Advances in Difference Equations
Volume 2009 (2009), Article ID 143175, 14 pages
doi:10.1155/2009/143175
Research Article

On the Spectrum of Almost Periodic Solution of Second-Order Neutral Delay Differential Equations with Piecewise Constant of Argument

Li Wang and Chuanyi Zhang

Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China

Received 16 December 2008; Accepted 10 April 2009

Academic Editor: Ondrej Dosly

Copyright © 2009 Li Wang and Chuanyi Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The spectrum containment of almost periodic solution of second-order neutral delay differential equations with piecewise constant of argument (EPCA, for short) of the form ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) = 𝑞 𝑥 ( 2 [ ( 𝑡 + 1 ) / 2 ] ) + 𝑓 ( 𝑡 ) is considered. The main result obtained in this paper is different from that given by some authors for ordinary differential equations (ODE, for short) and clearly shows the differences between ODE and EPCA. Moreover, it is also different from that given for equation ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) = 𝑞 𝑥 ( [ 𝑡 ] ) + 𝑓 ( 𝑡 ) because of the difference between [ 𝑡 ] and 2 [ ( 𝑡 + 1 ) / 2 ] .

1. Introduction and Some Preliminaries

Differential equations with piecewise constant argument, which were firstly considered by Cooke and Wiener [1] and Shah and Wiener [2], combine properties of both differential and difference equations and usually describe hybrid dynamical systems and have applications in certain biomedical models in the work of Busenberg and Cooke [3]. Over the years, more attention has been paid to the existence, uniqueness, and spectrum containment of almost periodic solutions of this type of equations (see, e.g., [4–12] and reference there in).

If 𝑔 1 ( 𝑡 ) and 𝑔 2 ( 𝑡 ) are almost periodic, then the module containment property m o d ( 𝑔 1 ) ⊂ m o d ( 𝑔 2 ) can be characterized in several ways (see [13–16]). For periodic function this inclusion just means that the minimal period of 𝑔 1 ( 𝑡 ) is a multiple of the minimal period of 𝑔 2 ( 𝑡 ) . Some properties of basic frequencies (the base of spectrum) were discussed for almost periodic functions by Cartwright. In [17], Cartwright compared basic frequencies (the base of spectrum) of almost periodic differential equations (ODE) ̇ 𝑥 = 𝜓 ( 𝑥 , 𝑡 ) , 𝑥 ∈ 𝑅 𝑛 , with those of its unique almost periodic solution. For scalar equation, 𝑛 = 1 , Cartwright’s results in [17] implied that the number of basic frequencies of ̇ 𝑥 = 𝜓 ( 𝑥 , 𝑡 ) , 𝑥 ∈ 𝑅 , is the same as that of basic frequencies of its unique solution.

The spectrum containment of almost periodic solution of equation ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) = 𝑞 𝑥 ( [ 𝑡 ] ) + 𝑓 ( 𝑡 ) was studied in [9, 10]. Up to now, there have been no papers concerning the spectrum containment of almost periodic solution of equation ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) 2 = 𝑞 𝑥 𝑡 + 1 2 + 𝑓 ( 𝑡 ) , ( 1 . 1 ) where [ ⋅ ] denotes the greatest integer function, 𝑝 , 𝑞 are nonzero real constants, | 𝑝 | ≠ 1 , 𝑞 ≠ − 2 ( 𝑝 2 + 1 ) , and 𝑓 ( 𝑡 ) is almost periodic. In this paper, we investigate the existence, uniqueness, and spectrum containment of almost periodic solutions of (1.1). The main result obtained in this paper is different from that given in [17] for ordinary differential equations (ODE, for short). This clearly shows differences between ODE and EPCA. Moreover, it is also different from that given in [9, 10] for equation ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) = 𝑞 𝑥 ( [ 𝑡 ] ) + 𝑓 ( 𝑡 ) . This is due to the difference between [ 𝑡 ] and 2 [ ( 𝑡 + 1 ) / 2 ] . As well known, both solutions of (1.1) and equation ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) = 𝑞 𝑥 ( [ 𝑡 ] ) + 𝑓 ( 𝑡 ) can be constructed by the solutions of corresponding difference equations. However, noticing the difference between [ 𝑡 ] and 2 [ ( 𝑡 + 1 ) / 2 ] , the solution of difference equation corresponding to the latter can be obtained directly (see [4]), while the solution { 𝑥 𝑛 } of difference equation corresponding to the former (i.e., (1.1) cannot be obtained directly. In fact, { 𝑥 𝑛 } consists of two parts: { 𝑥 2 𝑛 } and { 𝑥 2 𝑛 + 1 } . We will first obtain { 𝑥 2 𝑛 } by solving a difference equation and then obtain { 𝑥 2 𝑛 + 1 } from { 𝑥 2 𝑛 } . (Similar technology can be seen in [8].) A detailed account will be given in Section 2.

Now, We give some preliminary notions, definitions, and theorem. Throughout this paper 𝑍 , 𝑅 , and 𝐶 denote the sets of integers, real, and complex numbers, respectively. The following preliminaries can be found in the books, for example, [13–16].

