Space of Quasi-Periodic Limit Functions and Its Applications

Abstract

We introduce a class consisting of what we call quasi-periodic limit functions and then establish the relation between quasi-periodic limit functions and asymptotically quasi-periodic functions. At last, these quasi-periodic limit functions are applied to study the existence of asymptotically quasi-periodic solutions of abstract Cauchy problems.

1. Introduction

The study of quasi-periodic functions may go back to P. Bohl in 1893 and E. Esclangon in 1904; see [1,2]. These quasi-periodic functions, which appeared in astronomical perturbation problems, are those that can be approximated uniformly on $\mathbb{R}$ by ${\sum }_{k=1}^n{c}_k{A}_k\left(t\right)$ with $c_k\in \mathbb{C}$ and $A_k\left(t\right)=e^{i(m_1{a}_1+m_2{a}_2+\cdots +m_n{a}_n)t}$, where $m_j\in \mathbb{N}$ and $a_j\in \mathbb{R}$. It turns out that any quasi-periodic function can be obtained from a periodic function depending on several variables [3]. Let f be a periodic function from ${\mathbb{R}}^n$ into $\mathbb{C}$. Regarding the variable $t_i$, if the periodic function $f(t_1,t_2,\cdots ,t_n)$ has a period $2\pi$, then $F\left(t\right)=f({\omega }_{1}t,{\omega }_{2}t,\cdots ,{\omega }_{n}t)$ with ${\omega }_j\in \mathbb{R}/\left\{0\right\}$ is quasi-periodic.

The application of quasi-periodic functions has been taken in several directions. One direction is the study of quasi-periodic solutions to nonlinear PDE. Up until now, the existence of quasi-periodic solutions of different kinds of nonlinear equations has been shown by the KAM (Kolmogorov–Arnold–Moser) theory or the C-W-B (Craig–Wayne–Bourgain) method. We briefly mention some recent work in this direction and refer the reader to [4,5,6,7,8,9,10,11] for some pioneering work. In [12], F. Giuliani focused on the generalized KdV equations and discussed the existence of Cantor families of quasi-periodic solutions, which extended the results in [13]. In [14], based on a modified KAM theorem, the existence of quasi-periodic response solutions of reversible systems that have Liouvillian frequencies was studied. Regarding the beam equations defined on compact Lie groups, in [15], the authors showed the existence of quasi-periodic solutions by the Nash–Moser iteration; meanwhile, based on the KAM theorem, in [16], Y. Wang proved the existence of quasi-periodic solutions to beam equations when the nonlinear term has the time and space variables. In [17], the authors discussed the Whitney smooth family of quasi-periodic solutions to beam equations. For other applications of quasi-periodic functions, we mention the work by Küpper and Yuan [18] on quasi-periodic solutions for differential equations with piecewise constant argument (see also, e.g., [19,20]).

On the other hand, consider the inhomogeneous abstract Cauchy problem:

$$\left\{\begin{array}{c}{x}^{\prime }\left(t\right)=Ax\left(t\right)+F\left(t\right),t\in {\mathbb{R}}^{+};\hfill \\ x\left(0\right)=x_0\in X,\hfill \end{array}\right.$$

(1)

where A is the infinitesimal generator of an exponentially stable $C_0$-semigroup ${\left(T\left(t\right)\right)}_{t\ge 0}$, that is there exist $M>0$ and $r>0$ such that $\parallel T\left(t\right)\parallel \le M{e}^{-rt}$ for all $t\in {\mathbb{R}}^{+}$. In order to study the existence of asymptotically periodic solutions of (1), the authors in [21] proposed the concept of the $\omega$-periodic limit function, which is a generalization of the asymptotically $\omega$-periodic function. Later, in [22], the concept of the squared mean periodic limit process was proposed and the application to stochastic differential equations was studied. The theory of $\omega$-periodic limit functions has the advantage of considering the asymptotical periodicity as a whole. For other applications of the $\omega$-periodic limit function, we refer the reader to [23].

It follows from the method in Theorem 4.4 in [21] that the solution of (1) is asymptotically $\omega$-periodic when the coefficient F is the $\omega$-periodic limit. Moreover, if we assume $F=F_1+F_2$, $F_1$ is the ${\omega }_1$-periodic limit and $F_2$ is the ${\omega }_2$-periodic limit, then the solution of (1) is asymptotically quasi-periodic. Here, a question arises: To make sure the solution of (1) is asymptotically quasi-periodic, what kind of function can F be?

Motivated by the above discussions, in the present paper, we propose a class of functions, and we call them quasi-periodic limit functions, which could be regarded as a generalization of (asymptotically) quasi-periodic functions. We will show that the quasi-periodic limit functions contribute to studying the existence of asymptotically quasi-periodic solutions of (1). The main results are Theorem 2 and Theorem 4. To show this, we develop a very general method. We believe the method in this paper could contribute to studying the existence of asymptotically quasi-periodic solutions of some kinds of equations, such as fractional differential equations.

This paper is arranged in three main sections.

In Section 2, we define the notion of quasi-periodic limit functions and study their properties. Especially, we discuss the relation between quasi-periodic limit functions and asymptotically quasi-periodic functions. In Section 3, these quasi-periodic limit functions are applied to study the existence of asymptotically quasi-periodic solutions of abstract Cauchy problems. In Section 4, we propose some related questions. We believe that the questions found here are of interest in the theory of asymptotical quasi-periodicity and that their answers would certainly help to develop this field.

2. Space of Quasi-Periodic Limit Functions

In this paper, we denote the interval $[0,\infty )$ by ${\mathbb{R}}^{+}$. Let $(X,\parallel ·\parallel )$ be a Banach space, and let $C_b({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ be the space of bounded and continuous functions from ${\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$ into X, endowed with the uniform convergence norm ${\parallel ·\parallel }_{\infty }$. Assume that ${\omega }_1,{\omega }_2,\cdots ,{\omega }_n\in \mathbb{R}$. ${\omega }_1,{\omega }_2,\cdots ,{\omega }_n$ are called rationally independent if $k_1{\omega }_1+k_2{\omega }_2+\cdots +k_n{\omega }_n\ne 0$ for all $k_1,k_2,\cdots ,k_n\in Q\setminus \left\{0\right\}$, where Q is the set of all rational numbers. In this paper, we always assume ${\omega }_1,{\omega }_2,\cdots ,{\omega }_n$ are rationally independent and ${\omega }_1,{\omega }_2,\cdots ,{\omega }_n>0$. Let f be a periodic function from ${\mathbb{R}}^n$ into X. If the periodic function $f(t_1,t_2,\cdots ,t_n)$ in $t_1,t_2,\cdots ,t_n$ with the same periodic $2\pi$, then $F\left(t\right)=f(\frac{2\pi }{{\omega }_1}t,\frac{2\pi }{{\omega }_2}t,\cdots ,\frac{2\pi }{{\omega }_n}t)$ is said to be quasi-periodic. If we define $g(t_1,t_2,\cdots ,t_n)=f(\frac{2\pi }{{\omega }_1}{t}_1,\frac{2\pi }{{\omega }_2}{t}_2,\cdots ,\frac{2\pi }{{\omega }_n}{t}_n)$, then:

$$\begin{array}{cc}\hfill & g(t_1,t_2,\cdots ,t_k+{\omega }_k,\cdots ,t_n)\hfill \\ \hfill =& f\left(\frac{2\pi }{{\omega }_1}{t}_1,\frac{2\pi }{{\omega }_2}{t}_2,\cdots ,\frac{2\pi }{{\omega }_k}(t_k+{\omega }_k),\cdots ,\frac{2\pi }{{\omega }_n}{t}_n\right)\hfill \\ \hfill =& f\left(\frac{2\pi }{{\omega }_1}{t}_1,\frac{2\pi }{{\omega }_2}{t}_2,\cdots ,\frac{2\pi }{{\omega }_k}{t}_k+2\pi ,\cdots ,\frac{2\pi }{{\omega }_n}{t}_n\right)\hfill \\ \hfill =& f\left(\frac{2\pi }{{\omega }_1}{t}_1,\frac{2\pi }{{\omega }_2}{t}_2,\cdots ,\frac{2\pi }{{\omega }_k}{t}_k,\cdots ,\frac{2\pi }{{\omega }_n}{t}_n\right)\hfill \\ \hfill =& g(t_1,t_2,\cdots ,t_k,\cdots ,t_n)\hfill \end{array}$$

and we can define the same quasi-periodic function by g, that is $F\left(t\right)=g(t,t,\cdots ,t)$. A continuous function $f:{\mathbb{R}}^{+}\to X$ is called asymptotically quasi-periodic if it admits a decomposition $f\left(t\right)=g\left(t\right)+h\left(t\right)$, where $g:\mathbb{R}\to X$ is a quasi-periodic function and $h:{\mathbb{R}}^{+}\to X$ is a continuous function with ${lim}_{t\to +\infty }\parallel h\left(t\right)\parallel =0$. Let $\omega >0$. A bounded continuous function $f:{\mathbb{R}}^{+}\to X$ is called the $\omega$-periodic limit if there is a function $g:{\mathbb{R}}^{+}\to X$ such that ${lim}_{n\to \infty }f(t+n\omega )=g\left(t\right)$. The set of $\omega$-periodic limit functions will be denoted by $P_{\omega }L({\mathbb{R}}^{+},X)$.

