Pseudo almost periodic functions and pseudo almost periodic solutions to dynamic equations on time scales

Abstract

In this paper, we first introduce a concept of the mean-value of uniformly almost periodic functions on time scales and give some of its basic properties. Then, we propose a concept of pseudo almost periodic functions on time scales and study some basic properties of pseudo almost periodic functions on time scales. Finally, we establish some results about the existence of pseudo almost periodic solutions to dynamic equations on time scales.

1 Introduction

The theory of dynamic equations on time scales has been developed over the last several decades, it has been created in order to unify the study of differential and difference equations. Many papers have been published on the theory of dynamic equations on time scales [1–14]. In addition, the existence of almost periodic, asymptotically almost periodic, pseudo-almost periodic solutions is among the most attractive topics in the qualitative theory of differential equations and difference equations due to their applications, especially in biology, economics and physics [15–34]. Recently, in [14, 35], the almost periodic functions and the uniformly almost periodic functions on time scales were presented and investigated, as applications, the existence of almost periodic solutions to a class of functional differential equations and neural networks were studied effectively (see [13, 14, 35]). However, there is no concept of pseudo-almost periodic functions on time scales so that it is impossible for us to study pseudo almost periodic solutions for dynamic equations on time scales.

Motivated by the above, our main purpose of this paper is firstly to introduce a concept of mean-value of uniformly almost periodic functions and give some useful and important properties of it. Then we propose a concept of pseudo almost periodic functions which is a new generalization of uniformly almost periodic functions on time scales and present some relative results. Finally, we establish some results about the existence and uniqueness of pseudo almost periodic solutions to dynamic equations on time scales.

The organization of this paper is as follows: In Section 2, we introduce some notations, definitions and state some preliminary results needed in the later sections. In Section 3, we introduce a concept of mean-value of uniformly almost periodic functions and establish some useful and important results. In Section 4, we propose a concept of pseudo almost periodic functions on time scales and present some relative results. In Section 5, we establish some results about the existence and uniqueness of pseudo almost periodic solutions to dynamic equations on time scales. As applications of our results, in Section 6, we study the existence of pseudo almost periodic solutions to quasi-linear dynamic equations on time scales.

2 Preliminaries

Let be a nonempty closed subset (time scale) of ℝ. The forward and backward jump operators $\sigma ,\rho :\mathbb{T}\to \mathbb{T}$ and the graininess $\mu :\mathbb{T}\to {ℝ}^{+}$ are defined, respectively, by

$$\sigma \left(t\right)=\mathsf{\text{inf}}\left\{s\in \mathbb{T}:s>t\right\},\rho \left(t\right)=\mathsf{\text{sup}}\left\{s\in \mathbb{T}:s<t\right\},\mu \left(t\right)=\sigma \left(t\right)-t.$$

A point $t\in \mathbb{T}$ is called left-dense if $t>\mathsf{\text{inf}}\mathbb{T}$ and ρ(t) = t, left-scattered if ρ(t) < t, right-dense if $t<\mathsf{\text{sup}}\mathbb{T}$ and σ(t) = t, and right-scattered if σ(t) > t. If has a left-scattered maximum m, then ${\mathbb{T}}^k=\mathbb{T}\backslash \left\{m\right\}$; otherwise ${\mathbb{T}}^k=\mathbb{T}$. If has a right-scattered minimum m, then ${\mathbb{T}}_k=\mathbb{T}\backslash \left\{m\right\}$; otherwise ${\mathbb{T}}_k=\mathbb{T}$.

A function $f:\mathbb{T}\to ℝ$ is right-dense continuous provided that it is continuous at right-dense point in and its left-side limits exist at left-dense points in . If f is continuous at each right-dense point and each left-dense point, then f is said to be a continuous function on .

For $y:\mathbb{T}\to ℝ$ and $t\in {\mathbb{T}}^k$, we define the delta derivative of y(t), y^Δ(t), to be the number (if it exists) with the property that for a given ε > 0, there exists a neighborhood U of t such that

$$|\left[y\left(\sigma \left(t\right)\right)-y\left(s\right)\right]-y^{\mathrm{\Delta }}\left(t\right)\left[\sigma \left(t\right)-s\right]|<\epsilon |\sigma \left(t\right)-s|$$

for all s ∈ U.

Let y be right-dense continuous, if Y ^Δ(t) = y(t), then we define the delta integral by

$$\underset{a}{\overset{t}{\int }}y\left(s\right)\mathrm{\Delta }s=Y\left(t\right)-Y\left(a\right).$$

A function $p:\mathbb{T}\to ℝ$ is called regressive provided 1+µ(t)p(t) ≠ 0 for all $t\in {\mathbb{T}}^k$. The set of all regressive and rd-continuous functions $p:\mathbb{T}\to ℝ$ will be denoted by $\mathcal{R}=\mathcal{R}\left(\mathbb{T}\right)=\mathcal{R}\left(\mathbb{T},ℝ\right)$. We define the set ${\mathcal{R}}^{+}={\mathcal{R}}^{+}\left(\mathbb{T},ℝ\right)=\left\{p\in \mathcal{R}:1+\mu \left(t\right)p\left(t\right)>0,\forall t\in \mathbb{T}\right\}$.

A n × n-matrix-valued function A on a time scale is called regressive provided I + µ(t)A(t) is invertible for all $t\in \mathbb{T}$, and the class of all such regressive and rd-continuous functions is denoted, similar to the above scalar case, by $\mathcal{R}=ℝ\left(\mathbb{T}\right)=\mathcal{R}\left(\mathbb{T},{ℝ}^{n\times n}\right)$.

If r is a regressive function, then the generalized exponential function e _r is defined by

$$e_r\left(t,s\right)=\mathsf{\text{exp}}\left\{\underset{s}{\overset{t}{\int }}{\xi }_{\mu \left(\tau \right)}\left(r\left(\tau \right)\right)\mathrm{\Delta }\tau \right\}$$

for all s, $t\in \mathbb{T}$, with the cylinder transformation

$${\xi }_h\left(z\right)=\left\{\begin{array}{l}\frac{\mathsf{\text{Log}}\left(1+\mathsf{\text{hz}}\right)}{h},\mathsf{\text{if}}h\ne 0,\\ z,\mathsf{\text{if}}h=0.\end{array}\right.$$

Definition 2.1 ([1, 3]). Let p, $q:\mathbb{T}\to ℝ$ be two regressive functions, define

$$p\oplus q=p+q+\mu pq,\ominus p=-\frac{p}{1+\mu p},p\ominus q=p\oplus \left(\ominus q\right).$$

Lemma 2.1 ([1, 3]). Assume that p, $q:\mathbb{T}\to ℝ$ are two regressive functions, then

• (i) e₀(t, s) ≡ 1 and e _p (t, t) ≡ 1;

• (ii) e _p (σ(t), s) = (1 + µ(t)p(t))e _p (t, s);

• (iii) $e_p\left(t,s\right)=\frac{1}{e_p\left(s,t\right)}=e_{\ominus p}\left(s,t\right);e_p\left(t,s\right)e_p\left(s,r\right)=e_p\left(t,r\right)$ ;

• (iv) (e_⊖p(t, s))^Δ = (⊖p)(t)e_⊖p(t, s);

• (v) If a, b, $c\in \mathbb{T}$, then ${\int }_a^{b}p\left(t\right)e_p\left(c,\sigma \left(t\right)\right)\mathrm{\Delta }t=e_p\left(c,a\right)-e_p\left(c,b\right)$.

Definition 2.2 ([36]). For every x, $y\in ℝ$, $\left[x,y\right)=\left\{t\in ℝ:x\le t<y\right\}$, define a countably additive measure m₁ on the set

$${\mathcal{F}}_1=\left\{\left[ã,\tilde{b}\right)\cap \mathbb{T}:ã,\tilde{b}\in \mathbb{T},ã\le \tilde{b}\right\},$$

that assigns to each interval $\left[ã,\tilde{b}\right)\cap \mathbb{T}$ its length, that is,

$$m_{\mathsf{\text{1}}}\mathsf{\text{([}}ã\mathsf{\text{,}}\tilde{b}\mathsf{\text{)) = }}\tilde{b}-ã\mathsf{\text{.}}$$

The interval $\left[ã,a\right)$ is understood as the empty set. Using m₁, they generate the outer measure $m_1^{*}$ on $\mathcal{P}\left(\mathbb{T}\right)$, defined for each $E\in \mathcal{P}\left(\mathbb{T}\right)$ as

$$m_1^{*}\left(E\right)=\left\{\begin{array}{cc}\underset{\tilde{\mathcal{R}}}{\mathsf{\text{inf}}}\left\{\sum _{i\in I_{\tilde{\mathcal{R}}}}\left({\tilde{b}}_i-{ã}_i\right)\right\}\in {ℝ}^{+},\hfill & b\notin E,\hfill \\ +\infty ,\hfill & b\in E,\hfill \end{array}\right.$$

with

$$\tilde{\mathcal{R}}=\left\{{\left\{\left[{ã}_i,{\tilde{b}}_i\right)\cap \mathbb{T}\in {\mathcal{F}}_1\right\}}_{i\in I_{\tilde{\mathcal{R}}}}:I_{\tilde{\mathcal{R}}}\subset N,E\subset \underset{i\in I_{\tilde{\mathcal{R}}}}{\cup }\left(\left[a_i,b_i\right)\cap \mathbb{T}\right)\right\}.$$

A set $A\subset \mathbb{T}$ is said to be Δ-measurable if the following equality:

$$m_1^{*}\left(E\right)=m_1^{*}\left(E\cap A\right)+m_1^{*}\left(E\cap \left(\mathbb{T}\backslash A\right)\right)$$

holds true for all subset E of . Define the family

$$\mathcal{M}\left(m_1^{*}\right)=\left\{A\subset \mathbb{T}:Ais\mathrm{\Delta }-measurable\right\},$$

the Lebesgue Δ-measure, denoted by µ_Δ, is the restriction of $m_1^{*}$ to $\mathcal{M}\left(m_1^{*}\right)$.

Definition 2.3 ([35]). A time scale is called an almost periodic time scale if

$$\Pi :=\left\{\tau \in ℝ:t\pm \tau \in \mathbb{T},\forall t\in \mathbb{T}\right\}\ne \left\{0\right\}.$$

Remark 2.1. In the following, we always use to denote an almost periodic time scale.

Throughout this paper, ${\mathbb{E}}^n$ denotes ℝⁿ or ${ℂ}^n$, D denotes an open set in ${\mathbb{E}}^n$ or $\mathbb{D}={\mathbb{E}}^n$, S denotes an arbitrary compact subset of D.

