xi−1 , . . . , xi−k ), ω ω−1

(QN x)n =

i=0

so QNx = θ2 .

n ∈ Z,

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The isomorphism J : Im Q → Ker L is defined by (J (nα))n = α for α ∈ R and n ∈ Z. Then 1 f (xi , xi−1 , . . . , xi−k ) = 0, ω ω−1

(J QN x)n =

n ∈ Z.

(55)

i=0

¯ n∈Z , then In particular, we see that if x = {D} 1 ¯ D, ¯ . . . , D) ¯ > 0, f (D, ω ω−1

(J QN x)n =

n ∈ Z,

(56)

i=0

¯ n∈Z , then and if x = {−D} 1 ¯ −D, ¯ . . . , −D) ¯ < 0, (J QN x)n = f (−D, ω ω−1

n ∈ Z.

(57)

i=0

This shows that deg(J QN x, Ω ∩ Ker L, θ1 ) = 0. By Theorem A, we see that equation Lx = N x has at least one solution in Ω¯ ∩ Dom L. In other words, (4) has an ω-periodic solution. Thus, by using Theorem 2.1 we see that (3) has an ω-periodic solution. The proof of Theorem 3.1 is complete. 2 4. An example Consider the equation         1/5   x [t − 1] 2/5 x [t − 2} 2/5 + 2, x(t) + 5x(t − 1) = x [t]

(58)

1/5 2/5 2/5

where f (x0 , x1 , x2 ) = x0 x1 x2 + 2, c = 5, σ = −1. Let D > 4 and β = 1, α = 2, then the 1 ωβ = 78 < 1. Therefore, by Theorem 3.1 we conditions (a) , (b1 ) and (c1 ) are satisfied and 2|1−|c|| know that (58) has a 7-periodic solution. Furthermore, this solution is nontrivial since f (0, 0, 0) is not identically zero. References [1] K.L. Cooke, J. Wiener, Retarded differential equations with piecewise constant delays, J. Math. Anal. Appl. 99 (1984) 265–297. [2] S.M. Shah, J. Wiener, Advanced differential equations with piecewise constant argument deviations, Int. J. Math. Math. Sci. 6 (1983) 671–703. [3] A.R. Aftabizadeh, J. Wiener, J. Xu, Oscillatory and periodic solutions of delay differential equations with piecewise constant argument, Proc. Amer. Math. Soc. 99 (1987) 673–679. [4] S. Busenberg, L. Cooke, Models of vertically transmitted diseases with sequential-continuous dynamics, in: V. Lakshmikantham (Ed.), Nonlinear Phenomena in Mathematical Sciences, Academic Press, New York, 1982, pp. 179– 187. [5] J. Wiener, Boundary-value problems for partial differential equations with piecewise constant delay, Int. J. Math. Math. Sci. 14 (1991) 485–496. [6] K.L. Cooke, J. Wiener, An equation alternately of retarded and advanced type, Proc. Amer. Math. Soc. 99 (1987) 726–732. [7] J. Wiener, K.L. Cooke, Oscillations in systems of differential equations with piecewise constant argument, J. Math. Anal. Appl. 129 (1988) 243–255.

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