For the sake of simplicity, we establish the concept by a function with two variables.

Definition 1.

Let $f\in C_b({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$. If $g(t,s)={lim}_{n\to \infty }f(t+n{\omega }_1,s)$ and $h(t,s)={lim}_{n\to \infty }f(t,s+n{\omega }_2)$ are well defined for each pair $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$, where $n\in \mathbb{N}$, then f is said to be the $({\omega }_1,{\omega }_2)$-periodic limit.

Hypothesis 1 (H1).

For each $t\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in {\mathbb{R}}^{+}$.

Hypothesis 2 (H2).

For each $s\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t,s+n{\omega }_2)=h(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$.

Definition 2.

Let f be a $({\omega }_1,{\omega }_2)$-periodic limit function. Assume H1 and H2 hold, then $F\left(t\right)=f(t,t)$ is said to be the $({\omega }_1,{\omega }_2)$-quasi-periodic limit. The collection of all $({\omega }_1,{\omega }_2)$-periodic limit functions that satisfies H1 and H2 (respectively, $({\omega }_1,{\omega }_2)$-quasi-periodic limit functions) will be denoted by $P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$$\left(QP{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+},X)\right)$.

Remark 1.

The functions g and h in Definition 2 is measurable, but not necessarily continuous.

Now, we present some examples of $({\omega }_1,{\omega }_2)$-quasi-periodic limit functions.

Example 1.

Let $f(t,s)=f_1\left(t\right)+f_2\left(s\right)$, where $f_1\in P_{{\omega }_1}L({\mathbb{R}}^{+},X)$ and $f_2\in P_{{\omega }_2}L({\mathbb{R}}^{+},X)$. By the definition of the ω-periodic limit function, we have ${lim}_{n\to \infty }{f}_1(t+n{\omega }_1)=g_1\left(t\right)$ and ${lim}_{n\to \infty }{f}_2(t+n{\omega }_2)=g_2\left(t\right)$. Let $g(t,s)=g_1\left(t\right)+f_2\left(s\right)$ and $h(t,s)=f_1\left(t\right)+g_2\left(s\right)$. Thus, for each $t\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in {\mathbb{R}}^{+}$, and for each $s\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t,s+n{\omega }_2)=h(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$. Then, $F\left(t\right)=f(t,t)=f_1\left(t\right)+f_2\left(t\right)$ is the $({\omega }_1,{\omega }_2)$-quasi-periodic limit.

Example 2.

Let $f(t,s)=f_1\left(t\right)f_2\left(s\right)$, where $f_1\in P_{{\omega }_1}L({\mathbb{R}}^{+},\mathbb{C})$ and $f_2\in P_{{\omega }_2}L({\mathbb{R}}^{+},X)$. Then, $F\left(t\right)=f(t,t)=f_1\left(t\right)f_2\left(t\right)$ is the $({\omega }_1,{\omega }_2)$-quasi-periodic limit.

Example 3.

For $n\ge 0$, define a function f on $[n{\omega }_1,(n+1){\omega }_1]\times [m{\omega }_2\times (m+1){\omega }_2]$ as follows:

$$f(t,s)=\left\{\begin{array}{cc}1,\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ f_{n,m}^1(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3},n{\omega }_1+\frac{{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ f_{n,m}^2(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{2{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ f_{n,m}^3(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3},m{\omega }_2+\frac{{\omega }_2}{3}],\hfill \\ f_{n,m}^4(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{2{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3}],\hfill \\ f_{n,m}^5(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3},n{\omega }_1+\frac{{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3},m{\omega }_2+\frac{{\omega }_2}{3}],\hfill \\ f_{n,m}^6(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{2{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3}]\times [m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3},m{\omega }_2+\frac{{\omega }_2}{3}],\hfill \\ f_{n,m}^7(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3},n{\omega }_1+\frac{{\omega }_1}{3}]\times [m{\omega }_2+\frac{2{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3}],\hfill \\ f_{n,m}^8(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{2{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3}]\times [m{\omega }_2+\frac{2{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3}],\hfill \\ 0,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where $f_{n,m}^1(t,s)=\frac{n+3}{{\omega }_1}[t-(n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3})]$, $f_{n,m}^2(t,s)=-\frac{n+3}{{\omega }_1}[t-(n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3})]$, $f_{n,m}^3(t,s)=\frac{m+3}{{\omega }_2}[s-(m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3})]$, $f_{n,m}^4(t,s)=-\frac{m+3}{{\omega }_2}[s-(m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3})]$, $f_{n,m}^5(t,s)=\frac{\frac{n+3}{{\omega }_1}[t-(n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3})]}{\frac{{\omega }_2}{m+3}}[s-(m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3})]$, $f_{n,m}^6(t,s)=-\frac{\frac{n+3}{{\omega }_1}[t-(n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3})]}{\frac{{\omega }_2}{m+3}}[s-(m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3})]$, $f_{n,m}^7(t,s)=-\frac{\frac{n+3}{{\omega }_1}[t-(n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3})]}{\frac{{\omega }_2}{m+3}}[s-(m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3})]$, $f_{n,m}^8(t,s)=\frac{\frac{n+3}{{\omega }_1}[t-(n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3})]}{\frac{{\omega }_2}{m+3}}[s-(m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3})]$. The graph of the function f in each rectangle $[n{\omega }_1,(n+1){\omega }_1]\times [m{\omega }_1\times (m+1){\omega }_2]$ $(n\ge 1)$ consists of ten parts, and $f(t,s):{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to [0,1]$ is continuous. If we define the function g and h on $[n{\omega }_1,(n+1){\omega }_1]\times [m{\omega }_2\times (m+1){\omega }_2]$ by:

$$g(t,s)=\left\{\begin{array}{cc}1,\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ f_{n,m}^3(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3}-\frac{{\omega }_2}{m+3},m{\omega }_2+\frac{{\omega }_2}{3}],\hfill \\ f_{n,m}^4(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{2{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}+\frac{{\omega }_2}{m+3}],\hfill \\ 0,\hfill & \mathit{otherwise}\hfill \end{array}\right.$$

and:

$$h(t,s)=\left\{\begin{array}{cc}1,\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ f_{n,m}^1(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{{\omega }_1}{3}-\frac{{\omega }_1}{n+3},n{\omega }_1+\frac{{\omega }_1}{3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ f_{n,m}^2(t,s),\hfill & (t,s)\in [n{\omega }_1+\frac{2{\omega }_1}{3},n{\omega }_1+\frac{2{\omega }_1}{3}+\frac{{\omega }_1}{n+3}]\times [m{\omega }_2+\frac{{\omega }_2}{3},m{\omega }_2+\frac{2{\omega }_2}{3}],\hfill \\ 0,\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Then, for each $t\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in {\mathbb{R}}^{+}$, and for each $s\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t,s+n{\omega }_2)=h(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$. Thus, $F\left(t\right)=f(t,t)$ is the $({\omega }_1,{\omega }_2)$-quasi-periodic limit.

Next, we present the following properties of $({\omega }_1,{\omega }_2)$-quasi-periodic limit functions.

Proposition 1.

Let $F,F_1$ and $F_2$ be the $({\omega }_1,{\omega }_2)$-quasi-periodic limit. Assume $g(t,s)={lim}_{n\to \infty }f(t+n{\omega }_1,s)$, $h(t,s)={lim}_{n\to \infty }f(t,s+n{\omega }_2)$ are well defined for each pair $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$ and $F\left(t\right)=f(t,t)$. Then, the following statements are true:

(1) 

$F_1+F_2$ is the $({\omega }_1,{\omega }_2)$-quasi-periodic limit;

(2) 

$cF$ is the $({\omega }_1,{\omega }_2)$-quasi-periodic limit for any $c\in \mathbb{C}$;

(3) 

$g(t+{\omega }_1,s)=g(t,s)$, $h(t,s+{\omega }_2)=h(t,s)$ for each pair $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$;

(4) 

g and h are bounded on ${\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$; moreover, ${\parallel g\parallel }_{\infty }\le {\parallel f\parallel }_{\infty }$ and ${\parallel h\parallel }_{\infty }\le {\parallel f\parallel }_{\infty }$;

(5) 

F is bounded on ${\mathbb{R}}^{+}$; moreover, ${\parallel F\parallel }_{\infty }\le {\parallel f\parallel }_{\infty }$.