Definition 2.4 ([35]). Let be an almost periodic time scale. A function $f\in C\left(\mathbb{T}\times \mathbb{D},{\mathbb{E}}^n\right)$ is called an almost periodic function in $t\in \mathbb{T}$ uniformly for x ∈ D if the ε-translation set of f

$$E\left\{\epsilon ,f,S\right\}=\left\{\tau \in \Pi :|f\left(t+\tau ,x\right)-f\left(t,x\right)|<\epsilon ,forall\left(t,x\right)\in \mathbb{T}\times S\right\}$$

is a relatively dense set in for all ε > 0 and for each compact subset S of D; that is, for any given ε > 0 and each compact subset S of D, there exists a constant l(ε, S) > 0 such that each interval of length l(ε, S) contains a τ(ε, S) ∈ E{ε, f, S} such that

$$|f\left(t+\tau ,x\right)-f\left(t,x\right)|<\epsilon ,forallt\in \mathbb{T}\times S.$$

τ is called the ε-translation number of f and and l(ε, S) is called the inclusion length of E{ε, f, S}.

3 The mean-value of uniformly almost periodic functions on time scales

Let $f\in C\left(\mathbb{T}\times \mathbb{D},{\mathbb{E}}^n\right)$ and f(t, x) be almost periodic in t uniformly for x ∈ D, we denote

$$a\left(f,\lambda ,x\right):=\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)e^{-i\lambda t}\mathrm{\Delta }t,\mathsf{\text{where}}{t}_0\in \mathbb{T},T\in \Pi ,$$

(3.1)

where $\lambda \in ℝ$, $i=\sqrt{-1}.$ Obviously, for a fixed (f, λ, x), $a\left(f,\lambda ,x\right)\in {\mathbb{E}}^n$.

Definition 3.1. a(f(t, 0, x)) is called mean-value of f(t, x) if

$$0<a\left(f,0,x\right)=\underset{T\to \infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t<+\infty .$$

Theorem 3.1. For any $\lambda \in ℝ$, a(f, λ, x) defined by (3.1) exists uniformly for x ∈ S and is uniformly continuous on S with respect to x, where S is an arbitrary compact subset of D.

Proof. For any t₁ ∈ Π, t₁ > 0, we can make a sequence ${\left\{t_i\right\}}_{i\in {\mathbb{Z}}^{+}}\subset \prod$, where t _i = it₁. We will prove that the sequence ${\left\{\frac{1}{t_i}{\int }_{t_0}^{t_0+t_i}f\left(t,x\right)\mathrm{\Delta }t\right\}}_{i\in {\mathbb{Z}}^{+}}$ converges uniformly with respect to x ∈ S.

For any integers m, n and x ∈ S, taking t _m , t _n , we have

$$\begin{array}{l}\left|\frac{1}{t_n}\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\frac{1}{t_m}\underset{t_0}{\overset{t_0+t_m}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ \le \frac{1}{t_m{t}_n}\left|t_m\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+t_{mn}}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ +\frac{1}{t_m{t}_n}\left|\underset{t_0}{\overset{t_0+t_{mn}}{\int }}f\left(t,x\right)\mathrm{\Delta }t-t_n\underset{t_0}{\overset{t_0+t_m}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ \le \left|\frac{t_1}{t_m{t}_n}\right|\left[\sum _{k=1}^m\left|\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_{\left(k-1\right)n}}{\overset{t_0+t_{kn}}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\right.\\ \left.+\sum _{k=1}^n\left|\underset{{}^t\left(k-1\right)m}{\overset{t_0+t_{km}}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+t_m}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\right].\end{array}$$

(3.2)

Consider the following integral form:

$$\underset{t_a}{\overset{t_0+t_{a+s}}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+t_s}{\int }}f\left(t,x\right)\mathrm{\Delta }t,$$

(3.3)

where s = n, a = (k − 1)n, k = 1, 2, ..., m or s = m, a = (k − 1)m, k = 1, 2, ..., n. For arbitrary a, s, we can evaluate (3.3):

For any ε > 0, let $l=l\left(\frac{\epsilon }{4},S\right)$ be an inclusion length of $E\left(f,\frac{\epsilon }{4},S\right)$ and $\tau \in E\left(f,\frac{\epsilon }{4},S\right)\cap \left[t_a-t_0,t_a-t_0+l\right]$, then, for all x ∈ S, we get***

$$\begin{array}{l}\left|\underset{t_a}{\overset{t_0+t_{a+s}}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+t_s}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ =\left|\left(\underset{t_0+\tau }{\overset{t_{\mathsf{\text{0}}}+\tau +t_s}{\int }}-\underset{t_0}{\overset{t_0+t_s}{\int }}+\underset{t_0+\tau +t_s}{\overset{t_0+t_{a+s}}{\int }}+\underset{t_a}{\overset{t_0+\tau }{\int }}\right)f\left(t,x\right)\mathrm{\Delta }t\right|\\ \le \underset{t_0}{\overset{t_0+t_s}{\int }}|f\left(t+\tau ,x\right)-f\left(t,x\right)|\mathrm{\Delta }t+\underset{t_0+\tau +t_s}{\overset{t_0+t_{a+s}}{\int }}\left|f\left(t,x\right)\right|\mathrm{\Delta }t+\underset{t_a}{\overset{t_0+\tau }{\int }}\left|f\left(t,x\right)\right|\mathrm{\Delta }t\\ \le \frac{\epsilon t_s}{4}+2lG,\end{array}$$

(3.4)

where $G=\underset{\left(t,x\right)\in \mathbb{T}\times S}{\mathsf{\text{sup}}}\left|f\left(t,x\right)\right|$. According to (3.4), we can reduce (3.2) to the following:

$$\begin{array}{ll}\hfill \left|\frac{1}{t_n}\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t\right)\mathrm{\Delta }t-\frac{1}{t_m}\underset{t_0}{\overset{t_0+t_m}{\int }}f\left(t\right)\mathrm{\Delta }t\right|& <\frac{t_1}{t_m{t}_n}\left[m\left(\frac{\epsilon t_n}{4}+2lG\right)+n\left(\frac{\epsilon t_m}{4}+2lG\right)\right]\\ =\frac{\epsilon }{2}+\frac{2lG}{t_1}\left(\frac{1}{m}+\frac{1}{n}\right)\to 0,t_m,t_n\to +\infty .\end{array}$$

By the Cauchy convergence criterion, the sequence ${\left\{\frac{1}{t_i}{\int }_{t_0}^{t_0+t_i}f\left(t,x\right)\mathrm{\Delta }t\right\}}_{i\in \mathbb{N}}$ converges uniformly with respect to x ∈ S.

For any sufficiently large 0 < T ∈ Π, there exist 0 < t _n ∈ Π such that 0 < t _n < T ≤ t_n+1, so for all x ∈ S, we have

$$\left|\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\le G\left(T-t_n\right)\le G{t}_1.$$

Therefore,

$$\begin{array}{ll}\hfill \left|\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\frac{1}{t_n}\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|& <\frac{1}{T}\left|\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ +\left(\frac{1}{t_n}-\frac{1}{T}\right)\underset{t_0}{\overset{t_0+t_n}{\int }}\left|f\left(t,x\right)\right|\mathrm{\Delta }t\\ \le \frac{G{t}_1}{T}+\left(\frac{1}{t_n}-\frac{1}{T}\right)t_{n}G\\ <\frac{2G}{n}\to 0,t_n\to +\infty .\end{array}$$

Hence,

$$a\left(f,0,x\right)=\underset{n\to +\infty }{\mathsf{\text{lim}}}\frac{1}{t_n}\underset{t_0}{\overset{t_0+t_n}{\int }}f\left(t,x\right)\mathrm{\Delta }t\mathsf{\text{ uniformly for}}x\in S.$$

Besides, for $\frac{1}{T}{\int }_{t_0}^{t_0+T}f\left(t,x\right)\mathrm{\Delta }t$ is continuous with respect to x ∈ S, where S is an arbitrary compact set in ${\mathbb{E}}^n$, a(f, 0, x) is uniformly continuous on S.

It is oblivious that f(t, x)e^−iλt is almost periodic in t uniformly for x ∈ D and a(f, λ, x) = a(f(t, x)e^−iλt, 0, x), so it is easy to see that a(f, λ, x) exists uniformly for x ∈ S and is uniformly continuous on S with respect to x. This completes the proof. □

Theorem 3.2. Assume that T ∈ Π and $f\left(t,x\right)\in C\left(\mathbb{T}\times \mathbb{D},{\mathbb{E}}^n\right)$ is almost periodic in t uniformly for x ∈ D, then

$$\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{\alpha }{\overset{\alpha +T}{\int }}f\left(t,x\right)e^{-i\lambda t}\mathrm{\Delta }t:=m\left(f\left(t,x\right),\lambda ,x\right)$$

uniformly exists for $\alpha \in \mathbb{T}$ and

$$m\left(f\left(t,x\right),\lambda ,x\right)=a\left(f\left(t+\alpha ,x\right)e^{-i\lambda \alpha },\lambda ,x\right).$$

Proof. For m(f, λ, x) = m(f(t, x)e^−iλt, 0, x), it suffices to show that, for x ∈ S, $\forall \alpha \in \mathbb{T}$, the following uniformly exists:

$$m\left(f,0,x\right)=\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{\alpha }{\overset{\alpha +T}{\int }}f\left(t,x\right)\mathrm{\Delta }t.$$

(3.5)

Take $l=l\left(\frac{\epsilon }{4},S\right)$ and $\tau \in E\left\{\frac{\epsilon }{4},f,S\right\}\cap \left[\alpha -t_0,\alpha -t_0+l\right]$, $G=\underset{\left(t,x\right)\in \mathbb{T}\times S}{\mathsf{\text{sup}}}\left|f\left(t,x\right)\right|$, for x ∈ S, we obtain

$$\begin{array}{l}\left|\frac{1}{T}\underset{\alpha }{\overset{\alpha +T}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ =\frac{1}{T}\left|\left(\underset{t_0+\tau }{\overset{t_{\mathsf{\text{0}}}+\tau +T}{\int }}-\underset{t_0}{\overset{t_0+T}{\int }}+\underset{t_0+\tau +T}{\overset{\alpha +T}{\int }}+\underset{\alpha }{\overset{t_0+\tau }{\int }}\right)f\left(t,x\right)\mathrm{\Delta }t\right|\\ \le \frac{1}{T}\left(\underset{t_0}{\overset{t_0+T}{\int }}|f\left(t+\tau ,x\right)-f\left(t,x\right)|\mathrm{\Delta }t+\underset{t_0+\tau +T}{\overset{\alpha +T}{\int }}\left|f\left(t,x\right)\right|\mathrm{\Delta }t+\underset{\alpha }{\overset{t_0+\tau }{\int }}\left|f\left(t,x\right)\right|\mathrm{\Delta }t\right)\\ \le \frac{1}{T}\left(\frac{\epsilon T}{4}+\frac{2lG}{T}\right)=\frac{\epsilon }{4}+\frac{2lG}{T}\end{array}$$