Definition 1.1. ( 1 ) A subset 𝑃 of 𝑅 is said to be relatively dense in 𝑅 if there exists a number 𝑝 > 0 such that 𝑃 ∩ [ 𝑡 , 𝑡 + 𝑝 ] ≠ ∅ for all 𝑡 ∈ 𝑅 .
( 2 ) A continuous function 𝑓 ∶ 𝑅 → 𝑅 is called almost periodic (abbreviated as 𝒜 𝒫 ( 𝑅 ) ) if the 𝜖 -translation set of 𝑓 𝑇 | | 𝑓 | | ( 𝑓 , 𝜖 ) = 𝜏 ∈ 𝑅 ∶ ( 𝑡 + 𝜏 ) − 𝑓 ( 𝑡 ) < 𝜖 , ∀ 𝑡 ∈ 𝑅 ( 1 . 2 ) is relatively dense for each 𝜖 > 0 .

Definition 1.2. Let 𝑓 be a bounded continuous function. If the limit l i m 𝑇 → ∞ 1 2 𝑇 𝑇 − 𝑇 𝑓 ( 𝑡 ) 𝑑 𝑡 0 𝑥 0 2 0 0 𝑑 ( 1 . 3 ) exists, then we call the limit mean of 𝑓 and denote it by 𝑀 ( 𝑓 ) .

If 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) , then the limit l i m 𝑇 → ∞ 1 2 𝑇 𝑇 + 𝑠 − 𝑇 + 𝑠 𝑓 ( 𝑡 ) 𝑑 𝑡 0 𝑥 0 2 0 0 𝑑 ( 1 . 4 ) exists uniformly with respect to 𝑠 ∈ 𝑅 . Furthermore, the limit is independent of 𝑠 .

For any 𝜆 ∈ 𝑅 and 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) since the function 𝑓 𝑒 − 𝑖 𝜆 ⋅ is in 𝒜 𝒫 ( 𝑅 ) , the mean exists for this function. We write 𝑎 ( 𝜆 ; 𝑓 ) = 𝑀 𝑓 𝑒 − 𝑖 𝜆 ⋅ , ( 1 . 5 ) then there exists at most a countable set of 𝜆 ’s for which 𝑎 ( 𝜆 ; 𝑓 ) ≠ 0 . The set Λ 𝑓 = { 𝜆 ∶ 𝑎 ( 𝜆 ; 𝑓 ) ≠ 0 } ( 1 . 6 ) is called the frequency set (or spectrum) of 𝑓 . It is clear that if ∑ 𝑓 ( 𝑡 ) = 𝑛 𝑘 = 1 𝑐 𝑘 𝑒 𝑖 𝜆 𝑘 𝑡 0 𝑥 0 2 0 0 𝑑 , then 𝑎 ( 𝜆 ; 𝑓 ) = 𝑐 𝑘 if 𝜆 = 𝜆 𝑘 , for some 𝑘 = 1 , … , 𝑛 ; and 𝑎 ( 𝜆 ; 𝑓 ) = 0 if 𝜆 ≠ 𝜆 𝑘 , for any 𝑘 = 1 , … , 𝑛 . Thus, Λ 𝑓 = { 𝜆 𝑘 , 𝑘 = 1 , … , 𝑛 } .

Members of Λ 𝑓 are called the Fourier exponents of 𝑓 , and 𝑎 ( 𝜆 ; 𝑓 ) ’s are called the Fourier coefficients of 𝑓 . Obviously, Λ 𝑓 is countable. Let Λ 𝑓 = { 𝜆 𝑘 } and 𝐴 𝑘 = 𝑎 ( 𝜆 𝑘 ; 𝑓 ) . Thus 𝑓 can associate a Fourier series: 𝑓 ( 𝑡 ) ∼ ∞ 𝑘 = 1 𝐴 𝑘 𝑒 𝑖 𝜆 𝑘 𝑡 0 𝑥 0 2 0 0 𝑑 . ( 1 . 7 )

The Approximation Theorem
Let 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) and Λ 𝑓 = { 𝜆 𝑘 } . Then for any 𝜖 > 0 there exists a sequence { 𝜎 𝜖 } of trigonometric polynomials 𝜎 𝜖 ( 𝑡 ) = 𝑛 ( 𝜖 ) 𝑘 = 1 𝑏 𝑘 , 𝜖 𝑒 𝑖 𝜆 𝑘 𝑡 0 𝑥 0 2 0 0 𝑑 ( 1 . 8 ) such that ‖ 𝜎 𝜖 − 𝑓 ‖ ≤ 𝜖 , ( 1 . 9 ) where 𝑏 𝑘 , 𝜖 is the product of 𝑎 ( 𝜆 𝑘 ; 𝑓 ) and certain positive number (depending on 𝜖 and 𝜆 𝑘 ) and l i m 𝜖 → 0 𝑏 𝑘 , 𝜖 = 𝑎 ( 𝜆 𝑘 ; 𝑓 ) .