Proposition 2.

Let f be a $({\omega }_1,{\omega }_2)$-periodic limit function.

(1) 

Assume H1 holds, then $k(t,s)={lim}_{n\to \infty }g(t,s+n{\omega }_2)$ is well defined for each pair $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$.

(2) 

Assume H2 holds, then $\overline{k}(t,s)={lim}_{n\to \infty }h(t+n{\omega }_1,s)$ is well defined for each pair $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$.

(3) 

Assume H1 and H2 hold, then $k(t,s)=\overline{k}(t,s)$.

(4) 

Assume H1 and H2 hold, then $k(t+{\omega }_1,s)=k(t,s)=k(t,s+{\omega }_2)$.

Proof. 

(1) We only need to show that ${\left\{g(t,s+n{\omega }_2)\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence for each pair $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$. Let $\epsilon >0$. By Hypothesis of H1, for fixed $t\in {\mathbb{R}}^{+}$, there exists $N_1\in \mathbb{N}$ such that $\parallel f(t+p{\omega }_1,s)-g(t,s)\parallel <\frac{\epsilon }{4}$ uniformly for $s\in {\mathbb{R}}^{+}$ when $p\ge N_1$. For the above t, choose $p\ge N_1$, and fix $s\in {\mathbb{R}}^{+}$. By Definition 1, there exists $N_2\in \mathbb{N}$ such that $\parallel f(t+p{\omega }_1,s+n{\omega }_2)-h(t+p{\omega }_1,s)\parallel <\frac{\epsilon }{4}$ when $n\ge N_2$. Therefore,

$$\begin{array}{cc}\hfill & \parallel g(t,s+n{\omega }_2)-g(t,s+m{\omega }_2)\parallel \hfill \\ \hfill \le & \parallel g(t,s+n{\omega }_2)-f(t+p{\omega }_1,s+n{\omega }_2)\parallel +\parallel f(t+p{\omega }_1,s+n{\omega }_2)-h(t+p{\omega }_1,s)\parallel \hfill \\ \hfill & +\parallel h(t+p{\omega }_1,s)-f(t+p{\omega }_1,s+m{\omega }_2)\parallel +\parallel f(t+p{\omega }_1,s+m{\omega }_2)-g(t,s+m{\omega }_2)\parallel \hfill \\ \hfill <& \epsilon \hfill \end{array}$$

when $m,n\ge N_2$.

(2) In a similar way as (1), one can show (2).

(3) Let $\epsilon >0$, and fix $(t,s)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$. By Hypothesis H1, there exists $N_1\in \mathbb{N}$ such that:

$$\parallel f(t+n{\omega }_1,s^{\prime })-g(t,s^{\prime })\parallel <\frac{\epsilon }{4}$$

(2)

uniformly for $s^{\prime }\in {\mathbb{R}}^{+}$ when $n\ge N_1$. By Hypothesis H2, there exists $N_2\in \mathbb{N}$ such that:

$$\parallel f(t^{\prime },s+n{\omega }_2)-h(t^{\prime },s)\parallel <\frac{\epsilon }{4}$$

(3)

uniformly for $t^{\prime }\in {\mathbb{R}}^{+}$ when $n\ge N_2$. By the conclusion of (1), there exists $N_3\in \mathbb{N}$ such that:

$$\parallel g(t,s+n{\omega }_2)-k(t,s)\parallel <\frac{\epsilon }{4}$$

(4)

when $n\ge N_3$. By the conclusion of (2), there exists $N_4\in \mathbb{N}$ such that:

$$\parallel h(t+n{\omega }_1,s)-\overline{k}(t,s)\parallel <\frac{\epsilon }{4}$$

(5)

when $n\ge N_4$. Select $N_5=max\{N_1,N_2,N_3,N_4\}$. (2), (3), (4), and (5) imply:

$$\begin{array}{cc}\hfill & \parallel k(t,s)-\overline{k}(t,s)\parallel \hfill \\ \hfill \le & \parallel k(t,s)-g(t,s+N_5{\omega }_2)\parallel +\parallel g(t,s+N_5{\omega }_2)-f(t+N_5{\omega }_1,s+N_5{\omega }_2)\parallel \hfill \\ \hfill & +\parallel f(t+N_5{\omega }_1,s+N_5{\omega }_2)-h(t+N_5{\omega }_1,s)\parallel +\parallel h(t+N_5{\omega }_1,s)-\overline{k}(t,s)\parallel \hfill \\ \hfill <& \epsilon ,\hfill \end{array}$$

which shows $k(t,s)=\overline{k}(t,s)$.

(4) $k(t,s)={lim}_{n\to \infty }g(t,s+n{\omega }_2)={lim}_{n\to \infty }g(t+{\omega }_1,s+n{\omega }_2)=k(t+{\omega }_1,s)$. Similarly, $k(t,s)=k(t,s+{\omega }_2)$. □

Hypothesis 3 (H3).

f is uniformly continuous.

Proposition 3.

Let f be a $({\omega }_1,{\omega }_2)$-periodic limit function. Assume H3 holds, then g and h are uniformly continuous.

Proof. 

For $\epsilon >0$ given, there exists $\delta >0$ such that $\parallel f(t,s)-f(t^{\prime },s^{\prime })\parallel <\frac{\epsilon }{3}$ when $|t-t^{\prime }|<\delta$, $|s-s^{\prime }|<\delta$ for $t,t^{\prime },s,s^{\prime }\in {\mathbb{R}}^{+}$. Choose $t,t^{\prime },s,s^{\prime }\in {\mathbb{R}}^{+}$ such that $|t-t^{\prime }|<\delta$, $|s-s^{\prime }|<\delta$, then there exists $N_1\in \mathbb{N}$ such that $\parallel g(t,s)-f(t+n{\omega }_1,s)\parallel <\frac{\epsilon }{3}$ when $n\ge N_1$, and there exists $N_2\in \mathbb{N}$ such that $\parallel g(t^{\prime },s^{\prime })-f(t^{\prime }+n{\omega }_1,s^{\prime })\parallel <\frac{\epsilon }{3}$ when $n\ge N_2$. Choose $N_3=max\{N_1,N_2\}$. Then, one has:

$$\begin{array}{cc}\hfill & \parallel g(t,s)-g(t^{\prime },s^{\prime })\parallel \hfill \\ \hfill \le & \parallel g(t,s)-f(t+N_3{\omega }_1,s)\parallel +\parallel f(t+N_3{\omega }_1,s)-f(t^{\prime }+N_3{\omega }_1,s^{\prime })\parallel +\parallel f(t^{\prime }+N_3{\omega }_1,s^{\prime })-g(t^{\prime },s^{\prime })\parallel \hfill \\ \hfill <& \epsilon .\hfill \end{array}$$

Therefore, g is uniformly continuous. Similarly, h is uniformly continuous. □

In the following propositions, if F is a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function, then $f,g,h$ are defined in Definition 2, and k is defined in Proposition 2.

Proposition 4.

Let F be a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function. Assume H3 holds, then k is uniformly continuous.

Proof. 

Let $\epsilon >0$. Since f is uniformly continuous, g is uniformly continuous by Proposition 3. Therefore, there exists $\delta >0$ such that $\parallel g(t,s)-g(t^{\prime },s^{\prime })\parallel <\frac{\epsilon }{3}$ when $|t-t^{\prime }|<\delta$, $|s-s^{\prime }|<\delta$ for $t,t^{\prime },s,s^{\prime }\in {\mathbb{R}}^{+}$. Choose $t,t^{\prime },s,s^{\prime }\in {\mathbb{R}}^{+}$ such that $|t-t^{\prime }|<\delta$, $|s-s^{\prime }|<\delta$, then by Proposition 2 (1), there exists $N_1\in \mathbb{N}$ such that $\parallel k(t,s)-g(t,s+n{\omega }_2)\parallel <\frac{\epsilon }{3}$ when $n\ge N_1$, and there exists $N_2\in \mathbb{N}$ such that $\parallel k(t^{\prime },s^{\prime })-g(t^{\prime },s^{\prime }+n{\omega }_2)\parallel <\frac{\epsilon }{3}$ when $n\ge N_2$. Choose $N_3=max\{N_1,N_2\}$. Then, one has:

$$\begin{array}{cc}\hfill & \parallel k(t,s)-k(t^{\prime },s^{\prime })\parallel \hfill \\ \hfill \le & \parallel k(t,s)-g(t,s+N_3{\omega }_2)\parallel +\parallel g(t,s+N_3{\omega }_2)-g(t^{\prime },s^{\prime }+N_3{\omega }_2)\parallel +\parallel g(t^{\prime },s^{\prime }+N_3{\omega }_2)-k(t^{\prime },s^{\prime })\parallel \hfill \\ \hfill <& \epsilon .\hfill \end{array}$$

Thus, k is uniformly continuous. □

Proposition 5.