(3.6)

and

$$\begin{array}{l}\left|\frac{1}{nT}\underset{t_0}{\overset{t_0+nT}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\\ =\frac{1}{n}\left|\sum _{k=1}^n\frac{1}{T}\left[\underset{t_0+\left(k-1\right)T}{\overset{t_0+kT}{\int }}f\left(t,x\right)\mathrm{\Delta }t-\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right]\right|\\ \le \frac{\epsilon }{4}+\frac{2lG}{T}.\end{array}$$

(3.7)

From (3.7), let n → ∞, for x ∈ S, we have

$$\left|a\left(f,0,x\right)-\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t\right|\le \frac{\epsilon }{4}+\frac{2lG}{T}.$$

(3.8)

Using trigonometric inequality, according (3.6) and (3.8), we can take $T>\frac{8lG}{\epsilon }$ such that

$$\left|\frac{1}{T}\underset{\alpha }{\overset{\alpha +T}{\int }}f\left(t,x\right)\mathrm{\Delta }t-a\left(f,0,x\right)\right|\le \frac{\epsilon }{2}+\frac{4lG}{T}<\epsilon .$$

Hence, we can easily obtain that (3.5) uniformly exists for $\alpha \in \mathbb{T}$ and m(f, 0, x) = a(f, 0, x) = a(f(t, x), 0, x). Furthermore,

$$\frac{1}{T}\underset{\alpha }{\overset{\alpha +T}{\int }}f\left(t,x\right)\mathrm{\Delta }t=\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t+\alpha ,x\right)\mathrm{\Delta }t.$$

Therefore, a(f(t + α, x), 0, x) uniformly exists for $\alpha \in \mathbb{T}$ and m(f(t, x), 0, x) = a(f(t + α, x), 0, x). It is easy to see that f(t, x)e^−iλt is almost periodic in t uniformly for x ∈ D, thus, we have

$$\begin{array}{ll}\hfill m\left(f\left(t,x\right),\lambda ,x\right)& =m\left(f\left(t,x\right)e^{-i\lambda t},0,x\right)\\ =a\left(f\left(t+\alpha ,x\right)e^{-i\lambda \left(t+\alpha \right)},0,x\right)\\ =a\left(f\left(t+\alpha ,x\right)e^{-i\lambda \alpha },\lambda ,x\right).\end{array}$$

Hence, m(f(t, x), λ, x) uniformly exists for $\alpha \in \mathbb{T}$. This completes the proof. □

In Theorem 3.1 and Theorem 3.2, if we take λ = 0, then we have

$$a\left(f\left(t,x\right),0,x\right)=\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t,x\right)\mathrm{\Delta }t:=m_t\left(f\left(t,x\right)\right)$$

(3.9)

and

$$a\left(f\left(t+\alpha \right),0,x\right)=\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}f\left(t+\alpha ,x\right)\mathrm{\Delta }t:=m_t\left(f\left(t+\alpha ,x\right)\right)$$

(3.10)

uniformly converge for x ∈ S and for x ∈ S, $\alpha \in \mathbb{T}$, respectively.

Definition 3.2. (3.9) and (3.10) are called the mean value and the strong mean-value of f(t, x), respectively.

Lemma 3.1. Let T ∈ Π, then for any real number λ ≠ 0,

$$m_t\left(e^{i\lambda t}\right)=\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}{e}^{i\lambda t}\mathrm{\Delta }t=0,where{t}_0\in \mathbb{T}.$$

(3.11)

Proof. First note that for any fixed T > 0, by Lemma 3.1 and Theorem 5.2 in [36], [t₀, t₀ + T ] contains only finitely many right scattered points. Assume that $\left[t_0,t_0+T\right]={\bigcup }_{i=0}^n\left[\sigma \left(t_i\right),t_{i+1}\right]$, where

$$t_0<t_1<t_2<\cdots <t_n=t_0+T$$

are right scattered points. Then

$$\begin{array}{ll}\hfill m_t\left(e^{i\lambda t}\right)& =\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}{e}^{i\lambda t}\mathrm{\Delta }t\\ =\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\sum _{i=0}^{n-1}\left(\underset{t_i}{\overset{\sigma \left(t_i\right)}{\int }}{e}^{i\lambda t}\mathrm{\Delta }t+\underset{\sigma \left(t_i\right)}{\overset{t_{i+1}}{\int }}{e}^{i\lambda t}\mathsf{\text{d}}t\right)\\ =\underset{T\to +\infty }{\mathsf{\text{lim}}}\frac{1}{T}\sum _{i=0}^{n-1}\left(\mu \left(t_i\right)\left(\mathsf{\text{cos}}\lambda t_i+i\mathsf{\text{sin}}\lambda t_i\right)+\frac{1}{\lambda }\left[\right(\mathsf{\text{sin}}\lambda t_{i+1}-\mathsf{\text{sin}}\lambda \sigma \left(t_i\right))\right.\\ \left.+i\left(\mathsf{\text{cos}}\lambda \sigma \right(t_i)-\mathsf{\text{cos}}\lambda t_{i+1})]\right),\end{array}$$

since sin t and cos t are bounded for t ∈ ℝ, one can easily see that (3.11) holds. The proof is complete. □

Theorem 3.3. Let $f\left(t,x\right)\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right)$ be almost periodic in t uniformly for x ∈ D, then for any finite set of distinct real numbers λ₁, λ₂, ..., λ _N and any finite set of real or complex n-dimensional vectors b₁, b₂, ..., b _N ,

$$m_t\left({\left|f\left(t,x\right)-\sum _{k=1}^N{b}_k{e}^{i{\lambda }_{k}t}\right|}^2\right)=m_t\left(|f\left(t,x\right){|}^2\right)-\sum _{k=1}^N|a\left(f,{\lambda }_k,x\right){|}^2+\sum _{k=1}^N|b_k-a\left(f,{\lambda }_k,x\right){|}^2.$$

(3.12)

Proof. Note that $|f\left(t,x\right){|}^2=〈f\left(t,x\right),\overline{f\left(t,x\right)}〉$ is almost periodic in t for x ∈ D where 〈,〉 denotes the usual inner product in ${\mathbb{E}}^n$ and $\overline{f\left(t,x\right)}$ denotes the conjugate of f(t, x), so it has a mean-value, thus

$$\begin{array}{ll}\hfill m_t\left({\left|f\left(t,x\right)-\sum _{k=1}^N{b}_k{e}^{i{\lambda }_{k}t}\right|}^2\right)& =m_t\left(〈f\left(t,x\right)-\sum _{k=1}^N{b}_k{e^{i{\lambda }_k}}^t,\overline{f\left(t,x\right)}-\sum _{k=1}^N{b}_k{e}^{-i{\lambda }_{k}t}〉\right)\\ =m_t\left(|f\left(t,x\right){|}^2\right)-\sum _{k=1}^N〈\overline{b_k},a\left(f,{\lambda }_k,x\right)〉\\ -\sum _{k=1}^N〈b_k,\overline{a(f,{\lambda }_k,x)〉}+\sum _{l=1}^N\sum _{j=1}^N〈b_l,\overline{b_j}〉m_t\left(e^{i\left({\lambda }_l-{\lambda }_j\right)t}\right),\end{array}$$

by Lemma 3.1, it is easy to obtain that

$$\begin{array}{ll}\hfill m_t\left({\left|f\left(t,x\right)-\sum _{k=1}^N{b}_k{e}^{i{\lambda }_{k}t}\right|}^2\right)& =m_t\left(|f\left(t,x\right){|}^2\right)-\sum _{k=1}^N〈\overline{b_k},a\left(f,{\lambda }_k,x\right)〉\\ -\sum _{k=1}^N〈b_k,\overline{a(f,{\lambda }_k,x)〉}+\sum _{j=1}^N〈b_j,b_j〉\\ =m_t\left(|f\left(t,x\right){|}^2\right)-\sum _{k=1}^N|a\left(f,{\lambda }_k,x\right){|}^2+\sum _{k=1}^N|b_k-a\left(f,{\lambda }_k,x\right){|}^2.\end{array}$$

The proof is complete. □

In Theorem 3.3, if we take b _k = a(f, λ _k , x)(k = 1, 2, ..., N), then we have the following corollary:

Corollary 3.1. The best approximation of uniformly almost periodic function f(t, x) on time scales satisfies the following:

$$m_t\left({\left|f\left(t,x\right)-\sum _{k=1}^N{b}_k{e}^{i{\lambda }_{k}t}\right|}^2\right)=m_t\left(|f\left(t,x\right){|}^2\right)-\sum _{k=1}^N|a\left(f,{\lambda }_k,x\right){|}^2.$$

By Corollary 3.1, one can easily get the following corollary:

Corollary 3.2. Let $f\left(t,x\right)\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right)$ be almost periodic in t uniformly for x ∈ D, then

$$\sum _{k=1}^N|a\left(f,{\lambda }_k,x\right){|}^2\le m_t\left(|f\left(t,x\right){|}^2\right).$$

Theorem 3.4. Let $f\left(t,x\right)\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right)$be almost periodic in t uniformly for x ∈ D, then there is a countable set of real numbers Λ such that a(f, λ, x) ≡ 0 on S if λ ∉ Λ.

Proof. Since f(t, x) is uniformly almost periodic, then for all $\left(t,x\right)\in \mathbb{T}\times S$, there exists M > 0 such that |f(t, x)| ≤ M. Therefore, for any n ∈ ℕ, the real number set $\left\{\lambda \in ℝ:\left|a\left(f,\lambda ,x\right)\right|>\frac{1}{n}\right\}$ is finite (If it is infinite, then ${\sum }_{k=1}^{\infty }|a\left(f,{\lambda }_k,x\right){|}^2>{\sum }_{k=1}^{\infty }\frac{1}{n}\to +\infty$, this contradicts Corollary 3.2). Hence, for any fixed x ∈ S, one can obtain the real number set {λ ∈ ℝ: a(f, λ, x) ≠ 0} is countable. Furthermore, by Corollary 3.2, one can see that

$$\sum _{k=1}^N\underset{x\in S}{\mathsf{\text{sup}}}|a\left(f,{\lambda }_k,x\right){|}^2\le M^2.$$

Thus, there is a countable set of real numbers Λ such that a(f, λ, x) ≡ 0 on S if λ ∉ Λ. The proof is complete. □

Theorem 3.5. If $f:\mathbb{T}\times D\to {ℝ}^n$ is a non-negative almost periodic function in t uniformly for x ∈ D and f ≢ 0, then a(f, 0, x) > 0.