Definition 1.3. ( 1 ) For a sequence { 𝑔 ( 𝑛 ) ∶ 𝑛 ∈ 𝑍 } , define [ 𝑔 ( 𝑛 ) , 𝑔 ( 𝑛 + 𝑝 ) ] = { 𝑔 ( 𝑛 ) , … , 𝑔 ( 𝑛 + 𝑝 ) } and call it sequence interval with length 𝑝 ∈ 𝑍 . A subset 𝑃 of 𝑍 is said to be relatively dense in 𝑍 if there exists a positive integer 𝑝 such that 𝑃 ∩ [ 𝑛 , 𝑛 + 𝑝 ] ≠ ∅ for all 𝑛 ∈ 𝑍 .
( 2 ) A bounded sequence 𝑔 ∶ 𝑍 → 𝑅 is called an almost periodic sequence (abbreviated as 𝒜 𝒫 𝒮 ( 𝑅 ) ) if the 𝜖 -translation set of 𝑔 𝑇 | | 𝑔 | | ( 𝑔 , 𝜖 ) = 𝜏 ∈ 𝑍 ∶ ( 𝑛 + 𝜏 ) − 𝑔 ( 𝑛 ) < 𝜖 , ∀ 𝑛 ∈ 𝑍 ( 1 . 1 0 ) is relatively dense for each 𝜖 > 0 .

For an almost periodic sequence { 𝑔 ( 𝑛 ) } , it follows from the lemma in [13] that 𝑎 ( 𝑧 ; 𝑔 ) = l i m 𝑁 → ∞ 1 2 𝑁 𝑁 𝑘 = − 𝑁 𝑧 − 𝑘 𝑔 ( 𝑘 ) , ∀ 𝑧 ∈ 𝑆 1 = | | 𝑧 ∈ 𝐶 ∶ 𝑧 | | = 1 ( 1 . 1 1 ) exists. The set 𝜎 𝑏 ( 𝑔 ) = 𝑧 ∶ 𝑎 ( 𝑧 ; 𝑔 ) ≠ 0 , 𝑧 ∈ 𝑆 1 ( 1 . 1 2 ) is called the Bohr spectrum of { 𝑔 ( 𝑛 ) } . Obviously, for almost periodic sequence ∑ 𝑔 ( 𝑛 ) = 𝑚 𝑘 = 1 𝑟 𝑘 𝑧 𝑛 𝑘 0 𝑥 0 2 0 0 𝑑 , 𝑎 ( 𝑧 ; 𝑔 ) = 𝑟 𝑘 if 𝑧 = 𝑧 𝑘 , for some 𝑘 = 1 , … , 𝑚 ; 𝑎 ( 𝑧 ; 𝑔 ) = 0 if 𝑧 ≠ 𝑧 𝑘 , for any 𝑘 = 1 , … , 𝑚 . So, 𝜎 𝑏 ( 𝑔 ) = { 𝑧 𝑘 , 𝑘 = 1 , … , 𝑚 } .

2. The Statement of Main Theorem

We begin this section with a definition of the solution of (1.1).

Definition 2.1. A continuous function 𝑥 ∶ 𝑅 → 𝑅 is called a solution of (1.1) if the following conditions are satisfied:(i) 𝑥 ( 𝑡 ) satisfies (1.1) for 𝑡 ∈ 𝑅 , 𝑡 ≠ 𝑛 ∈ 𝑍 ;(ii)the one-sided second-order derivatives ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) exist at 𝑛 , 𝑛 ∈ 𝑍 .

In [8], the authors pointed out that if 𝑥 ( 𝑡 ) is a solution of (1.1), then ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) are continuous at 𝑡 ∈ 𝑅 , which guarantees the uniqueness of solution of (1.1) and cannot be omitted.

To study the spectrum of almost periodic solution of (1.1), we firstly study the solution of (1.1). Let 𝑓 𝑛 ( 1 ) = 𝑛 𝑛 + 1 𝑠 𝑛 𝑓 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 , 𝑓 𝑛 ( 2 ) = 𝑛 𝑛 − 1 𝑠 𝑛 𝑓 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 , ℎ 𝑛 = 𝑓 𝑛 ( 1 ) + 𝑓 𝑛 ( 2 ) . ( 2 . 1 )

Suppose that 𝑥 ( 𝑡 ) is a solution of (1.1), then ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) exist and are continuous everywhere on 𝑅 . By a process of integrating (1.1) two times in 𝑡 ∈ [ 2 𝑛 − 1 , 2 𝑛 + 1 ) or 𝑡 ∈ [ 2 𝑛 , 2 𝑛 + 2 ) as in [7, 8, 18], we can easily get 𝑥 ( 2 𝑛 + 1 ) + ( 𝑝 − 2 − 𝑞 ) 𝑥 ( 2 𝑛 ) + ( 1 − 2 𝑝 ) 𝑥 ( 2 𝑛 − 1 ) + 𝑝 𝑥 ( 2 𝑛 − 2 ) = ℎ 2 𝑛 , 𝑞 1 − 2 𝑞 𝑥 ( 2 𝑛 + 2 ) + ( 𝑝 − 2 ) 𝑥 ( 2 𝑛 + 1 ) + 1 − 2 𝑝 − 2 𝑥 ( 2 𝑛 ) + 𝑝 𝑥 ( 2 𝑛 − 1 ) = ℎ 2 𝑛 + 1 . ( 2 . 2 )