Let F be a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function. Assume H3 holds, and denote $r(t,s)=f(t,s)-k(t,s)$, then $r\in C_0({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$, that is for any $\epsilon >0$, there exists $M>0$ such that $\parallel r(t,s)\parallel <\epsilon$ when $t>M$, $s>M$.

Proof. 

It is equivalent to show that for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $\parallel f(t+n{\omega }_1,s+m{\omega }_2)-k(t+n{\omega }_1,s+m{\omega }_2)\parallel <\epsilon$ uniformly for $t\in [0,{\omega }_1]$, $s\in [0,{\omega }_2]$ when $n,m\ge N$. By Proposition 2 (4), one has $k(t,s)=k(t+n{\omega }_1,s+m{\omega }_2)$ for any $m,n\in \mathbb{N}$. Note that:

$$\begin{array}{cc}\hfill & f(t+n{\omega }_1,s+m{\omega }_2)-k(t+n{\omega }_1,s+m{\omega }_2)\hfill \\ \hfill =& f(t+n{\omega }_1,s+m{\omega }_2)-g(t,s+m{\omega }_2)+g(t,s+m{\omega }_2)-k(t,s).\hfill \end{array}$$

Let $\epsilon >0$. By Proposition 3 and Proposition 4, g and k are uniformly continuous. Thus, there exists $\delta >0$ such that $\parallel f(t,s)-f(t^{\prime },s^{\prime })\parallel <\epsilon$, $\parallel g(t,s)-g(t^{\prime },s^{\prime })\parallel <\epsilon$, $\parallel k(t,s)-k(t^{\prime },s^{\prime })\parallel <\epsilon$ when $|t-t^{\prime }|<\delta$, $|s-s^{\prime }|<\delta$ for $t,t^{\prime },s,s^{\prime }\in {\mathbb{R}}^{+}$.

Divide $[0,{\omega }_1]$ into p equal intervals such that $\frac{{\omega }_1}{p}<\delta$, $p\in \mathbb{N}$. Define $t_0=0,t_1=\frac{{\omega }_1}{p},t_2=\frac{2{\omega }_1}{p},\cdots ,t_{p-1}=\frac{(p-1){\omega }_1}{p},t_p={\omega }_1$. Then, there exists $N_1\in \mathbb{N}$ such that $\parallel f(t_i+n{\omega }_1,s+m{\omega }_2)-g(t_i,s+m{\omega }_2)\parallel <\epsilon$ uniformly for $s\in [0,{\omega }_2]$ and $m\in \mathbb{N}$, $i=0,1,2,\cdots ,p$ when $n\ge N_1$. Choose any $t\in [0,{\omega }_1]$, one can pick $t_{i_0}\in \left\{t_i\right\}$ such that $|t-t_{i_0}|<\delta$. Then, one has:

$$\begin{array}{cc}\hfill & \parallel f(t+n{\omega }_1,s+m{\omega }_2)-g(t,s+m{\omega }_2)\parallel \hfill \\ \hfill \le & \parallel f(t+n{\omega }_1,s+m{\omega }_2)-f(t_{i_0}+n{\omega }_1,s+m{\omega }_2)\parallel +\parallel f(t_{i_0}+n{\omega }_1,s+m{\omega }_2)-g(t_{i_0},s+m{\omega }_2)\parallel \hfill \\ \hfill & +\parallel g(t_{i_0},s+m{\omega }_2)-g(t,s+m{\omega }_2)\parallel \hfill \\ \hfill <& 3\epsilon \hfill \end{array}$$

when $n\ge N_1$ uniformly for $s\in [0,{\omega }_2]$ and $m\in \mathbb{N}$.

Fix $t\in {\mathbb{R}}^{+}$. In a similar way, by dividing the interval $[0,{\omega }_2]$ and using the uniform continuousness of $g,k$, one can show there exists $N_2\in \mathbb{N}$ such that:

$$\parallel g(t,s+m{\omega }_2)-k(t,s)\parallel <3\epsilon$$

(6)

when $m\ge N_2$ uniformly for $s\in [0,{\omega }_2]$.

In a similar way again, by dividing the interval $[0,{\omega }_1]$ and using (6), one can show there exists $N_3\in \mathbb{N}$ such that $\parallel g(t,s+m{\omega }_2)-k(t,s)\parallel <5\epsilon$ when $m\ge N_3$ uniformly for $s\in [0,{\omega }_2]$ and $t\in [0,{\omega }_1]$. Note that $g(t+n{\omega }_1,s)=g(t,s)$, $k(t+n{\omega }_1,s)=k(t,s)$. Therefore, for any $t\in {\mathbb{R}}^{+}$, one has $\parallel g(t,s+m{\omega }_2)-k(t,s)\parallel <5\epsilon$ when $m\ge N_3$ uniformly for $s\in [0,{\omega }_2]$.

Choose $N_4=max\{N_1,N_3\}$; one has:

$$\begin{array}{cc}\hfill & \parallel f(t+n{\omega }_1,s+m{\omega }_2)-k(t+n{\omega }_1,s+m{\omega }_2)\parallel \hfill \\ \hfill \le & \parallel f(t+n{\omega }_1,s+m{\omega }_2)-g(t,s+m{\omega }_2)\parallel +\parallel g(t,s+m{\omega }_2)-k(t,s)\parallel \hfill \\ \hfill <& 8\epsilon \hfill \end{array}$$

when $n,m\ge N_4$ uniformly for $t\in [0,{\omega }_1]$ and $s\in [0,{\omega }_2]$. □

Proposition 6.

Let F be a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function. Assume H3 holds, then F is asymptotically quasi-periodic.

Proof. 

Denote $K\left(t\right)=k(t,t)$, $R\left(t\right)=r(t,t)$. By Proposition 5, $F\left(t\right)=K\left(t\right)+R\left(t\right)$, $R\in C_0({\mathbb{R}}^{+},X)$. By Proposition 2 (4) and Proposition 4, K is quasi-periodic. Therefore, F is asymptotically quasi-periodic. □

Let us introduce the following conditions.

Hypothesis 4 (H4).

${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$.

Hypothesis 5 (H5).

${lim}_{n\to \infty }f(t,s+n{\omega }_2)=h(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$.

Proposition 7.

Let f be a $({\omega }_1,{\omega }_2)$-periodic limit function. Assume H4 and H5 hold, then:

(1) 

$k(t,s)={lim}_{n\to \infty }g(t,s+n{\omega }_2)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$;

(2) 

$\overline{k}(t,s)={lim}_{n\to \infty }h(t+n{\omega }_1,s)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$;

(3) 

$k(t,s)=\overline{k}(t,s)$;

(4) 

$k(t+{\omega }_1,s)=k(t,s)=k(t,s+{\omega }_2)$.

Proof. 

(1) We only need to show that ${\left\{g(t,s+n{\omega }_2)\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$. Let $\epsilon >0$. By Hypothesis H4, there exists $N_1\in \mathbb{N}$ such that $\parallel f(t+p{\omega }_1,s)-g(t,s)\parallel <\frac{\epsilon }{4}$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$ when $p\ge N_1$. By Hypothesis H5, there exists $N_2\in \mathbb{N}$ such that $\parallel f(t,s+n{\omega }_2)-h(t,s)\parallel <\frac{\epsilon }{4}$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$ when $n\ge N_2$.

Therefore,

$$\begin{array}{cc}\hfill & \parallel g(t,s+n{\omega }_2)-g(t,s+m{\omega }_2)\parallel \hfill \\ \hfill \le & \parallel g(t,s+n{\omega }_2)-f(t+p{\omega }_1,s+n{\omega }_2)\parallel +\parallel f(t+p{\omega }_1,s+n{\omega }_2)-h(t+p{\omega }_1,s)\parallel \hfill \\ \hfill & +\parallel h(t+p{\omega }_1,s)-f(t+p{\omega }_1,s+m{\omega }_2)\parallel +\parallel f(t+p{\omega }_1,s+m{\omega }_2)-g(t,s+m{\omega }_2)\parallel \hfill \\ \hfill <& \epsilon \hfill \end{array}$$

uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$ when $m,n\ge N_2$.

(2) In a similar way as (1), one can show (2).

(3) and (4) are the conclusions of Proposition 2 (3) (4). □

Proposition 8.

Let F be a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function. Assume H4 and H5 hold, then F is asymptotically quasi-periodic.