Proof. Let $f\left(t_0^{\prime },x\right)=M>0$ and pick δ > 0 so that $f\left(t,x\right)\ge \frac{2M}{3}$ on $\left(t_0^{\prime }-\delta ,t_0^{\prime }+\delta \right)\times S$. Let l ∈ Π be an inclusion length of $E\left\{\frac{M}{3},f,S\right\}$ and take l > 2δ (In fact, one can choose 0 < τ₀ ∈ Π such that nτ₀ = l ∈ Π, n is some positive integer). If h ∈ Π, $t_0\in \mathbb{T}$, find $\tau \in E\left\{\frac{M}{3},f,S\right\}\cap \left[h+\delta -t_0^{\prime },h+\delta -t_0^{\prime }+l\right]$. Then $t_0^{\prime }-\delta +\tau \in \left[h,h+l\right]$. Either $t_0^{\prime }+\tau$ or $t_0^{\prime }-2\delta +\tau \in \left[h,h+l\right]$ since l > 2δ. In the first case if $t\in \left(t_0^{\prime }-\delta +\tau ,t_0^{\prime }+\tau \right)$ then

$$\left|f\left(t,x\right)\right|\ge \left|f\left(t+\tau ,x\right)\right|-|f\left(t+\tau ,x\right)-f\left(t,x\right)|\ge \frac{2M}{3}-\frac{M}{3}=\frac{M}{3}.$$

The second case can be handled similarly. In either case ${\int }_{t_0+h}^{t_0+h+l}f\left(t,x\right)\mathrm{\Delta }t>\frac{M}{3}\delta$ since on a subinterval of length $\delta ,f\left(t,x\right)\ge \frac{M}{3}$. Now write h = (n -1) l to get ${\int }_{t_0+\left(n-1\right)l}^{t_0+nl}f\left(t,x\right)\mathrm{\Delta }t>\frac{M}{3}\delta$. Hence

$$\frac{1}{Nl}\underset{t_0}{\overset{t_0+Nl}{\int }}f\left(t,x\right)\mathrm{\Delta }t=\frac{1}{Nl}\sum _{n=1}^N\underset{t_0+\left(n-1\right)l}{\overset{t_0+nl}{\int }}f\left(t,x\right)\mathrm{\Delta }t>\frac{M\delta }{3l}.$$

Letting N → ∞ one can get $a\left(f,0,x\right)\ge \frac{M\delta }{3l}>0$. The proof is complete. □

4. Pseudo almost periodic functions on time scales

Let $\mathbb{B}C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right)$ denote the space of all bounded continuous functions from $\mathbb{T}\times D$ to ${\mathbb{E}}^n$. Set

$$\begin{array}{l}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n=\left\{g\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right):g\mathsf{\text{ is almost periodic in }}t\mathsf{\text{ uniformly for }}x\in D\right\},\\ \mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n=\{g\in C(\mathbb{T},{\mathbb{E}}^n):g\mathsf{\text{ is almost periodic} ,}}\end{array}$$

$$\begin{array}{ll}\hfill \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n& =\left\{\phi \in \mathbb{B}C\left(\mathbb{T},{\mathbb{E}}^n\right):\phi \mathsf{\text{ is }}\mathrm{\Delta }\mathsf{\text{ - measurable such that }}\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s\right)\right|\mathrm{\Delta }s=0,\right.\\ \left.\mathsf{\text{where }}{t}_0\in \mathbb{T},r\in \Pi \right\}\end{array}$$

and

$$\begin{array}{l}\tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n\\ =\left\{\phi \in \mathbb{B}C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right):\phi \left(\cdot ,x\right)\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\right)\mathsf{\text{ for each }}x\in D\mathsf{\text{ and}}\right.\\ \left.\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\right|\phi \left(s,x\right)\left|\right|\mathrm{\Delta }s=0\mathsf{\text{ uniformly for }}x\in D,\mathsf{\text{ where }}{t}_0\in \mathbb{T},r\in \Pi \right\}.\end{array}$$

Remark 4.1. $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\right)$ does not require $\underset{\left|t\right|\to \infty }{\mathsf{\text{lim}}}\phi \left(t\right)$ exists. Consider, for example, let $\mathbb{T}={\bigcup }_{n=1}^{\infty }\left[n,n+\frac{1}{n}\right]$ and

$$\phi \left(t\right)=\left\{\begin{array}{cc}\hfill \frac{1}{\sqrt{n}},\hfill & \hfill n\le t\le n+\frac{1}{n},\hfill \\ \hfill 0,\hfill & \hfill telsewhere.\hfill \end{array}\right.$$

Obviously, for any fixed $n_0\in \mathbb{N}$ and $t\in \mathbb{T}$, one can easily see that $t\pm n_0\in \mathbb{T}$, thus n₀ ∈ ∏, that is, is an almost periodic time scale. It is clear that lim_|t|→∞φ(t) does not exist, noting that ${\left\{n+\frac{1}{n}\right\}}_{n\in \mathbb{N}}$ are right scattered points, so

$$\begin{array}{ll}\hfill \underset{r\to \infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s\right)\right|\mathrm{\Delta }s& =\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{n}\left[\sum _{k=1}^n\underset{k}{\overset{k+\frac{1}{k}}{\int }}\frac{1}{\sqrt{k}}\mathsf{\text{d}}s+\sum _{k=1}^n\mu \left(k+\frac{1}{k}\right)\frac{1}{\sqrt{k}}\right]\\ =\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{n}\sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\cdot \frac{1}{k}+\left(1-\frac{1}{k}\right)\frac{1}{\sqrt{k}}\right)\\ =\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{n}\sum _{k=1}^n\frac{1}{\sqrt{k}}=0.\end{array}$$

Hence $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\right)$.

Definition 4.1. A function $f\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right)$ is called pseudo almost periodic in t uniformly for x ∈ D if f = g + φ, where $g\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ and $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$.

Remark 4.2. Note that g and φ are uniquely determined. Indeed, since

$$N\left(\phi \right)=\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\right|\phi \left(s,x\right)\left|\right|\mathrm{\Delta }s=0,$$

if f = g₁ + φ₁ = g₂ + φ₂, then one has N(g₁ − g₂) = 0, which implies that g₁ = g₂, thus, φ₁ = φ₂. g and φ are called the almost periodic component and the ergodic perturbation of the function f, respectively. Denote by $\tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ the set of pseudo almost periodic functions uniformly for x ∈ D.

Example 4.1. Let $\mathbb{T}={\bigcup }_{k=1}^{\infty }\left[2k,2k+1\right],$

f(t) = g(t) + φ(t), where g(t) = sin t + sin πt, $\phi \left(t\right)=-\frac{1}{t\sigma \left(t\right)},t\in \mathbb{T}$

and

F (t, x) = f(t) cos x, $t\in \mathbb{T}$.

Since

$$\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s\right)\right|\mathrm{\Delta }s=\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\frac{1}{s\sigma \left(s\right)}\mathrm{\Delta }s=\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\cdot {\left.\frac{1}{s}\right|}_{t_0-r}^{t_0+r}=0,$$

so, $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\right)$. Therefore, $f\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}\left(\mathbb{T}\right)$, $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}\left(\mathbb{T}\times D\right)$.

Theorem 4.1. If $f\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$, then

$$\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}f\left(s,x\right)\mathrm{\Delta }s:=M\left(f\right)$$

exists and is finite. It is the mean value of f. Moreover M(f) = M(g).

Proof. Indeed

$$\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}f\left(s,x\right)\mathrm{\Delta }s=\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}g\left(s,x\right)\mathrm{\Delta }s+\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\phi \left(s,x\right)\mathrm{\Delta }s.$$

Since $g\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ then

$$\underset{r\to +\infty }{\mathsf{\text{lim}}}\underset{t_0-r}{\overset{t_0+r}{\int }}g\left(s,x\right)\mathrm{\Delta }s$$

exists and is finite by Theorem 3.1. Furthermore, one has

$$-\left|\phi \left(s,x\right)\right|\le \phi \left(s,x\right)\le \left|\phi \left(s,x\right)\right|.$$

Then

$$-\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s\le \underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\phi \left(s,x\right)\mathrm{\Delta }s\le \underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}|\phi \left(s,x\right)\mathrm{\Delta }s.$$

Since $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$,

$$\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s=0=M\left(\phi \right).$$

Hence

$$\underset{r\to +\infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\phi \left(s,x\right)\mathrm{\Delta }s=0.$$

Therefore M(f) = M(g). The proof is complete. □

By Definition 4.1 and the definition of a(·, λ, x), one can easily have

Corollary 4.1. If $f\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ then a(f, λ, x) = a(g, λ, x).

Furthermore, from Definition 4.1, one can easily show that

Theorem 4.2. If $f\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ and g is the almost periodic component of f, then we have

$$g\left(\mathbb{T}\times S\right)\subset \overline{f\left(\mathbb{T}\times S\right)}$$

and

$$∥f∥\ge ∥g∥\ge \underset{\left(t,x\right)\in \mathbb{T}\times S}{\mathsf{\text{inf}}}\left|g\left(t,x\right)\right|\ge \underset{\left(t,x\right)\in \mathbb{T}\times S}{\mathsf{\text{inf}}}\left|f\left(t,x\right)\right|,$$

where $f\left(\mathbb{T}\times S\right)$ and $g\left(\mathbb{T}\times S\right)$ denote the value field of f and g on $\mathbb{T}\times S$, respectively, $\overline{f\left(\mathbb{T}\times S\right)}$ denotes the closure of $f\left(\mathbb{T}\times S\right)$, where S is an arbitrary compact subset of D.

Definition 4.2. A closed subset C of is said to be an ergodic zero set in if

$$\frac{{\mu }_{\mathrm{\Delta }}\left(C\cap \left(\left[t_0-r,t_0+r\right]\cap \mathbb{T}\right)\right)}{2r}\to 0asr\to \infty ,where{t}_0\in \mathbb{T}.$$

By the definition of $\tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$, the proof of the following theorem is straightforward.

Theorem 4.3. A function $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$ if and only if for ε > 0, the set $C_{\epsilon }=\left\{t\in \mathbb{T}:\left|\phi \left(t,x\right)\right|\ge \epsilon \right\}$ is an ergodic zero subset in .

Theorem 4.4. (i) A function $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\times D\right)$ if and only if ${\left|\phi \right|}^2\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\times D\right)$.

(ii) $\Phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$ if and only if the norm function $\left|\Phi \left(\cdot ,x\right)\right|\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\times D\right)$.

Proof. (i) The sufficiency follows since

$$\begin{array}{ll}\hfill \frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s& \le \frac{1}{2r}{\left[\underset{t_0-r}{\overset{t_0+r}{\int }}{\left|\phi \left(t,x\right)\right|}^2\mathrm{\Delta }s\right]}^{1/2}{\left[\underset{t_0-r}{\overset{t_0+r}{\int }}1\mathrm{\Delta }s\right]}^{1/2}\\ ={\left[\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}{\left|\phi \left(s,x\right)\right|}^2\mathrm{\Delta }s\right]}^{1/2}.\end{array}$$

The necessity follows from the fact that

$$\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}{\left|\phi \left(s,x\right)\right|}^2\mathrm{\Delta }s\le ∥\phi ∥\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(t,x\right)\right|\mathrm{\Delta }s,$$

since φ is bounded on . Therefore, one can easily see that (i) is satisfied.