These lead to the difference equations 𝑝 𝑥 2 𝑛 − 2 + ( 1 − 2 𝑝 ) 𝑥 2 𝑛 − 1 + ( 𝑝 − 2 − 𝑞 ) 𝑥 2 𝑛 + 𝑥 2 𝑛 + 1 = ℎ 2 𝑛 , ( 2 . 3 ) 𝑝 𝑥 2 𝑛 − 1 + 𝑞 1 − 2 𝑝 − 2 𝑥 2 𝑛 + ( 𝑝 − 2 ) 𝑥 2 𝑛 + 1 + 𝑞 1 − 2 𝑥 2 𝑛 + 2 = ℎ 2 𝑛 + 1 . ( 2 . 4 )

Suppose that | 𝑝 | ≠ 1 . First, multiply the two sides of (2.3) and (2.4) by 𝑝 and ( 2 𝑝 − 1 ) , respectively, then add the resulting equations to get 𝑥 2 𝑛 + 1 = 1 2 ( 𝑝 − 1 ) 2 𝑝 ℎ 2 𝑛 − 𝑝 ( 𝑝 − 2 − 𝑞 ) 𝑥 2 𝑛 − 𝑝 2 𝑥 2 𝑛 − 2 + ( 2 𝑝 − 1 ) ℎ 2 𝑛 + 1 − 1 2 ( 𝑝 − 1 ) 2 ( 𝑞 2 𝑝 − 1 ) 1 − 2 𝑥 2 𝑛 + 2 𝑞 + ( 2 𝑝 − 1 ) 1 − 2 𝑝 − 2 𝑥 2 𝑛 . ( 2 . 5 )

Similarly, one gets 𝑥 2 𝑛 − 1 = 1 2 ( 𝑝 − 1 ) 2 ( 2 − 𝑝 ) ℎ 2 𝑛 − ( 2 − 𝑝 ) ( 𝑝 − 2 − 𝑞 ) 𝑥 2 𝑛 − ( 2 − 𝑝 ) 𝑝 𝑥 2 𝑛 − 2 + 1 2 ( 𝑝 − 1 ) 2 ℎ 2 𝑛 + 1 − 𝑞 1 − 2 𝑥 2 𝑛 + 2 − 𝑞 1 − 2 𝑝 − 2 𝑥 2 𝑛 . ( 2 . 6 )

Replacing 2 𝑛 by ( 2 𝑛 + 2 ) in (2.6) and comparing with (2.5), one gets 𝑞 1 − 2 𝑥 2 𝑛 + 4 − 𝑝 2 𝑥 − 2 𝑝 𝑞 + 3 𝑞 + 2 2 𝑛 + 2 + 2 𝑝 2 𝑞 + 2 𝑝 𝑞 − 2 𝑥 + 1 2 𝑛 − 𝑝 2 𝑥 2 𝑛 − 2 = ℎ 2 𝑛 + 3 + ( 2 − 𝑝 ) ℎ 2 𝑛 + 2 + ( 1 − 2 𝑝 ) ℎ 2 𝑛 + 1 − 𝑝 ℎ 2 𝑛 . ( 2 . 7 )

The corresponding homogeneous equation is 𝑞 1 − 2 𝑥 2 𝑛 + 4 − 𝑝 2 𝑥 − 2 𝑝 𝑞 + 3 𝑞 + 2 2 𝑛 + 2 + 2 𝑝 2 𝑞 + 2 𝑝 𝑞 − 2 𝑥 + 1 2 𝑛 − 𝑝 2 𝑥 2 𝑛 − 2 = 0 . ( 2 . 8 )

We can seek the particular solution as 𝑥 2 𝑛 = 𝜉 𝑛 for this homogeneous difference equation. At this time, 𝜉 will satisfy the following equation: 𝑝 1 𝑞 ( 𝜉 ) = 1 − 2 𝜉 3 − 𝑝 2 𝜉 − 2 𝑝 𝑞 + 3 𝑞 + 2 2 + 2 𝑝 2 𝑞 + 2 𝑝 𝑞 − 2 + 1 𝜉 − 𝑝 2 = 0 . ( 2 . 9 )

From the analysis above one sees that if 𝑥 ( 𝑡 ) is a solution of (1.1) and | 𝑝 | ≠ 1 , then one gets (2.3) and (2.4). In fact, a solution of (1.1) is constructed by the common solution { 𝑥 𝑛 } of (2.3) and (2.4). Moreover, it is clear that { 𝑥 𝑛 } consists of two parts: { 𝑥 2 𝑛 } and { 𝑥 2 𝑛 + 1 } . { 𝑥 2 𝑛 } can be obtained by solving (2.7), and { 𝑥 2 𝑛 + 1 } can be obtained by substituting { 𝑥 2 𝑛 } into (2.5) or (2.6). Without loss of generality, we consider (2.5) only. These will be shown in Lemmas 2.5 and 2.6.