Proof. 

Denote $r(t,s)=f(t,s)-k(t,s)$; we first show that $r\in C_0({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$.

We only need to show that for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $\parallel f(t+n{\omega }_1,s+m{\omega }_2)-k(t+n{\omega }_1,s+m{\omega }_2)\parallel <\epsilon$ uniformly for $t\in [0,{\omega }_1]$, $s\in [0,{\omega }_2]$ when $n,m\ge N$. By Proposition 7 (4), one has $k(t,s)=k(t+n{\omega }_1,s+m{\omega }_2)$ for any $m,n\in \mathbb{N}$.

By Hypothesis H4, there exists $N_1\in \mathbb{N}$ such that $\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel <\frac{\epsilon }{2}$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$ when $n\ge N_1$. By Proposition 7, there exists $N_2\in \mathbb{N}$ such that $\parallel g(t,s+m{\omega }_2)-k(t,s)\parallel <\frac{\epsilon }{2}$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$ when $m\ge N_2$. Choose $N=max\{N_1,N_2\}$. Therefore,

$$\begin{array}{cc}\hfill & \parallel f(t+n{\omega }_1,s+m{\omega }_2)-k(t+n{\omega }_1,s+m{\omega }_2)\parallel \hfill \\ \hfill \le & \parallel f(t+n{\omega }_1,s+m{\omega }_2)-g(t,s+m{\omega }_2)\parallel +\parallel g(t,s+m{\omega }_2)-k(t,s)\parallel \hfill \\ \hfill <& \epsilon \hfill \end{array}$$

uniformly for $t\in [0,{\omega }_1]$, $s\in [0,{\omega }_2]$ when $n,m\ge N$.

Next, we show that k is continuous. Take any $t_0,s_0\in {\mathbb{R}}^{+}$. Since f is continuous, there exists $\delta >0$ such that:

$$\parallel f(t_0+N_1{\omega }_1,s_0+N_2{\omega }_2)-f(t^{\prime }+N_1{\omega }_1,s^{\prime }+N_2{\omega }_2)\parallel <\epsilon$$

when $|t_0-t^{\prime }|<\delta$, $|s_0-s^{\prime }|<\delta$. Therefore,

$$\begin{array}{cc}\hfill & \parallel k(t_0,s_0)-k(t^{\prime },s^{\prime })\parallel \hfill \\ \hfill =& \parallel k(t_0,s_0)-g(t_0,s_0+N_2{\omega }_2)\parallel +\parallel g(t_0,s_0+N_2{\omega }_2)-f(t_0+N_1{\omega }_1,s_0+N_2{\omega }_2)\parallel \hfill \\ \hfill & +\parallel f(t_0+N_1{\omega }_1,s_0+N_2{\omega }_2)-f(t^{\prime }+N_1{\omega }_1,s^{\prime }+N_2{\omega }_2)\parallel \hfill \\ \hfill & +\parallel f(t^{\prime }+N_1{\omega }_1,s^{\prime }+N_2{\omega }_2)-g(t^{\prime },s^{\prime }+N_2{\omega }_2)\parallel +\parallel g(t^{\prime },s^{\prime }+N_2{\omega }_2)-k(t^{\prime },s^{\prime })\parallel \hfill \\ \hfill <& 3\epsilon \hfill \end{array}$$

when $|t_0-t^{\prime }|<\delta$, $|s_0-s^{\prime }|<\delta$.

Denote $K\left(t\right)=k(t,t)$, $R\left(t\right)=r(t,t)$. Then, $F\left(t\right)=K\left(t\right)+R\left(t\right)$, $R\in C_0({\mathbb{R}}^{+},X)$, and K is quasi-periodic. Therefore, F is asymptotically quasi-periodic. □

The collection of all $({\omega }_1,{\omega }_2)$-periodic limit functions that satisfies H4 and H5 will be denoted by $AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$.

Theorem 1.

$AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ is a Banach space.

Proof. 

Let $\left\{f_k\right\}\subset AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ such that:

$$\underset{k\to \infty }{lim}{f}_k(t,s)=f(t,s)$$

uniformly in $t,s\in {\mathbb{R}}^{+}$.

Since $\left\{f_k\right\}\subset AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$, for each $k\in \mathbb{N}$, one has:

$$\underset{n\to \infty }{lim}{f}_k(t+n{\omega }_1,s)=g_k(t,s)$$

uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$.

At the same time, for each $k\in \mathbb{N}$, one has:

$$\underset{n\to \infty }{lim}{f}_k(t,s+n{\omega }_2)=h_k(t,s)$$

uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$.

Then, the sequence of $\left\{g_k(t,s)\right\}$ is a Cauchy sequence in X uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$ because the following inequality:

$$\parallel g_k(t,s)-g_h(t,s)\parallel \le \parallel g_k(t,s)-f_k(t+n{\omega }_1,s)\parallel +\parallel f_k(t+n{\omega }_1,s)-f_h(t+n{\omega }_1,s)\parallel +\parallel f_h(t+n{\omega }_1,s)-g_h(t,s)\parallel .$$

Thus, the sequence $\left\{g_k(t,s)\right\}$ converges to a function $g(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$.

Now, we only need to show:

$$\underset{n\to \infty }{lim}f(t+n{\omega }_1,s)=g(t,s)$$

uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$. However, we can get it from the following inequality.

$$\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel \le \parallel f(t+n{\omega }_1,s)-f_k(t+n{\omega }_1,s)\parallel +\parallel f_k(t+n{\omega }_1,s)-g_k(t,s)\parallel +\parallel g_k(t,s)-g(t,s)\parallel .$$

In a similar way, we can show that there exists a function h such that:

$$\underset{n\to \infty }{lim}f(t,s+n{\omega }_2)=h(t,s)$$

uniformly for $t\in {\mathbb{R}}^{+}$ and $s\in {\mathbb{R}}^{+}$. Therefore, $f\in AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$. □

3. Existence of Asymptotically Quasi-Periodic Solutions of Abstract Cauchy Problems

In this section, we first introduce the following definition:

Definition 3.

A function $x\in C_b({\mathbb{R}}^{+},X)$ is said to be a mild solution of Problem (1) if:

$$x\left(t\right)=T\left(t\right)x_0+{\int }_0^{t}T(t-s)F\left(s\right)ds,t\in {\mathbb{R}}^{+},$$

where ${\left(T\left(t\right)\right)}_{t\ge 0}$ is an exponentially stable $C_0$-semigroup.

Lemma 1.

Let F be a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function and ${\left(T\left(t\right)\right)}_{t\ge 0}$ be an exponentially stable $C_0$-semigroup, then $U\left(t\right)={\int }_0^{t}T(t-s)F\left(s\right)ds$ is asymptotically quasi-periodic.

Proof. 

By the definition of the $({\omega }_1,{\omega }_2)$-quasi-periodic limit function, there exists a $f\in C_b({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ such that for each $t\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in {\mathbb{R}}^{+}$, for each $s\in {\mathbb{R}}^{+}$, ${lim}_{n\to \infty }f(t,s+n{\omega }_2)=h(t,s)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $F\left(t\right)=f(t,t)$. Assume $\parallel f(t,s)\parallel \le K$ for $t,s\in {\mathbb{R}}^{+}$. Then, $\parallel g(t,s)\parallel \le K$ and $\parallel h(t,s)\parallel \le K$ for $t,s\in {\mathbb{R}}^{+}$.

Define:

$$u(x,y)=\left\{\begin{array}{c}{\int }_0^{x}T(x-s)f(s,s+y-x)ds,y\ge x,x,y\in {\mathbb{R}}^{+},\hfill \\ {\int }_0^{y}T(y-s)f(s+x-y,s)ds,x\ge y,x,y\in {\mathbb{R}}^{+}.\hfill \end{array}\right.$$

(7)

Clearly, u is continuous and bounded.

Note that $U\left(t\right)=u(t,t)={\int }_0^{t}T(t-s)f(s,s)ds={\int }_0^{t}T(t-s)F\left(s\right)ds$. By Proposition 8, we only need to show that H4 and H5 hold for u. Next, we only show that H4 holds for u because the case for H5 is similar.