(ii) By (i), $\Phi =\left({\phi }_1,{\phi }_2,\dots ,{\phi }_n\right)\in \mathcal{P}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$ if and only if ${\phi }_i{\bar{\phi }}_i\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\times D\right),i=1,2,\dots ,n$. The latter is equivalent to ${\left|\Phi \left(\cdot ,x\right)\right|}^2=\sum _{i=1}^n{\left|\phi \left(\cdot ,x\right)\right|}^2\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\times D\right)$, which again by (i), is equivalent to $\left|\Phi \left(\cdot ,x\right)\right|\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\times D\right)$. □

The proof is complete.

For $H=\left(h_1,h_2,\dots ,h_n\right)\in {\mathbb{E}}^n$, suppose that H(t) ∈ D for all $t\in \mathbb{T}$. Define $H\times \iota :\mathbb{T}\to \mathbb{T}\times D$ by

$$H\times \iota \left(t\right)=\left(t,h_1\left(t\right),h_2\left(t\right),\dots ,h_n\left(t\right)\right).$$

For $F=\left(f_1,f_2,\dots ,f_n\right)\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$, let G = (g₁, g₂, ..., g _n ) and Φ = (φ₁, φ₂, ..., φ _n ), where g _i and φ _i are the almost periodic component and the ergodic perturbation of f _i (i = 1, 2, ..., n), respectively.

Definition 4.3. Let S be a compact subset of D. A function $f\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right)$ is said to be continuous in x ∈ S uniformly for $t\in \mathbb{T}$ if for given x ∈ S and ε > 0, there exists a δ(x, ε) > 0 such that x' ∈ S and |x − x' | < δ(x, ε) imply that |f(t, x') − f(t, x)| < ε for all $t\in \mathbb{T}$.

Theorem 4.5. Suppose that the function $f\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ is continuous in x ∈ S uniformly for $t\in \mathbb{T}$ and $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ such that $F\left(\mathbb{T}\right)\subset D$, then $f\circ \left(F\times \iota \right)\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$, where $F\left(\mathbb{T}\right)$ denotes the value field of F and S is an arbitrary compact subset of D.

Proof. Let f = g + φ and F = G + Φ with $G=\left(g_1,g_2,\dots ,g_n\right)\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ as above. Note that

$$\begin{array}{ll}\hfill f\circ \left(F\times \iota \right)& =g\circ \left(F\times \iota \right)+\phi \circ \left(F\times \iota \right)\\ =g\circ \left(G\times \iota \right)+\left[g\circ \left(F\times \iota \right)-g\circ \left(G\times \iota \right)+\phi \circ \left(F\times \iota \right)\right].\end{array}$$

It follows from Theorem 4.2 that $G\left(\mathbb{T}\right)\subset \overline{F\left(\mathbb{T}\right)}\subset D$. By Theorem 3.15 in [35], we have $g\circ \left(G\times \iota \right)\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. To finish the proof, we need to show that the function h = g ο (F × ι) - g ο (G × ι) + φ ο (F × ι) is in $\tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$.

First we show that $g\circ \left(F\times \iota \right)-g\circ \left(G\times \iota \right)\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$.

It is trivial in the case that g = 0. So we assume that g ≠ 0. Set $D_1=\overline{F\left(\mathbb{T}\right)}.$ By Theorem 3.1 in [35], the function g is uniformly continuous on $\mathbb{T}\times D_1$. For ε > 0, there exists a δ > 0 such that

$$\left|g\left(t,x_1\right)-g\left(t,x_2\right)\right|<\frac{\epsilon }{2},x_1,x_2\in D_1,\left|x_1-x_2\right|<\delta ,t\in \mathbb{T}.$$

(4.1)

Set

$$C_{\delta }=\left\{t\in \mathbb{T}:\left|F\left(t\right)-G\left(t\right)\right|=\left|\Phi \left(t\right)\right|\ge \delta \right\}.$$

(4.2)

It follows from Theorem 4.3 and (ii) of Theorem 4.4 that C _δ is an ergodic zero subset of . We can find T > 0 such that when r ≥ T,

$$\frac{{\mu }_{\mathrm{\Delta }}\left(\left(\left[t_0-r,t_0+r\right]\cap \mathbb{T}\right)\cap C_{\delta }\right)}{2r}<\frac{\epsilon }{4∥g∥}.$$

(4.3)

By (4.1), (4.2) and (4.3), we have

$$\begin{array}{l}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|g\left(s,F\left(s\right)\right)-g\left(s,G\left(s\right)\right)\right|\mathrm{\Delta }s\\ =\frac{1}{2r}\left\{\underset{\left(\left[t_0-r,t_0+r\right]\cap \mathbb{T}\right)\backslash C_{\delta }}{\int }+\underset{\left(\left[t_0-r,t_0+r\right]\cap \mathsf{\text{T}}\right)\cap C_{\delta }}{\int }\left|g\left(s,F\left(s\right)\right)-g\left(s,G\left(s\right)\right)\right|\mathrm{\Delta }s\right\}\\ \le \frac{\epsilon }{2}+2∥g∥\frac{{\mu }_{\mathrm{\Delta }}\left(\left(\left[t_0-r,t_0+r\right]\cap \mathbb{T}\right)\cap C_{\delta }\right)}{2r}<\epsilon .\end{array}$$

Therefore, $g\circ \left(F\times \iota \right)-g\circ \left(G\times \iota \right)\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$.

Finally, we show that $\phi \circ \left(F\times \iota \right)\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$. Note that f = g + φ and g is uniformly continuous on $\mathbb{T}\times D_1$. By the hypothesis, f is continuous in S ⊂ D₁ uniformly for $t\in \mathbb{T}$; so is φ. Since D₁ is compact in ${\mathbb{E}}^n$, one can find, say m, open balls O _k with center x^k ∈ D₁, k = 1, 2, ..., m, and radius δ(x^k, ε/2) such that $D_1\subset \bigcup _{k=1}^m{O}_k$ and

$$\left|\phi \left(t,x\right)-\phi \left(t,x^k\right)\right|<\frac{\epsilon }{2},x\in O_k,t\in \mathbb{T}.$$

(4.4)

The set

$${\mathbb{B}}_k=\left\{t\in \mathbb{T}:F\left(t\right)\in O_k\right\}$$

(4.5)

is open and $\mathbb{T}={\bigcup }_{k=1}^m{\mathbb{B}}_k$. Let $E_k={\mathbb{B}}_k\backslash {\bigcup }_{j=1}^{k-1}{\mathbb{B}}_j$, then E _k ∩ E _j = ∅ when k ≠ j, 1 ≤ k, j ≤ m.

Since for each $\phi \left(\cdot ,x^{\left(k\right)}\right)\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$, there is a number T₀ > 0 such that

$$\sum _{k=1}^m\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x^{\left(k\right)}\right)\right|\mathrm{\Delta }s<\frac{\epsilon }{2},r\ge {\mathbb{T}}_0.$$

(4.6)

It follows from (4.4), (4.5) and (4.6) that

$$\begin{array}{ll}\hfill \frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,F\left(s\right)\right)\right|\mathrm{\Delta }s& \le \frac{1}{2r}\sum _{k=1}^m\underset{E_k\cap \left(\left[t_0-r,t_0+r\right]\cap \mathbb{T}\right)}{\int }\left(\left|\phi \left(s,F\left(s\right)\right)-\phi \left(s,x^{\left(k\right)}\right)\right|+\left|\phi (s,x^{\left(k\right)}\right|\right)+\mathrm{\Delta }s\\ \le \frac{\epsilon }{2}+\sum _{k=1}^m\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x^{\left(k\right)}\right)\right|\mathrm{\Delta }s<\epsilon .\end{array}$$

This shows that $\phi \circ \left(F\times \iota \right)\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$. The proof is complete. □

Define

$$E_0{\left(\mathbb{T}\times D\right)}_n=\left\{f\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right):f\left(t,x\right)\to 0,\mathsf{\text{uniformly}}\mathsf{\text{in}}x\in D,\mathsf{\text{as}}\left|t\right|\to \infty \right\}.$$

$$E_0{\left(\mathbb{T}\right)}_n=\left\{f\in C\left(\mathbb{T},{\mathbb{E}}^n\right):f\left(t\right)\to 0,\mathsf{\text{as}}\left|t\right|\to \infty \right\}.$$

Definition 4.4. Let $\mathcal{A}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ denote all the functions f of the form f = g + φ, where $g\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ and $\phi \in E_0{\left(\mathbb{T}\times D\right)}_n$. The members of $\mathcal{A}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ are called asymptotically almost periodic functions in t uniformly for x ∈ D.

It is obvious that $E_0{\left(\mathbb{T}\times D\right)}_n\subset \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$ and $\mathcal{A}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n\subset \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$.

Corollary 4.2. If $f\in \mathcal{A}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ and $F\in \mathcal{A}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ such that $F\left(\mathbb{T}\right)\subset D$, then $f\circ \left(F\times \iota \right)\in \mathcal{A}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$.

Proof. Obliviously,

$$\begin{array}{ll}\hfill f\circ \left(F\times \iota \right)& =g\circ \left(F\times \iota \right)+\phi \circ \left(F\times \iota \right)\\ =g\circ \left(G\times \iota \right)+\left[g\circ \left(F\times \iota \right)-g\circ \left(G\times \iota \right)+\phi \circ \left(F\times \iota \right)\right],\end{array}$$

where $g\circ \left(G\times \iota \right)\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. By the hypothesis that $\Phi =F-G\in E_0{\left(\mathbb{T}\right)}_n$ and $\phi \in E_0{\left(\mathbb{T}\times D\right)}_n$, it follows that $g\circ \left(F\times \iota \right)-g\circ \left(G\times \iota \right)\in E_0{\left(\mathbb{T}\right)}_n$ since the uniform continuity of g and $\phi \circ \left(F\times \iota \right)\in E_0{\left(\mathbb{T}\right)}_n$ since $\phi \left(t,F\left(t\right)\right)\le \underset{x\in D}{\mathsf{\text{sup}}}\phi \left(t,x\right)$. The proof is complete. □

Theorem 4.6. Suppose that $g\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ satisfies that for every ε > 0,

$$\frac{{\mu }_{\mathrm{\Delta }}\left\{t:g\left(t,x\right)>-\epsilon ,t\in \left[t_0-r,t_0+r\right]\cap \mathbb{T}\right\}}{2r}\to 1,asr\to +\infty ,where{t}_0\in \mathbb{T},r\in \Pi .$$

Then g ≥ 0 for all $\mathbb{T}\times S$, where S is an arbitrary compact subset of D.