Lemma 2.2. If 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) , then { 𝑓 𝑛 ( 𝑖 ) } , { ℎ 𝑛 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) , 𝑖 = 1 , 2 .

Lemma 2.3. Suppose that | 𝑝 | ≠ 1 and 𝑞 ≠ − 2 ( 𝑝 2 + 1 ) , then the roots of polynomial 𝑝 1 ( 𝜉 ) are of moduli different from 1.

Lemma 2.4. Suppose that 𝑋 is a Banach space, ℒ ( 𝑋 ) denotes the set of bounded linear operators from 𝑋 to 𝑋 , 𝐴 ∈ ℒ ( 𝑋 ) , and ‖ 𝐴 ‖ < 1 , then 𝐼 𝑑 − 𝐴 is bounded invertible and ( 𝐼 𝑑 − 𝐴 ) − 1 = ∞ 𝑛 = 0 𝐴 𝑛 ‖ ‖ 0 𝑥 0 2 0 0 𝑑 , ( 𝐼 − 𝐴 ) − 1 ‖ ‖ ≤ 1 , ( 1 − ‖ 𝐴 ‖ ) ( 2 . 1 0 ) where 𝐴 0 = 𝐼 𝑑 , and 𝐼 𝑑 is an identical operator.

The proofs of Lemmas 2.2, 2.3, and 2.4 are elementary, and we omit the details.

Lemma 2.5. Suppose that | 𝑝 | ≠ 1 and 𝑞 ≠ − 2 ( 𝑝 2 + 1 ) , then (2.7) has a unique solution { 𝑥 2 𝑛 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) .

Proof. As the proof of Theorem 9 in [8], define 𝐴 ∶ 𝑋 → 𝑋 by 𝐴 { 𝑥 2 𝑛 } = { 𝑥 2 𝑛 + 2 } , where 𝑋 is the Banach space consisting of all bounded sequences { 𝑥 𝑛 } in 𝐶 with ‖ { 𝑥 𝑛 } ‖ = s u p 𝑛 ∈ 𝑍 | 𝑥 𝑛 | . It follows from Lemmas 2.2–2.4 that (2.7) has a unique solution { 𝑥 2 𝑛 } = 𝑃 ( 𝐴 ) − 1 { ℎ 2 𝑛 + 5 + ( 2 − 𝑝 ) ℎ 2 𝑛 + 4 + ( 1 − 2 𝑝 ) ℎ 2 𝑛 + 3 − 𝑝 ℎ 2 𝑛 + 2 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) .
Substituting 𝑥 2 𝑛 into (2.5), we obtain 𝑥 2 𝑛 + 1 . Easily, we can get { 𝑥 2 𝑛 + 1 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) . Consequently, the common solution { 𝑥 𝑛 } of (2.3) and (2.4) can be obtained. Furthermore, we have that { 𝑥 𝑛 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) is unique.

Lemma 2.6. Suppose that | 𝑝 | ≠ 1 and 𝑞 ≠ − 2 ( 𝑝 2 + 1 ) , 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) . Let { 𝑥 𝑛 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) be the common solution of (2.3) and (2.4). Then (1.1) has a unique solution 𝑥 ( 𝑡 ) ∈ 𝒜 𝒫 ( 𝑅 ) such that 𝑥 ( 𝑛 ) = 𝑥 𝑛 , 𝑛 ∈ 𝑍 . In this case the solution 𝑥 ( 𝑡 ) is given for 𝑡 ∈ 𝑅 by ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝑥 ( 𝑡 ) = ∞ 𝑘 = 0 ( − 𝑝 ) 𝑘 | | 𝑝 | | − 𝜔 ( 𝑡 − 𝑘 ) 0 𝑥 0 2 0 0 𝑑 , < 1 , ∞ 𝑘 = 1 ( − 𝑝 ) − 𝑘 | | 𝑝 | | 𝜔 ( 𝑡 + 𝑘 ) 0 𝑥 0 2 0 0 𝑑 , > 1 , ( 2 . 1 1 ) where 𝜔 ( 𝑡 ) = 𝑥 2 𝑛 + 𝑝 𝑥 2 𝑛 − 1 + 𝑦 2 𝑛 ( 𝑡 − 2 𝑛 ) + 𝑞 𝑥 2 𝑛 ( 𝑡 − 2 𝑛 ) 2 2 + 𝑡 2 𝑛 𝑠 2 𝑛 𝑦 𝑓 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 0 𝑥 0 2 0 0 𝑑 , 2 𝑛 = 𝑥 2 𝑛 + 1 + 𝑞 𝑝 − 1 − 2 𝑥 2 𝑛 − 𝑝 𝑥 2 𝑛 − 1 − 𝑓 ( 1 ) 2 𝑛 , ( 2 . 1 2 ) for 𝑡 ∈ [ 2 𝑛 − 1 , 2 𝑛 + 1 ) , 𝑛 ∈ 𝑍 ; { 𝑦 2 𝑛 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) , 𝜔 ( 𝑡 ) ∈ 𝒜 𝒫 ( 𝑅 ) .