Define $f(t,s)=f(t,0)$ and $g(t,s)=g(t,0)$ when $s<0$. Then, we obtain ${lim}_{n\to \infty }f(t+n{\omega }_1,s)={lim}_{n\to \infty }f(t+n{\omega }_1,0)=g(t,0)=g(t,s)$ when $s<0$. Moreover, ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in \mathbb{R}$. Note that ${\int }_0^{y}T(y-s)f(s+x-y,s)ds={\int }_{x-y}^{x}T(x-s)f(s,s+y-x)ds$ when $x\ge y$. Then, one has:

$$u(x,y)=\left\{\begin{array}{c}{\int }_0^{x}T(x-s)f(s,s+y-x)ds,y\ge x,x,y\in {\mathbb{R}}^{+},\hfill \\ {\int }_0^{x}T(x-s)f(s,s+y-x)ds-{\int }_0^{x-y}T(x-s)f(s,s+y-x)ds,x\ge y,x,y\in {\mathbb{R}}^{+}.\hfill \end{array}\right.$$

Denote $u_1(x,y)={\int }_0^{x}T(x-s)f(s,s+y-x)ds$ and:

$$u_2(x,y)=\left\{\begin{array}{c}0,y\ge x,\hfill \\ {\int }_0^{x-y}T(x-s)f(s,s+y-x)ds,x\ge y,\hfill \end{array}\right.=\left\{\begin{array}{c}0,y\ge x,\hfill \\ {\int }_0^{x-y}T(x-s)f(s,0)ds,x\ge y.\hfill \end{array}\right.$$

To show ${lim}_{n\to \infty }u(x+n{\omega }_1,y)=v(x,y)$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$, we only need to show ${lim}_{n\to \infty }{u}_1(x+n{\omega }_1,y)=v_1(x,y)$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$ and ${lim}_{n\to \infty }{u}_2(x+n{\omega }_1,y)=v_2(x,y)$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

Step 1. Let $x\in {\mathbb{R}}^{+}$, $y\in {\mathbb{R}}^{+}$.

$$\begin{array}{cc}\hfill u_1(x+n{\omega }_1,y)=& {\int }_0^{x+n{\omega }_1}T(x+n{\omega }_1-s)f(s,s+y-x-n{\omega }_1)ds\hfill \\ \hfill =& {\int }_{-n{\omega }_1}^{x}T(x-s)f(s+n{\omega }_1,s+y-x)ds\hfill \\ \hfill =& {\int }_{-n{\omega }_1}^{0}T(x-s)f(s+n{\omega }_1,s+y-x)ds+{\int }_0^{x}T(x-s)f(s+n{\omega }_1,s+y-x)ds\hfill \\ \hfill =& {\int }_0^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,y-x-s)ds+{\int }_0^{x}T(x-s)f(s+n{\omega }_1,s+y-x)ds\hfill \\ \hfill =& I_1(x,y,n)+I_2(x,y,n).\hfill \end{array}$$

We discuss the terms $I_i(x,y,n)(i=1,2)$ separately. First, we show that $I_1(x,y,n)$ is a Cauchy sequence in X for each $x\in {\mathbb{R}}^{+}$ and each $y\in {\mathbb{R}}^{+}$.

For any $p\in \mathbb{N}$, $n\in \mathbb{N}$, one has:

$$\begin{array}{cc}\hfill & I_1(x,y,n+p)-I_1(x,y,n)\hfill \\ \hfill =& {\int }_0^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,y-x-s)ds\hfill \\ \hfill & -{\int }_0^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,y-x-s)ds\hfill \\ \hfill =& {\int }_{n{\omega }_1}^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,y-x-s)ds\hfill \\ \hfill & +{\int }_0^{n{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,y-x-s)-f(n{\omega }_1-s,y-x-s)]ds\hfill \\ \hfill =& I_3(x,y,n,p)+I_4(x,y,n,p).\hfill \end{array}$$

We see that:

$$\begin{array}{cc}\hfill \parallel I_3(x,y,n,p)\parallel \le & {\int }_{n{\omega }_1}^{(n+p){\omega }_1}\parallel T(x+s)\parallel \parallel f((n+p){\omega }_1-s,y-x-s)\parallel ds\hfill \\ \hfill \le & KM{\int }_{n{\omega }_1}^{(n+p){\omega }_1}{e}^{-r(x+s)}ds\hfill \\ \hfill \le & KM{\int }_{n{\omega }_1}^{\infty }{e}^{-r(x+s)}ds\hfill \\ \hfill \le & \frac{KM}{r}{e}^{-rn{\omega }_1}.\hfill \end{array}$$

Let $\epsilon >0$. We can choose $N_1\in \mathbb{N}$ such that $\frac{KM}{r}{e}^{-rn{\omega }_1}<\epsilon$ when $n\ge N_1$. Then, one gets $\parallel I_3(x,y,n,p)\parallel <\epsilon$ when $n\ge N_1$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

For $n\ge N_1$, we consider:

$$\begin{array}{cc}\hfill I_4(x,y,n,p)=& {\int }_0^{N_1{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,y-x-s)-f(n{\omega }_1-s,y-x-s)]ds\hfill \\ \hfill & +{\int }_{N_1{\omega }_1}^{n{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,y-x-s)-f(n{\omega }_1-s,y-x-s)]ds\hfill \\ \hfill =& I_5(x,y,n,p)+I_6(x,y,n,p).\hfill \end{array}$$

Now, we estimate the term $I_5(x,y,n,p)$.

$$\begin{array}{cc}\hfill & \parallel I_5(x,y,n,p)\parallel \hfill \\ \hfill \le & {\int }_0^{N_1{\omega }_1}\parallel T(x+s)\parallel \parallel f((n+p){\omega }_1-s,y-x-s)-f(n{\omega }_1-s,y-x-s)\parallel ds\hfill \\ \hfill =& {\int }_0^{N_1{\omega }_1}\parallel T(x+N_1{\omega }_1-s)\parallel \parallel f((n-N_1+p){\omega }_1+s,s-N_1{\omega }_1+y-x)\hfill \\ \hfill & -f((n-N_1){\omega }_1+s,s-N_1{\omega }_1+y-x)\parallel ds\hfill \\ \hfill \le & M{\int }_0^{N_1{\omega }_1}{e}^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1+p){\omega }_1+s,s-N_1{\omega }_1+y-x)-g(s,s-N_1{\omega }_1+y-x)\parallel ds\hfill \\ \hfill +& M{\int }_0^{N_1{\omega }_1}{e}^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1){\omega }_1+s,s-N_1{\omega }_1+y-x)-g(s,s-N_1{\omega }_1+y-x)\parallel ds.\hfill \end{array}$$

For each $s\in [0,N_1{\omega }_1]$, we have:

$$e^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1){\omega }_1+s,s-N_1{\omega }_1+y-x)-g(s,s-N_1{\omega }_1+y-x)\parallel \le 2K{e}^{-r(N_1{\omega }_1-s)}$$

and:

$${\int }_0^{N_1{\omega }_1}2K{e}^{-r(N_1{\omega }_1-s)}ds=\frac{2K}{r}(1-e^{-r{N}_1{\omega }_1}).$$

Since ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in \mathbb{R}$, for each $s\in [0,N_1{\omega }_1]$, one has:

$$e^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1){\omega }_1+s,s-N_1{\omega }_1+y-x)-g(s,s-N_1{\omega }_1+y-x)\parallel \to 0$$

as $n\to \infty$. By Lebesgue’s dominated convergence theorem, we obtain:

$$\underset{n\to \infty }{lim}{\int }_0^{N_1{\omega }_1}{e}^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1){\omega }_1+s,s-N_1{\omega }_1+y-x)-g(s,s-N_1{\omega }_1+y-x)\parallel ds=0.$$

(8)

Since ${lim}_{n\to \infty }f(t+n{\omega }_1,s)=g(t,s)$ uniformly for $s\in \mathbb{R}$, (8) holds uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$. Moreover,

$$\underset{n\to \infty }{lim}{\int }_0^{N_1{\omega }_1}{e}^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1+p){\omega }_1+s,s-N_1{\omega }_1+y-x)-g(s,s-N_1{\omega }_1+y-x)\parallel ds=0$$

uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$. Thus, we can select $N_2\in \mathbb{N}$ $(N_2>N_1)$ such that $\parallel I_5(x,y,n,p)\parallel <\epsilon$ when $n\ge N_2$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

Next, we estimate the term $I_6(x,y,n,p)$.

$$\begin{array}{cc}\hfill \parallel I_6(x,y,n,p)\parallel \le & {\int }_{N_1{\omega }_1}^{n{\omega }_1}\parallel T(x+s)\parallel \parallel f((n+p){\omega }_1-s,y-x-s)-f(n{\omega }_1-s,y-x-s)\parallel ds\hfill \\ \hfill \le & 2KM{\int }_{N_1{\omega }_1}^{n{\omega }_1}{e}^{-r(x+s)}ds\hfill \\ \hfill \le & 2KM{\int }_{N_1{\omega }_1}^{\infty }{e}^{-r(x+s)}ds\hfill \\ \hfill \le & \frac{2KM}{r}{e}^{-r{N}_1{\omega }_1}\hfill \\ \hfill <& 2\epsilon \hfill \end{array}$$

uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

Thus, $\parallel I_1(x,y,n+p)-I_1(x,y,n)\parallel \le \parallel I_3(x,y,n,p)\parallel +\parallel I_5(x,y,n,p)\parallel +\parallel I_6(x,y,n,p)\parallel <4\epsilon$ when $n\ge N_2$. This shows that $I_1(x,y,n)$ is a Cauchy sequence, and we denote $l_1(x,y)={lim}_{n\to \infty }{I}_1(x,y,n)$. Besides, from the proof, we also know that ${lim}_{n\to \infty }{I}_1(x,y,n)=l_1(x,y)$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