Proof. Suppose that the conclusion does not hold. This implies that $g\left(t_0^{\prime },x\right)<0$ for some $t_0^{\prime }$. Choose ε > 0, $\epsilon <-g\left(t_0^{\prime },x\right)$.

By continuity, there exists δ > 0 so that $\left|t-{t^{\prime }}_0\right|\le \delta$ implies g(t, x) < −ε. In view of Definition 2.4, there exists l(ε, S) > 0 so that in each interval I of length l, one can find $\frac{\epsilon }{2}$-almost period τ with the property that

$$\left|g\left(t+\tau ,x\right)-g\left(t,x\right)\right|<\frac{\epsilon }{2}.$$

Choose a sequence τ _k of almost periods, ${\tau }_k\in \left[t_0+kl,t_0+\left(k+1\right)l\right]$, we have

$$g\left(t+{\tau }_k,x\right)<-\frac{\epsilon }{2},\mathsf{\text{and}}t\in \left[t_0^{\prime }-\delta ,t_0^{\prime }+\delta \right]\cap \mathbb{T}\mathsf{\text{and every}}k\in \mathbb{N}.$$

Denote $M=\left|t_0^{\prime }\right|+\delta$, we have

$${\mu }_{\mathrm{\Delta }}\left\{t\in \left[t_0-kl-M,t_0+kl+M\right]\cap \mathbb{T}:g\left(t,x\right)<-\frac{\epsilon }{2}\right\}\ge 2k\delta .$$

Therefore,

$$\frac{{\mu }_{\mathrm{\Delta }}\left\{t\in \left[t_0-kl-M,t_0+kl+M\right]\cap \mathbb{T}:g\left(t,x\right)<-\frac{\epsilon }{2}\right\}}{2kl+2M}\ge \frac{2k\delta }{2kl+2M}.$$

The right hand side does not tend to zero as k → +∞. This contradicts the assumption made in the lemma. Therefore, g ≥ 0. The proof is complete. □

Theorem 4.7. If $f\in C\left(\mathbb{T}\times D,{\mathbb{E}}^n\right),f=g+\phi$, where $g\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ and $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$, then

• (i) If $\underset{\left|t\right|\to \infty }{\mathsf{\text{lim}}}\phi \left(t,x\right)$ exists, then $\underset{\left|t\right|\to \infty }{\mathsf{\text{lim}}}\phi \left(t,x\right)=0$.

• (ii) For all $\left(t,x\right)\in \mathbb{T}\times S$, if f ≥ 0 then g ≥ 0, where S is an arbitrary compact subset of D.

Proof. (i) Suppose that the property does not hold, then there exist a constant $\tilde{\alpha }>0$ and c ∈ Π such that $\phi \left(t,x\right)>\tilde{\alpha }$ for t ≥ c, which yields

$$\frac{1}{r}\underset{t_0}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s=\frac{1}{r}\left[\underset{t_0}{\overset{t_0+c}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s+\underset{t_0+c}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s\right]\ge \frac{1}{r}\tilde{\alpha }\left(r-c\right).$$

Passing to the limit as r → ∞, we obtain

$$\underset{r\to \infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s\ge \tilde{\alpha },$$

which contradicts the fact that $\phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\times D\right)}_n$.

(ii) Assuming f ≥ 0, we want to show that g ≥ 0. We have f = g + φ with

$$\underset{r\to \infty }{\mathsf{\text{lim}}}\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|\phi \left(s,x\right)\right|\mathrm{\Delta }s=0.$$

Thus, there exists {c _n }_n∈ℕ⊂ Π, c _n → +∞ as n → ∞ such that g(t + c _n , x) → g(t, x) for all $\left(t,x\right)\in \mathbb{T}\times S$. Furthermore, for any ε > 0 and r > 0, one can easily get

$${\mu }_{\mathrm{\Delta }}\left\{t\in \left[t_0-r,t_0+r\right]\cap \mathbb{T}:\phi \left(t,x\right)>\epsilon \right\}\to 0,\mathsf{\text{as}}r\to \infty ,$$

which implies that

$$\frac{{\mu }_{\mathrm{\Delta }}\left\{t:g\left(t,x\right)>-\epsilon ,t\in \left[t_0-r,t_0+r\right]\cap \mathbb{T}\right\}}{2r}\to 1,\mathsf{\text{as}}r\to +\infty ,\mathsf{\text{where}}{t}_0\in \mathbb{T},r\in \Pi .$$

By Theorem 4.6, one can have g(t, x) ≥ 0 for all $\left(t,x\right)\in \mathbb{T}\times S$.

The proof is complete. □

5 Pseudo almost periodic solutions of dynamic equations on time scales

Consider the non-autonomous equation

$$x^{\mathrm{\Delta }}=A\left(t\right)x+F\left(t\right)$$

(5.1)

and its associated homogeneous equation

$$x^{\mathrm{\Delta }}=A\left(t\right)x,$$

(5.2)

where the n × n coefficient matrix A(t) is continuous on and column vector F = (f₁, f₂, ..., f _n )^T is in ${\mathbb{E}}^n$. Define $∥F∥=\underset{t\in \mathbb{T}}{\mathsf{\text{sup}}}\left|F\left(t\right)\right|$. We will call A(t) almost periodic if all the entries are almost periodic.

Definition 5.1 ([37]). Equation (5.2) is said to admit an exponential dichotomy on if there exist positive constants K, α, projection P and the fundamental solution matrix X(t) of (5.2), satisfying

$$\left\{\begin{array}{c}\left|X\left(t\right)P{X}^{-1}\left(s\right)\right|\le K{e}_{\ominus \alpha }\left(t,s\right),s,t\in \mathbb{T},t\ge s,\hfill \\ \left|X\left(t\right)\left(I-P\right)X^{-1}\left(s\right)\right|\le K{e}_{\ominus \alpha }\left(s,t\right),s,t\in \mathbb{T},t\le s.\hfill \end{array}\right.$$

(5.3)

Lemma 5.1. Let α > 0, then for any fixed $s\in \mathbb{T}$ and s = −∞, one has the following:

$$e_{\ominus \alpha }\left(t,s\right)\to 0,t\to +\infty .$$

Proof. If µ(t) > 0, since $\alpha \in {\mathcal{R}}^{+}$, we have

$$1+\mu \left(t\right)\ominus \alpha =1+\mu \left(t\right)\frac{-\alpha }{1+\mu \left(t\right)\alpha }=\frac{1}{1+\mu \left(t\right)\alpha }<1.$$

Thus, $\ominus \alpha \in {\mathcal{R}}^{+}$ and it is easy to have

$$\mathsf{\text{Log}}\left(1+\mu \left(t\right)\ominus \alpha \right)\in ℝ\mathsf{\text{for}}\mathsf{\text{all}}t\in \mathbb{T}.$$

So

$${\xi }_{\mu \left(t\right)}\left(\ominus \alpha \right)=\frac{\mathsf{\text{Log}}\left(1+\mu \left(t\right)\ominus \alpha \right)}{\mu \left(t\right)}<0.$$

Hence

$$e_{\ominus \alpha }\left(t,s\right)=\mathsf{\text{exp}}\left\{\underset{s}{\overset{t}{\int }}{\xi }_{\mu \left(t\right)}\left(\ominus \alpha \right)\mathrm{\Delta }t\right\}\to 0\mathsf{\text{as}}t\to +\infty .$$

If μ(t) = 0, one can easily get the conclusion. If s = -∞, it is easy to see that ${\int }_s^t{\xi }_{\mu \left(t\right)}\left(\ominus \alpha \right)\mathrm{\Delta }t\to -\infty$ as t → +∞, thus, e_⊖α(t, s) → 0. The proof is complete. □

Theorem 5.1. Suppose that A(t) is almost periodic, (5.2) admits an exponential dichotomy and the function $F\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$. Then (5.1) has a unique bounded solution $x\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$.

Proof. Similar to the proof of Theorem 4.1 in [35], by checking directly, one can see that the function:

$$x\left(t\right)=\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s$$

(5.4)

is a solution of (5.1). Now, we show that the solution is bounded. It follows from (5.4) that

$$\begin{array}{ll}\hfill \left|x\left(t\right)\right|& =\underset{t\in \mathbb{T}}{\mathsf{\text{sup}}}\left|\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s\right|\\ \le \underset{t\in \mathbb{T}}{\mathsf{\text{sup}}}\left(\left|\underset{-\infty }{\overset{t}{\int }}{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)▵s\right|+\left|\underset{t}{\overset{+\infty }{\int }}{e}_{\ominus \alpha }\left(\sigma \left(s\right),t\right)\mathrm{\Delta }s\right|\right)K∥F∥\\ \le \left(\frac{1}{\alpha }-\frac{1}{\ominus \alpha }\right)K∥F∥=\frac{2+\mu \alpha }{\alpha }K∥F∥,\end{array}$$

where $∥\cdot ∥=\underset{t\in \mathbb{T}}{\mathsf{\text{sup}}}\left|\cdot \right|$. The solution x is bounded since F is bounded. By Lemma 4.13 in [35], the bounded solution is unique since the homogeneous equation (5.2) has no nontrivial bounded solution.