The proof is easy, we omit the details. Since the almost periodic solution 𝑥 ( 𝑡 ) of (1.1) is constructed by the common almost periodic solution of (2.3) and (2.4), easily, we have that ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) are continuous at 𝑡 ∈ 𝑅 . It must be pointed out that in many works only one of (2.3) and (2.4) is considered while seeking the unique almost periodic solution of (1.1), and it is not true for the continuity of ( 𝑥 ( 𝑡 ) + 𝑝 𝑥 ( 𝑡 − 1 ) ) on 𝑅 , consequently, it is not true for the uniqueness (see [8]).

The expressions of 𝑥 2 𝑛 , 𝑥 2 𝑛 + 1 , 𝑦 2 𝑛 , 𝜔 ( 𝑡 ) , and 𝑥 ( 𝑡 ) are important in the process of studying the spectrum containment of the almost periodic solution of (1.1). Before giving the main theorem, we list the following assumptions which will be used later.

(H1) | 𝑝 | ≠ 1 , 𝑞 ≠ − 2 ( 𝑝 2 + 1 ) .(H2) 𝑘 𝜋 ∉ Λ 𝑓 , for all 𝑘 ∈ 𝑍 .(H3)If 𝜆 ∈ Λ 𝑓 , then 𝜆 + 𝑘 𝜋 ∉ Λ 𝑓 , 0 ≠ 𝑘 ∈ 𝑍 .

Our result can be formulated as follows.

Main Theorem
Let 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) and ( H 1 ) be satisfied. Then (1.1) has a unique almost periodic solution 𝑥 ( 𝑡 ) and Λ 𝑥 ⊂ Λ 𝑓 + { 𝑘 𝜋 ∶ 𝑘 ∈ 𝑍 } . Additionally, if ( H 2 ) and ( H 3 ) are also satisfied, then Λ 𝑓 + { 𝑘 𝜋 ∶ 𝑘 ∈ 𝑍 } ⊂ Λ 𝑥 , that is, the following spectrum relation Λ 𝑥 = Λ 𝑓 + { 𝑘 𝜋 ∶ 𝑘 ∈ 𝑍 } holds, where the sum of sets 𝐴 and 𝐵 is defined as 𝐴 + 𝐵 = { 𝑎 + 𝑏 ∶ 𝑎 ∈ 𝐴 , 𝑏 ∈ 𝐵 } .

We postpone the proof of this theorem to the next section.

3. The Proof of Main Theorem

To show the Main Theorem, we need some more lemmas.

Lemma 3.1. Let 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) , then 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) , 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) , 𝜎 𝑏 ( ℎ 2 𝑛 ) , 𝜎 𝑏 ( ℎ 2 𝑛 + 1 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 , 𝑖 = 1 , 2 . If ( 𝐻 3 ) is satisfied, then 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) = 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) = 𝑒 𝑖 2 Λ 𝑓 , 𝑖 = 1 , 2 . Furthermore, if ( 𝐻 3 ) and ( 𝐻 2 ) are both satisfied, then 𝜎 𝑏 ( ℎ 2 𝑛 ) = 𝜎 𝑏 ( ℎ 2 𝑛 + 1 ) = 𝑒 𝑖 2 Λ 𝑓 .