Finally, we consider the term $I_2(x,y,n)$. Note that ${\int }_0^{x}T(x-s)g(s,s+y-x)ds$ is well defined for each $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$. For $m{\omega }_1\le x<(m+1){\omega }_1$, then one has:

$$\begin{array}{cc}\hfill & \parallel I_2(x,y,n)-{\int }_0^{x}T(x-s)g(s,s+y-x)ds\parallel \hfill \\ \hfill \le & {\int }_0^x\parallel T(x-s)\parallel \parallel f(s+n{\omega }_1,s+y-x)-g(s,s+y-x)\parallel ds\hfill \\ \hfill \le & M{\int }_0^x{e}^{-r(x-s)}\parallel f(s+n{\omega }_1,s+y-x)-g(s,s+y-x)\parallel ds\hfill \\ \hfill \le & M{\int }_0^{m{\omega }_1}{e}^{-r(x-s)}\parallel f(s+n{\omega }_1,s+y-x)-g(s,s+y-x)\parallel ds\hfill \\ \hfill & +M{\int }_{m{\omega }_1}^x\parallel f(s+n{\omega }_1,s+y-x)-g(s,s+y-x)\parallel ds\hfill \\ \hfill \le & M\sum _{k=0}^{m-1}{\int }_{k{\omega }_1}^{(k+1){\omega }_1}{e}^{-r(x-s)}\parallel f(s+n{\omega }_1,s+y-x)-g(s,s+y-x)\parallel ds\hfill \\ \hfill & +M{\int }_{m{\omega }_1}^{(m+1){\omega }_1}\parallel f(s+n{\omega }_1,s+y-x)-g(s,s+y-x)\parallel ds\hfill \end{array}$$

For each $t\in [0,{\omega }_1]$, we have $\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel \to 0$ as $n\to \infty$ uniformly for $s\in \mathbb{R}$ and $\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel \le 2K$. By Lebesgue’s dominated convergence theorem, we obtain:

$$\underset{n\to \infty }{lim}{\int }_0^{{\omega }_1}\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel dt=0$$

uniformly for $s\in \mathbb{R}$. For $\epsilon >0$ given, we select $N_3\in \mathbb{N}$ such that:

$${\int }_0^{{\omega }_1}\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel dt<\epsilon$$

when $n\ge N_3$ uniformly for $s\in \mathbb{R}$. For any $i\in \mathbb{N}$, one has:

$$\begin{array}{cc}\hfill & {\int }_{i{\omega }_1}^{(i+1){\omega }_1}\parallel f(t+n{\omega }_1,s)-g(t,s)\parallel dt\hfill \\ \hfill =& {\int }_0^{{\omega }_1}\parallel f(t+i{\omega }_1+n{\omega }_1,s)-g(t+i{\omega }_1,s)\parallel dt\hfill \\ \hfill =& {\int }_0^{{\omega }_1}\parallel f(t+i{\omega }_1+n{\omega }_1,s)-g(t,s)\parallel dt<\epsilon \hfill \end{array}$$

when $n\ge N_3$ uniformly for $s\in \mathbb{R}$.

Therefore,

$$\begin{array}{cc}\hfill & \parallel I_2(x,y,n)-{\int }_0^{x}T(x-s)g(s,s+y-x)ds\parallel \hfill \\ \hfill \le & M\sum _{k=0}^{m-1}{e}^{-r(x-(k+1){\omega }_1)}\epsilon +M\epsilon \hfill \\ \hfill \le & \left(\frac{1}{1-e^{-r{\omega }_1}}+1\right)M\epsilon \hfill \end{array}$$

Hence, ${lim}_{n\to \infty }{I}_2(x,y,n)={\int }_0^{x}T(x-s)g(s,s+y-x)ds$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

Therefore,

$$\underset{n\to \infty }{lim}{u}_1(x+n\omega ,y)=\underset{n\to \infty }{lim}{I}_1(x,y,n)+\underset{n\to \infty }{lim}{I}_2(x,y,n)=l_1(x,y)+{\int }_0^{x}T(x-s)g(s,s+y-x)ds$$

uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

Step 2. For any $x\in {\mathbb{R}}^{+}$. Firstly, let $y\in [0,x]$.

$$\begin{array}{cc}\hfill u_2(x+n{\omega }_1,y)=& {\int }_0^{x+n{\omega }_1-y}T(x+n{\omega }_1-s)f(s,0)ds\hfill \\ \hfill =& {\int }_{-n{\omega }_1}^{x-y}T(x-s)f(s+n{\omega }_1,0)ds\hfill \\ \hfill =& {\int }_{-n{\omega }_1}^{0}T(x-s)f(s+n{\omega }_1,0)ds+{\int }_0^{x-y}T(x-s)f(s+n{\omega }_1,0)ds\hfill \\ \hfill =& {\int }_0^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,0)ds+{\int }_0^{x-y}T(x-s)f(s+n{\omega }_1,0)ds.\hfill \\ \hfill =& I_7(x,n)+I_8(x,y,n).\hfill \end{array}$$

In a similar way as the case $I_1(x,y,n)$, we can show that $I_7(x,n)$ is a Cauchy sequence, and we denote $l_2\left(x\right)={lim}_{n\to \infty }{I}_7(x,n)$. Besides, we know $l_2\left(x\right)={lim}_{n\to \infty }{I}_7(x,n)$ uniformly for $x\in {\mathbb{R}}^{+}$.

Note that:

$$\begin{array}{cc}\hfill & \parallel I_8(x,y,n)-{\int }_0^{x-y}T(x-s)g(s,0)ds\parallel \hfill \\ \hfill \le & {\int }_0^{x-y}\parallel T(x-s)\parallel \parallel f(s+n{\omega }_1,0)-g(s,0)\parallel ds\hfill \\ \hfill \le & {\int }_0^x\parallel T(x-s)\parallel \parallel f(s+n{\omega }_1,0)-g(s,0)\parallel ds\hfill \\ \hfill \le & M{\int }_0^x{e}^{-r(x-s)}\parallel f(s+n{\omega }_1,0)-g(s,0)\parallel ds.\hfill \end{array}$$

Then, in a similar way as the case $I_2(x,y,n)$, we can show ${lim}_{n\to \infty }{I}_8(x,y,n)={\int }_0^{x-y}T(x-s)g(s,0)ds$ uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in [0,x]$.

Therefore,

$$\underset{n\to \infty }{lim}{u}_2(x+n\omega ,y)=\underset{n\to \infty }{lim}{I}_7(x,n)+\underset{n\to \infty }{lim}{I}_8(x,y,n)=l_2\left(x\right)+{\int }_0^{x-y}T(x-s)g(s,0)ds$$

uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in [0,x]$.

Secondly, let $y>x$. Let $\epsilon >0$, and choose $N_1\in \mathbb{N}$ such that $\frac{KM}{r}{e}^{-rn{\omega }_1}<\epsilon$ when $n\ge N_1$. We now prove that $u_2(x+n{\omega }_1,y)$ is a Cauchy sequence in X for each $y\in (x,+\infty )$.