In the following, we show that $x\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$. Let $I\left(t\right)={\int }_{-\infty }^{t}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s$ and $H\left(t\right)={\int }_t^{+\infty }X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s$. Then x = I + H. It follows from (5.3) and Theorem 2.15 in [38] that

$$\begin{array}{ll}\hfill \frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|I\left(t\right)\right|\mathrm{\Delta }t& \le \frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\mathrm{\Delta }t\underset{-\infty }{\overset{t}{\int }}\left|X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)\right|\left|F\left(s\right)\right|\mathrm{\Delta }s\\ \le \frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\mathrm{\Delta }t\underset{-\infty }{\overset{t}{\int }}K{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)\left|F\left(s\right)\right|\mathrm{\Delta }s\\ =\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\mathrm{\Delta }t\left(\underset{-\infty }{\overset{t_0-r}{\int }}+\underset{t_0-r}{\overset{t}{\int }}K{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)\left|F\left(s\right)\right|\right)\mathrm{\Delta }s\\ =\frac{1}{2r}\underset{-\infty }{\overset{t_0-r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{t_0-r}{\overset{t_0+r}{\int }}K{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)\mathrm{\Delta }t\\ +\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{s}{\overset{t_0+r}{\int }}K{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)\mathrm{\Delta }t=I_1+I_2.\end{array}$$

To show that $I\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$, we only need to show that both I₁ → 0 and I₂ → 0 when r → ∞. By Lemma 5.1, one can obtain

$$\begin{array}{ll}\hfill I_1& =\frac{1}{2r}\underset{-\infty }{\overset{t_0-r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{t_0-r}{\overset{t_0+r}{\int }}K{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)\mathrm{\Delta }t\\ =\frac{1}{2r}\underset{-\infty }{\overset{t_0-r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{t_0-r}{\overset{t_0+r}{\int }}\frac{K}{1+\mu \left(t\right)\ominus \alpha }{e}_{\ominus \alpha }\left(\sigma \left(t\right),\sigma \left(s\right)\right)\mathrm{\Delta }t\\ \le \frac{1}{2r}K\left(1+\bar{\mu }\alpha \right)\underset{-\infty }{\overset{t_0-r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{t_0-r}{\overset{t_0+r}{\int }}{e}_{\alpha }\left(\sigma \left(s\right),\sigma \left(t\right)\right)\mathrm{\Delta }t\\ =\frac{1}{2r}\frac{K\left(1+\bar{\mu }\alpha \right)}{\alpha }\underset{-\infty }{\overset{t_0-r}{\int }}\left|F\left(s\right)\right|\left[e_{\alpha }\left(\sigma \left(s\right),t_0-r\right)-e_{\alpha }\left(\sigma \left(s\right),t_0+r\right)\right]\mathrm{\Delta }s\\ \le \frac{1}{2r}\frac{K\left(1+\bar{\mu }\alpha \right)}{\alpha }∥F∥\left(\underset{-\infty }{\overset{t_0-r}{\int }}{e}_{\ominus \alpha }\left(t_0-r,\sigma \left(s\right)\right)\mathrm{\Delta }s-\underset{-\infty }{\overset{t_0-r}{\int }}{e}_{\ominus \alpha }\left(t_0+r,\sigma \left(s\right)\right)\mathrm{\Delta }s\right)\\ =\frac{1}{2r}\frac{K\left(1+\bar{\mu }\alpha \right)}{\alpha }\frac{1}{\ominus \alpha }\left(e_{\ominus \alpha }\right(t_0-r,-\infty )-e_{\ominus \alpha }\left(t_0-r,t_0-r\right)-e_{\ominus \alpha }\left(t_0+r,-\infty \right)\\ +e_{\ominus \alpha }(t_0+r,t_0-r))\to 0\mathsf{\text{as}}r\to +\infty ;\end{array}$$

$$\begin{array}{ll}\hfill I_2& =\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{s}{\overset{t_0+r}{\int }}K{e}_{\ominus \alpha }\left(t,\sigma \left(s\right)\right)\mathrm{\Delta }t\\ =\frac{1}{2r}\underset{t_0-r}{\overset{t_0+r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{s}{\overset{t_0+r}{\int }}\frac{K}{1+\mu \left(t\right)\ominus \alpha }{e}_{\ominus \alpha }\left(\sigma \left(t\right),\sigma \left(s\right)\right)\mathrm{\Delta }t\\ \le \frac{1}{2r}K\left(1+\bar{\mu }\alpha \right)\underset{t_0-r}{\overset{t_0+r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s\underset{s}{\overset{t_0+r}{\int }}{e}_{\alpha }\left(\sigma \left(s\right),\sigma \left(t\right)\right)\mathrm{\Delta }t\\ =\frac{1}{2r}\frac{K\left(1+\bar{\mu }\alpha \right)}{\alpha }\underset{t_0-r}{\overset{t_0+r}{\int }}\left|F\left(s\right)\right|\left[e_{\alpha }\left(\sigma \left(s\right),s\right)-e_{\alpha }\left(\sigma \left(s\right),t_0+r\right)\right]\mathrm{\Delta }s\\ \le \frac{1}{2r}\frac{K{\left(1+\bar{\mu }\alpha \right)}^2}{\alpha }\underset{t_0-r}{\overset{t_0+r}{\int }}\left|F\left(s\right)\right|\mathrm{\Delta }s.\end{array}$$

Therefore, by (ii) of Theorem 4.4, $\left|F\left(\cdot \right)\right|\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0\left(\mathbb{T}\right)$, so I₂ → 0 as r → +∞.

Similarly, one can show that $H\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$. The proof is complete. □

Theorem 5.2. Suppose that A(t) is almost periodic and (5.2) admits an exponential dichotomy. Then, for every $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. (5.1) has a unique bounded solution $x_F\in \mathcal{P}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. The mapping F → x _F is bounded and linear with

$$∥x_F∥\le \left(\frac{1}{\alpha }-\frac{1}{\ominus \alpha }\right)K∥F∥=\frac{2+\mu \alpha }{\alpha }K∥F∥.$$

(5.5)

Proof. Since $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n,F=G+\Phi$, where $G\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ and $\Phi \in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$.

According to the proof of Theorem 5.1, the function

$$\begin{array}{ll}\hfill x_F& =\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s\\ =\left(\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)G\left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)G\left(s\right)\mathrm{\Delta }s\right)\\ +\left(\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)\Phi \left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)\Phi \left(s\right)\mathrm{\Delta }s\right)\\ :=x_G+x_{\Phi }\end{array}$$

is the unique solution of (5.1), where

$$\begin{array}{ll}\hfill x_G& :=\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)G\left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)G\left(s\right)\mathrm{\Delta }s,\\ \hfill x_{\Phi }& :=\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)\Phi \left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)\Phi \left(s\right)\mathrm{\Delta }s.\end{array}$$

By Theorem 4.1 in [35], $x_G\in \mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. By Theorem 5.1, $x_{\Phi }\in \tilde{\mathcal{P}}\mathcal{A}{\mathcal{P}}_0{\left(\mathbb{T}\right)}_n$. Therefore, $x_F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. Obviously, the mapping F → x _F is linear. The proof is complete. □

Lemma 5.2. Let $c_i\left(t\right):\mathbb{T}\to {ℝ}^{+}$ be an almost periodic function, $-c_i\in {\mathcal{R}}^{+},T\in \Pi$ and

$$m\left(c_i\right)=\underset{T\to \infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t}{\overset{t+T}{\int }}{c}_i\left(s\right)\mathrm{\Delta }s>0,i=1,2,\dots ,n.$$

Then the following linear system

$$x^{\mathrm{\Delta }}\left(t\right)=\mathsf{\text{diag}}\left(-c_1\left(t\right),-c_2\left(t\right),\dots ,-c_n\left(t\right)\right)x\left(t\right)$$

(5.7)

admits an exponential dichotomy on , where m(c _i ) denote the mean-value of c _i , i = 1, 2, ..., n.

Proof. According to Theorem 2.77 in [3], the linear system (5.7) has a unique solution

$$x(t\mathsf{\text{)}}\mathsf{\text{ = }}{x}_0{e}_{-c}\left(t,t_0\right)\mathsf{\text{,}}$$

where x(t₀) = x₀, −c = diag(−c₁(t), −c₂(t), ..., −c _n (t)).

Now, we prove that x(t) admits an exponential dichotomy on .

According to Theorem 3.2 and Theorem 3.5, one has

$$m\left(c_i\right)=\underset{T\to \infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t}{\overset{t+T}{\int }}{c}_i\left(s\right)\mathrm{\Delta }s=\underset{T\to \infty }{\mathsf{\text{lim}}}\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}{c}_i\left(s\right)\mathrm{\Delta }s>0,t_0\in \mathbb{T},i=1,2,\dots ,n.$$

So there exists T₀ > 0, when T > T₀, one has

$$\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}{c}_i\left(s\right)\mathrm{\Delta }s>\frac{1}{2}m\left(c_i\right)=\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}\frac{1}{2}m\left(c_i\right)\mathrm{\Delta }s,i=1,2,\dots ,n,$$

that is

$$\frac{1}{T}\underset{t_0}{\overset{t_0+T}{\int }}\left(c_i\left(s\right)-\frac{1}{2}m\left(c_i\right)\right)\mathrm{\Delta }s>0,i=1,2,\dots ,n,$$

thus, for T > T₀, we have $c_i\left(t\right)>\frac{1}{2}m\left(c_i\right),i=1,2,...,n$.

Case 1. If $\mu \left(\eta \right)>0,\eta \in {\left[s,t\right]}_{\mathbb{T}},s,t\in \mathbb{T}$, we have

$$1-\frac{\mu \left(t\right)\frac{m\left(c_i\right)}{2}}{1+\mu \left(t\right)\frac{m\left(c_i\right)}{2}}>1-\mu \left(t\right)\frac{m\left(c_i\right)}{2}>1-\mu \left(t\right)c_i\left(t\right),i=1,2,\dots ,n,$$

then

$$\underset{s}{\overset{t}{\int }}\frac{\mathsf{\text{log}}\left(1-\mu \left(\eta \right)c_i\left(\eta \right)\right)}{\mu \left(\eta \right)}\mathrm{\Delta }\eta \le \underset{s}{\overset{t}{\int }}\frac{\mathsf{\text{log}}\left(1-\frac{\mu \left(\eta \right)\frac{m\left(c_i\right)}{2}}{1+\mu \left(\eta \right)\frac{m\left(c_i\right)}{2}}\right)}{\mu \left(\eta \right)}\mathrm{\Delta }\eta ,i=1,2,\dots ,n,$$

thus

$$\mathsf{\text{exp}}\left\{\underset{s}{\overset{t}{\int }}\frac{\mathsf{\text{log}}\left(1-\mu \left(\eta \right)c_i\left(\eta \right)\right)}{\mu \left(\eta \right)}\mathrm{\Delta }\eta \right\}\le \mathsf{\text{exp}}\left\{\underset{s}{\overset{t}{\int }}\frac{\mathsf{\text{log}}\left(1-\frac{\mu \left(\eta \right)\frac{m\left(c_i\right)}{2}}{1+\mu \left(\eta \right)\frac{m\left(c_i\right)}{2}}\right)}{\mu \left(\eta \right)}\mathrm{\Delta }\eta \right\},i=1,2,\dots ,n,$$

that is,

$$e_{-c_i}\left(t,s\right)\le e_{\ominus \frac{m\left(c_i\right)}{2}}\left(t,s\right),i=1,2,\dots ,n.$$

Case 2. If $\mu \left(\eta \right)=0,\eta \in {\left[s,t\right]}_{\mathbb{T}},s,t\in \mathbb{T}$, one cane easily obtain

$$e_{-c_i}\left(t,s\right)=\mathsf{\text{exp}}\left\{\underset{s}{\overset{t}{\int }}-c_i\left(\eta \right)\mathrm{\Delta }\eta \right\}\le \mathsf{\text{exp}}\left\{\underset{s}{\overset{t}{\int }}-\frac{m\left(c_i\right)}{2}\mathrm{\Delta }\eta \right\}=e_{\ominus \frac{m\left(c_i\right)}{2}}\left(t,s\right),i=1,2,\dots ,n.$$