Proof. Since 𝑓 ∈ 𝒜 𝒫 ( 𝑅 ) , by Lemma 2.2 we know that { 𝑓 ( 𝑖 ) 2 𝑛 } , { 𝑓 ( 𝑖 ) 2 𝑛 + 1 } , { ℎ 2 𝑛 } , { ℎ 2 𝑛 + 1 } ∈ 𝒜 𝒫 𝒮 ( 𝑅 ) , 𝑖 = 1 , 2 . It follows from The Approximation Theorem that, for any 𝑚 > 0 , 𝑚 ∈ 𝑍 , there exists 𝑃 𝑚 ∑ ( 𝑡 ) = 𝑛 ( 𝑚 ) 𝑘 = 1 𝑏 𝑘 , 𝑚 𝑒 𝑖 𝜆 𝑘 𝑡 0 𝑥 0 2 0 0 𝑑 , 𝜆 𝑘 ∈ Λ 𝑓 such that ‖ 𝑃 𝑚 − 𝑓 ‖ ≤ 1 / 𝑚 , where l i m 𝑚 → ∞ 𝑏 𝑘 , 𝑚 = 𝑎 ( 𝜆 𝑘 ; 𝑓 ) , and we can assume that 𝑏 𝑘 , 𝑚 𝑒 𝑖 𝜆 𝑘 𝑡 and 𝑏 𝑘 , 𝑚 𝑒 − 𝑖 𝜆 𝑘 𝑡 appear together in the trigonometric polynomial 𝑃 𝑚 ( 𝑡 ) . Define 𝑄 ( 1 ) 𝑚 , 2 𝑛 = 2 𝑛 + 1 2 𝑛 𝑠 2 𝑛 𝑃 𝑚 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 0 𝑥 0 2 0 0 𝑑 = 𝑛 ( 𝑚 ) 𝑘 = 1 𝑐 ( 1 ) 𝑘 , 𝑚 𝑒 𝑖 2 𝜆 𝑘 𝑛 𝑄 0 𝑥 0 2 0 0 𝑑 , ( 2 ) 𝑚 , 2 𝑛 = 2 𝑛 − 1 2 𝑛 𝑠 2 𝑛 𝑃 𝑚 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 0 𝑥 0 2 0 0 𝑑 = 𝑛 ( 𝑚 ) 𝑘 = 1 𝑐 ( 2 ) 𝑘 , 𝑚 𝑒 𝑖 2 𝜆 𝑘 𝑛 𝑄 0 𝑥 0 2 0 0 𝑑 , ( 1 ) 𝑚 , 2 𝑛 + 1 = 2 𝑛 + 2 2 𝑛 + 1 𝑠 2 𝑛 + 1 𝑃 𝑚 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 = 𝑛 ( 𝑚 ) 𝑘 = 1 𝑐 ( 1 ) 𝑘 , 𝑚 𝑒 𝑖 𝜆 𝑘 𝑒 𝑖 2 𝜆 𝑘 𝑛 𝑄 0 𝑥 0 2 0 0 𝑑 , ( 2 ) 𝑚 , 2 𝑛 + 1 = 2 𝑛 2 𝑛 + 1 𝑠 2 𝑛 + 1 𝑃 𝑚 ( 𝜎 ) 𝑑 𝜎 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 = 𝑛 ( 𝑚 ) 𝑘 = 1 𝑐 ( 2 ) 𝑘 , 𝑚 𝑒 𝑖 𝜆 𝑘 𝑒 𝑖 2 𝜆 𝑘 𝑛 0 𝑥 0 2 0 0 𝑑 , ( 3 . 1 ) where 𝑐 ( 1 ) 𝑘 , 𝑚 = ⎧ ⎪ ⎨ ⎪ ⎩ 𝑏 𝑘 , 𝑚 2 , 𝜆 𝑘 = 0 , − 𝑏 𝑘 , 𝑚 𝑒 𝑖 𝜆 𝑘 − 1 − 𝑖 𝜆 𝑘 𝜆 2 𝑘 , 𝜆 𝑘 𝑐 ≠ 0 , ( 2 ) 𝑘 , 𝑚 = ⎧ ⎪ ⎨ ⎪ ⎩ 𝑏 𝑘 , 𝑚 2 , 𝜆 𝑘 = 0 , − 𝑏 𝑘 , 𝑚 𝑒 − 𝑖 𝜆 𝑘 − 1 + 𝑖 𝜆 𝑘 𝜆 2 𝑘 , 𝜆 𝑘 ≠ 0 . ( 3 . 2 ) Obviously, 𝜎 𝑏 ( 𝑄 ( 𝑖 ) 𝑚 , 2 𝑛 ) , 𝜎 𝑏 ( 𝑄 ( 𝑖 ) 𝑚 , 2 𝑛 + 1 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 , 𝑖 = 1 , 2 , for all 𝑚 ∈ 𝑍 . For any 𝑧 ∈ 𝑆 1 , 𝑎 ( 𝑧 ; 𝑓 ( 𝑖 ) 2 𝑛 ) = l i m 𝑚 → ∞ 𝑎 ( 𝑧 ; 𝑄 ( 𝑖 ) 𝑚 , 2 𝑛 ) , 𝑎 ( 𝑧 ; 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) = l i m 𝑚 → ∞ 𝑎 ( 𝑧 ; 𝑄 ( 𝑖 ) 𝑚 , 2 𝑛 + 1 ) , thus, we have 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) , 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 , 𝑖 = 1 , 2 .
Since ℎ 2 𝑛 = 𝑓 ( 1 ) 2 𝑛 + 𝑓 ( 2 ) 2 𝑛 and ℎ 2 𝑛 + 1 = 𝑓 ( 1 ) 2 𝑛 + 1 + 𝑓 ( 2 ) 2 𝑛 + 1 , for all 𝑛 ∈ 𝑍 . For all 𝑧 ∈ 𝑆 1 , we have 𝑎 𝑧 ; ℎ 2 𝑛 = 𝑎 𝑧 ; 𝑓 ( 1 ) 2 𝑛 + 𝑎 𝑧 ; 𝑓 ( 2 ) 2 𝑛 𝑎 , ( 3 . 3 ) 𝑧 ; ℎ 2 𝑛 + 1 = 𝑎 𝑧 ; 𝑓 ( 1 ) 2 𝑛 + 1 + 𝑎 𝑧 ; 𝑓 ( 2 ) 2 𝑛 + 1 . ( 3 . 4 ) Thus, 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 and 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 imply 𝜎 𝑏 ( ℎ 2 𝑛 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 and 𝜎 𝑏 ( ℎ 2 𝑛 + 1 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 , respectively, 𝑖 = 1 , 2 .
If ( H 3 ) is satisfied, then for any 𝜆 𝑗 ∈ Λ 𝑓 , we have 𝑎 𝑒 𝑖 2 𝜆 𝑗 ; 𝑓 ( 1 ) 2 𝑛 = l i m 𝑚 → ∞ 𝑎 𝑒 𝑖 2 𝜆 𝑗 ; 𝑄 ( 1 ) 𝑚 , 2 𝑛 = l i m 𝑚 → ∞ 𝑐 ( 1 ) 𝑗 , 𝑚 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝑎 𝜆 𝑗 ; 𝑓 2 , 𝜆 𝑗 𝜆 = 0 , − 𝑎 𝑗 𝑒 ; 𝑓 𝑖 𝜆 𝑗 − 1 − 𝑖 𝜆 𝑗 𝜆 2 𝑗 , 𝜆 𝑗 𝑎 𝑒 ≠ 0 , 𝑖 2 𝜆 𝑗 ; 𝑓 ( 2 ) 2 𝑛 = l i m 𝑚 → ∞ 𝑎 𝑒 𝑖 2 𝜆 𝑗 ; 𝑄 ( 2 ) 𝑚 , 2 𝑛 = l i m 𝑚 → ∞ 𝑐 ( 2 ) 𝑗 , 𝑚 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝑎 𝜆 𝑗 ; 𝑓 2 , 𝜆 𝑗 𝜆 = 0 , − 𝑎 𝑗 𝑒 ; 𝑓 − 𝑖 𝜆 𝑗 − 1 + 𝑖 𝜆 𝑗 𝜆 2 𝑗 , 𝜆 𝑗 𝑎 𝑒 ≠ 0 , 𝑖 2 𝜆 𝑗 ; 𝑓 ( 1 ) 2 𝑛 + 1 = l i m 𝑚 → ∞ 𝑎 𝑒 𝑖 2 𝜆 𝑗 ; 𝑄 ( 1 ) 𝑚 , 2 𝑛 + 1 = l i m 𝑚 → ∞ 𝑒 𝑖 𝜆 𝑗 𝑐 ( 1 ) 𝑗 , 𝑚 , 𝑎 𝑒 𝑖 2 𝜆 𝑗 ; 𝑓 ( 2 ) 2 𝑛 + 1 = l i m 𝑚 → ∞ 𝑎 𝑒 𝑖 2 𝜆 𝑗 ; 𝑄 ( 2 ) 𝑚 , 2 𝑛 + 1 = l i m 𝑚 → ∞ 𝑒 𝑖 𝜆 𝑗 𝑐 ( 2 ) 𝑗 , 𝑚 . ( 3 . 5 ) Easily, we have 𝑎 ( 𝑒 𝑖 2 𝜆 𝑗 ; 𝑓 ( 𝑖 ) 2 𝑛 ) ≠ 0 and 𝑎 ( 𝑒 𝑖 2 𝜆 𝑗 ; 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) ≠ 0 , that is, 𝑒 𝑖 2 𝜆 𝑗 ⊂ 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) , 𝑒 𝑖 2 𝜆 𝑗 ⊂ 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) , 𝑖 = 1 , 2 . By the arbitrariness of 𝜆 𝑗 , we get 𝑒 𝑖 2 Λ 𝑓 ⊂ 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) and 𝑒 𝑖 2 Λ 𝑓 ⊂ 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) . So, 𝑒 𝑖 2 Λ 𝑓 = 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 ) = 𝜎 𝑏 ( 𝑓 ( 𝑖 ) 2 𝑛 + 1 ) , 𝑖 = 1 , 2 .
If ( H 3 ) and ( H 2 ) are both satisfied, suppose that there exists 𝑧 0 = 𝑒 𝑖 2 𝜆 𝑗 ∈ 𝑒 𝑖 2 Λ 𝑓 such that 𝑎 ( 𝑧 0 ; ℎ 2 𝑛 ) = 0 . ( H 2 ) implies 𝑒 𝑖 𝜆 𝑗 ≠ ± 1 . Moreover, since ( H 3 ) holds, we have 𝑎 ( 𝑧 0 ; 𝑓 ( 𝑖 ) 2 𝑛 ) ≠ 0 , 𝑖 = 1 , 2 . 𝑎 ( 𝑧 0 ; ℎ 2 𝑛 ) = 𝑎 ( 𝑧 0 ; 𝑓 ( 1 ) 2 𝑛 ) + 𝑎 ( 𝑧 0 ; 𝑓 ( 2 ) 2 𝑛 ) leads to 𝑒 𝑖 𝜆 𝑗 = 1 , which contradicts with 𝑒 𝑖 𝜆 𝑗 ≠ ± 1 . So, 𝑒 𝑖 2 Λ 𝑓 ⊂ 𝜎 𝑏 ( ℎ 2 𝑛 ) . Noticing that 𝜎 𝑏 ( ℎ 2 𝑛 ) ⊂ 𝑒 𝑖 2 Λ 𝑓 , we have 𝑒 𝑖 2 Λ 𝑓 = 𝜎 𝑏 ( ℎ 2 𝑛 ) . Similarly, we can get 𝑒 𝑖 2 Λ 𝑓 = 𝜎 𝑏 ( ℎ 2 𝑛 + 1 ) . The proof is completed.

Lemma 3.2. Suppose that ( 𝐻 1 ) is satisfied, then