Set P = I, we have

$$\left|X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)\right|=\left|x_0{e}_{-c}\left(t,t_0\right)I{x}_0^{-1}{e}_{\ominus -c}\left(s,t_0\right)\right|\le K{e}_{\ominus \frac{M}{2}}\left(t,s\right),$$

where K ≥ 1, $M=\underset{1\le i\le n}{\mathsf{\text{min}}}\left\{m\left(c_1\right),m\left(c_2\right),\dots ,m\left(c_n\right)\right\}$. Therefore, x(t) admits an exponential dichotomy with P = I on . This completes the proof. □

Example 5.1. Consider the following dynamic equation on an almost periodic time scale $\mathbb{T}={\bigcup }_{i=1}^{\infty }\left[2k,2k+1\right]$:

$$x^{\mathrm{\Delta }}\left(t\right)=Ax\left(t\right)+F\left(t\right),$$

(5.8)

where

$$A=\left(\begin{array}{cc}\hfill -\frac{1}{16}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -\frac{1}{16}\hfill \end{array}\right),F\left(t\right)=\left(\begin{array}{c}\hfill \mathsf{\text{sin}}\sqrt{3}t+\frac{1}{t\sigma \left(t\right)}\hfill \\ \hfill \mathsf{\text{cos}}\sqrt{2}t+\frac{1}{t\sigma \left(t\right)}\hfill \end{array}\right)and0\le \mu \left(t\right)<16.$$

Obviously, $-A\in {\mathcal{R}}^{+}$. By Lemma 5.2, it is easy to know that the homogeneous equation of (5.8) admits an exponential dichotomy with P = I on . Similar to Example 4.1, one easily to see that $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_2$. By Theorem 5.2 and Theorem 2.77 in [3], one can obtain that (5.8) has a unique pseudo almost periodic solution:

$$\begin{array}{ll}\hfill x\left(t\right)& =\underset{-\infty }{\overset{t}{\int }}X\left(t\right)P{X}^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s-\underset{t}{\overset{+\infty }{\int }}X\left(t\right)\left(I-P\right)X^{-1}\left(\sigma \left(s\right)\right)F\left(s\right)\mathrm{\Delta }s\\ =\underset{-\infty }{\overset{t}{\int }}{e}_{-\frac{1}{16}}\left(t,\sigma \left(s\right)\right)\left(\begin{array}{c}\hfill 10\hfill \\ \hfill 01\hfill \end{array}\right)\left(\begin{array}{c}\hfill \mathsf{\text{sin}}\sqrt{3}s+\frac{1}{s\sigma \left(s\right)}\hfill \\ \hfill \mathsf{\text{cos}}\sqrt{2}s+\frac{1}{s\sigma \left(s\right)}\hfill \end{array}\right)\mathrm{\Delta }s.\end{array}$$

6 Applications

Application 1. Consider the following quasi-linear system

$$x^{\mathrm{\Delta }}=A\left(t\right)x+F+{\mu }_{0}G\circ \left(x\times \iota \right),$$

(6.1)

where ${\mu }_0\in {\mathbb{E}}^n\backslash \left\{0\right\},A\left(t\right)$ is a n × n almost periodic matrix, $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ and $G\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$. We call the system

$$x^{\mathrm{\Delta }}=A\left(t\right)x+F$$

(6.2)

the generating system of (6.1).

By Theorem 5.2, system (6.2) has a unique solution $x^0\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ if (5.2) admits an, exponential dichotomy. Now we have the following theorem about (6.1).

Theorem 6.1. If $F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$, A(t) be almost periodic and (5.2) admits an exponential dichotomy. Let $x^0\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ be the unique solution of system (6.2) and denote $D=\left\{x\in {\mathbb{E}}^n:\left|x-x^0\left(t\right)\right|\le a,t\in \mathbb{T}\right\}$, where a > 0. Assume that

(i) $G\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$ and L > 0 such that

$$\left|G\left(t,x^{\prime }\right)-G\left(t,x^{″}\right)\right|\le L|x^{\prime }-x^{″}|,x^{\prime },x^{″}\in D,t\in \mathbb{T};$$

(6.3)

(ii) $0<\left|{\mu }_0\right|<\mathsf{\text{min}}\left\{\frac{\alpha }{\left(2+\bar{\mu }\alpha \right)KL},\frac{\alpha a}{\left(2+\bar{\mu }\alpha \right)K∥G∥}\right\}$, where K and α are the same as those in Theorem 5.2, $\bar{\mu }=\underset{t\in \mathbb{T}}{\mathsf{\text{sup}}}\mu \left(t\right)$.

Then system (6.1) has a unique solution $x\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ such that x ∈ D for all $t\in \mathbb{T}$. Furthermore, ||x - x⁰|| → 0 as μ₀ → 0.

Proof. We construct a sequence of approximations by induction, starting with x⁰ and taking x^k to be the bounded solution of the system

$${\left(x^k\right)}^{\mathrm{\Delta }}=A\left(t\right)x^k+F+{\mu }_{0}G\circ \left(x^{k-1}\times \iota \right).$$

(6.4)

First, we show that x^k exists, $x^k\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ and $x^k\left(\mathbb{T}\right)\subset D,k=0,1,2,\dots ,$ where $x^k\left(\mathbb{T}\right)$ denotes the value field of x^k. Obliviously, the conclusion holds for k = 0 by the hypothesis. Assume that the conclusion holds for k − 1. Then we shall show that the conclusion also holds for k. By Theorem 4.5, one can see that $G\circ \left(x^{k-1}\times \iota \right)\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$, so from Theorem 5.2, (6.4) has a unique solution $x^k\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. It follows from (6.2) and (6.4) that,

$${\left(x^k-x^0\right)}^{\mathrm{\Delta }}=A\left(t\right)\left(x^k-x^0\right)+{\mu }_{0}G\circ \left(x^{k-1}\times \iota \right).$$

By (5.5), we have

$$∥x^k-x^0∥\le \frac{2+\bar{\mu }\alpha }{\alpha }K\left|{\mu }_0\right|∥G∥.$$

Therefore, $x^k\left(\mathbb{T}\right)\subset D$, since

$$\left|{\mu }_0\right|\le \frac{\alpha a}{\left(2+\bar{\mu }\alpha \right)K∥G∥}.$$

Next, we show that {x^k} is Cauchy sequence in $\tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. Since

$${\left(x^{k+1}-x^k\right)}^{\mathrm{\Delta }}=A\left(t\right)\left(x^{k+1}-x^k\right)+{\mu }_0\left[G\circ \left(x^k\times \iota \right)-G\circ \left(x^{k-1}\times \iota \right)\right],$$

it follows from (5.5) and (6.3) that

$$\begin{array}{ll}\hfill ∥x^{k+1}-x^k∥& \le \frac{2+\bar{\mu }\alpha }{\alpha }K\left|{\mu }_0\right|∥G\circ \left(x^k\times \iota \right)-G\circ \left(x^{k-1}\times \iota \right)∥\\ \le \frac{2+\bar{\mu }\alpha }{\alpha }K\left|{\mu }_0\right|L∥x^k-x^{k-1}∥\\ =\theta ∥x^k-x^{k-1}∥\\ ⋮\\ \le {\theta }^k∥x^1-x^0∥,\end{array}$$

where $0<\theta =\frac{2+\bar{\mu }\alpha }{\alpha }K\left|{\mu }_0\right|L<1$. This shows that {x^k} is a Cauchy sequence in $\tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$. Since $\tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ is a Banach space, there is an $x\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ such that ||x^k − x|| → 0, when k → ∞. It follows from (6.4) that x is a solution of (6.1). It is clear that ||x − x⁰|| → 0, as µ₀ → 0.

To show the uniqueness, let x* be another solution of (6.1). Similar to the discussion above, we have

$$∥x-x^{*}∥\le \theta ∥x-x^{*}∥,$$

this is a contradiction. The proof is complete. □

Application 2. Let D be a ball in ${\mathbb{E}}^n$ with center at origin and radius r₀. Consider the following system

$$x^{\mathrm{\Delta }}=A\left(t\right)x+G\circ \left(x\times \iota \right),$$

(6.5)

where A(t) is a n × n almost periodic matrix and $G\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$. Set $\mathbb{B}=\left\{F\in \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n:F\left(\mathbb{T}\right)\subset D\right\}$, where $F\left(\mathbb{T}\right)$ denotes the value field of F.

is a closed subset of $\tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\times D\right)}_n$. Therefore, is a complete metric space.

Theorem 6.2. Let D, G, A(t), and be as those in the previous paragraph. Assume that, (5.2) admits an exponential dichotomy and the function G satisfies,

$$\frac{2+\bar{\mu }\alpha }{\alpha }K\underset{\left(t,x\right)\in \mathbb{T}\times D}{\mathsf{\text{sup}}}\left|G\left(t,x\right)\right|\le r_{0}where\bar{\mu }=\underset{t\in \mathbb{T}}{\mathsf{\text{sup}}}\mu \left(t\right)$$

and

$$\left|G\left(t,x^{\prime }\right)-G\left(t,x^{″}\right)\right|\le L\left|x^{\prime }-x^{″}\right|,x^{\prime },x^{″}\in D,t\in \mathbb{T}$$

(6.6)

with $\frac{2+\bar{\mu }\alpha }{\alpha }KL<1$. Then (6.5) has a unique solution in $\tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$.

Proof. By Theorem 5.2, one can define the mapping $\tilde{T}:\mathbb{B}\to \tilde{\mathcal{P}}\mathcal{A}\mathcal{P}{\left(\mathbb{T}\right)}_n$ by the fact that, for $F\in \mathbb{B}$, $\tilde{T}F$ is the unique pseudo almost periodic solution of the system

$$x^{\mathrm{\Delta }}=A\left(t\right)x+G\circ \left(F\times \iota \right).$$

(6.7)

We claim that $\tilde{T}\mathbb{B}\subset \mathbb{B}$ since by (5.5),

$$∥\tilde{T}F∥\le \frac{2+\bar{\mu }\alpha }{\alpha }K∥G\circ \left(F\times \iota \right)∥\le r_0.$$

The mapping is a contraction on . In fact, for F₁, $F_2\in \mathbb{B}$, it follows from (5.5) and (6.6) that

$$\begin{array}{ll}\hfill \left|\right|\tilde{T}{F}_1-\tilde{T}{F}_2\left|\right|& \le \frac{2+\bar{\mu }\alpha }{\alpha }K\left|\right|G\circ \left(F_1\times \iota \right)-G\circ \left(F_2\times \iota \right)\left|\right|\\ \le \frac{2+\bar{\mu }\alpha }{\alpha }KL\left|\right|F_1-F_2\left|\right|.\end{array}$$

Therefore, $\tilde{T}$ has a unique fixed point in , which is the unique pseudo almost periodic solution of (6.5). The proof is complete. □