Case 1: $y-x\in (0,N_1{\omega }_1]$. Let $n>N_1$, $p\in \mathbb{N}$.

$$\begin{array}{cc}\hfill & u_2(x+(n+p){\omega }_1,y)-u_2(x+n{\omega }_1,y)\hfill \\ \hfill =& {\int }_0^{x+(n+p){\omega }_1-y}T(x+(n+p){\omega }_1-s)f(s,0)ds-{\int }_0^{x+n{\omega }_1-y}T(x+n{\omega }_1-s)f(s,0)ds\hfill \\ \hfill =& {\int }_{-(n+p){\omega }_1}^{x-y}T(x-s)f(s+(n+p){\omega }_1,0)ds-{\int }_{-n{\omega }_1}^{x-y}T(x-s)f(s+n{\omega }_1,0)ds\hfill \\ \hfill =& {\int }_{y-x}^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,0)ds-{\int }_{y-x}^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,0)ds\hfill \\ \hfill =& {\int }_{y-x}^{n{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,0)-f(n{\omega }_1-s,0)]ds+{\int }_{n{\omega }_1}^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,0)ds\hfill \\ \hfill =& {\int }_{y-x}^{N_1{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,0)-f(n{\omega }_1-s,0)]ds\hfill \\ \hfill & +{\int }_{N_1{\omega }_1}^{n{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,0)-f(n{\omega }_1-s,0)]ds\hfill \\ \hfill & +{\int }_{n{\omega }_1}^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,0)ds\hfill \\ \hfill =& I_9(x,y,n,p)+I_{10}(x,y,n,p)+I_{11}(x,y,n,p).\hfill \end{array}$$

Now, we estimate the term $I_9(x,y,n,p)$.

$$\begin{array}{cc}\hfill \parallel I_9(x,y,n,p)\parallel \le & {\int }_0^{N_1{\omega }_1}T(x+s)[f((n+p){\omega }_1-s,0)-f(n{\omega }_1-s,0)]ds\hfill \\ \hfill =& {\int }_0^{N_1{\omega }_1}T(x+N_1{\omega }_1-s)[f((n-N_1+p){\omega }_1+s,0)-f((n-N_1){\omega }_1+s,0)]ds\hfill \\ \hfill \le & M{\int }_0^{N_1{\omega }_1}{e}^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1+p){\omega }_1+s,0)-g(s,0)\parallel ds\hfill \\ \hfill & +M{\int }_0^{N_1{\omega }_1}{e}^{-r(N_1{\omega }_1-s)}\parallel f((n-N_1){\omega }_1+s,0)-g(s,0)\parallel ds.\hfill \end{array}$$

By Lebesgue’s dominated convergence theorem, there exists $N_4\in \mathbb{N}$ $(N_4>N_1)$ such that $\parallel I_9(x,y,n,p)\parallel <\epsilon$ when $n\ge N_4$ uniformly for $y-x\in (0,N_1{\omega }_1]$ and $x\in {\mathbb{R}}^{+}$. Since $\frac{KM}{r}{e}^{-rn{\omega }_1}<\epsilon (n\ge N_1)$, then we have:

$$\begin{array}{cc}\hfill \parallel I_{10}(x,y,n,p)\parallel \le & 2KM{\int }_{N_1{\omega }_1}^{n{\omega }_1}{e}^{-r(x+s)}ds\hfill \\ \hfill \le & 2KM{\int }_{N_1{\omega }_1}^{\infty }{e}^{-r(x+s)}ds\hfill \\ \hfill \le & \frac{2KM}{r}{e}^{-r{N}_1{\omega }_1}\hfill \\ \hfill <& 2\epsilon \hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill \parallel I_{11}(x,y,n,p)\parallel \le & KM{\int }_{n{\omega }_1}^{(n+p){\omega }_1}{e}^{-r(x+s)}ds\hfill \\ \hfill \le & KM{\int }_{n{\omega }_1}^{\infty }{e}^{-r(x+s)}ds\hfill \\ \hfill \le & \frac{KM}{r}{e}^{-rn{\omega }_1}\hfill \\ \hfill <& \epsilon \hfill \end{array}$$

uniformly for $y-x\in (0,N_1{\omega }_1]$ and $x\in {\mathbb{R}}^{+}$.

Therefore, $\parallel u_2(x+(n+p){\omega }_1,y)-u_2(x+n{\omega }_1,y)\parallel <4\epsilon$ when $n\ge N_4$ uniformly for $y-x\in (0,N_1{\omega }_1]$ and $x\in {\mathbb{R}}^{+}$.

Case 2: $y-x\in (N_1{\omega }_1,N_4{\omega }_1]$. Note that $\frac{KM}{r}{e}^{-rn{\omega }_1}<\epsilon (n\ge N_1)$.

$$\begin{array}{cc}\hfill & \parallel u_2(x+(n+p){\omega }_1,y)-u_2(x+n{\omega }_1,y)\parallel \hfill \\ \hfill \le & \parallel {\int }_0^{x+(n+p){\omega }_1-y}T(x+(n+p){\omega }_1-s)f(s,0)ds\parallel +\parallel {\int }_0^{x+n{\omega }_1-y}T(x+n{\omega }_1-s)f(s,0)ds\parallel \hfill \\ \hfill =& \parallel {\int }_{-(n+p){\omega }_1}^{x-y}T(x-s)f(s+(n+p){\omega }_1,0)ds\parallel +\parallel {\int }_{-n{\omega }_1}^{x-y}T(x-s)f(s+n{\omega }_1,0)ds\parallel \hfill \\ \hfill =& \parallel {\int }_{y-x}^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,0)ds\parallel +\parallel {\int }_{y-x}^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,0)ds\parallel \hfill \\ \hfill \le & KM{\int }_{y-x}^{(n+p){\omega }_1}{e}^{-r(x+s)}ds+KM{\int }_{y-x}^{n{\omega }_1}{e}^{-r(x+s)}ds\hfill \\ \hfill \le & KM{\int }_{y-x}^{\infty }{e}^{-r(x+s)}ds+KM{\int }_{y-x}^{\infty }{e}^{-r(x+s)}ds\hfill \\ \hfill \le & \frac{KM}{r}{e}^{-r(y-x)}+\frac{KM}{r}{e}^{-r(y-x)}\hfill \\ \hfill <& 2\epsilon \hfill \end{array}$$

when $n\ge N_4$ uniformly for $y-x\in (N_1{\omega }_1,N_4{\omega }_1]$ and $x\in {\mathbb{R}}^{+}$.

Case 3: $y-x\in (N_4{\omega }_1,+\infty )$. Consider $n\ge N_4$. Note that:

$$u_2(x+n{\omega }_1,y)=\left\{\begin{array}{cc}{\int }_{y-x}^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,0)ds,\hfill & y\le x+n{\omega }_1,\hfill \\ 0,\hfill & y\ge x+n{\omega }_1\hfill \end{array}\right.$$

and:

$$u_2(x+(n+p){\omega }_1,y)=\left\{\begin{array}{cc}{\int }_{y-x}^{(n+p){\omega }_1}T(x+s)f((n+p){\omega }_1-s,0)ds,\hfill & y\le x+(n+p){\omega }_1,\hfill \\ 0,\hfill & y\ge x+(n+p){\omega }_1.\hfill \end{array}\right.$$

Then, one has:

$$\begin{array}{cc}\hfill & \parallel u_2(x+n{\omega }_1,y)\parallel \hfill \\ \hfill =& \parallel {\int }_{y-x}^{n{\omega }_1}T(x+s)f(n{\omega }_1-s,0)ds\parallel \hfill \\ \hfill \le & KM{\int }_{y-x}^{n{\omega }_1}{e}^{-r(x+s)}ds\hfill \\ \hfill \le & KM{\int }_{y-x}^{\infty }{e}^{-r(x+s)}ds\hfill \\ \hfill \le & \frac{KM}{r}{e}^{-r(y-x)}\hfill \\ \hfill <& \epsilon \hfill \end{array}$$

when $y-x\in (N_4{\omega }_1,n{\omega }_1]$.

Besides, $\parallel u_2(x+n{\omega }_1,y)\parallel =0$ when $y-x\in (n{\omega }_1,+\infty )$. Therefore, $\parallel u_2(x+n{\omega }_1,y)\parallel <\epsilon$ when $y-x\in (N_4{\omega }_1,+\infty )$. Similarly, $\parallel u_2(x+(n+p){\omega }_1,y)\parallel <\epsilon$ when $y-x\in (N_4{\omega }_1,+\infty )$. Thus, $\parallel u_2(x+(n+p){\omega }_1,y)-u_2(x+n{\omega }_1,y)\parallel <2\epsilon$ when $n\ge N_4$ uniformly for $y-x\in (N_4{\omega }_1,+\infty )$ and $x\in {\mathbb{R}}^{+}$.

Therefore, $\parallel u_2(x+(n+p){\omega }_1,y)-u_2(x+n{\omega }_1,y)\parallel <4\epsilon$ when $n\ge N_4$ uniformly for $y-x\in (0,+\infty )$ and $x\in {\mathbb{R}}^{+}$. This shows that $u_2(x+n{\omega }_1,y)$ is a Cauchy sequence, and we denote $l_3(x,y)={lim}_{n\to \infty }{u}_2(x+n{\omega }_1,y)$. Besides, from the proof, we also know that ${lim}_{n\to \infty }{u}_2(x+n{\omega }_1,y)=l_3(x,y)$ uniformly for $y\in (x,+\infty )$ and $x\in {\mathbb{R}}^{+}$.

Therefore,

$$\underset{n\to \infty }{lim}{u}_2(x+n{\omega }_1,y)=\left\{\begin{array}{cc}{l}_2\left(x\right)+{\int }_0^{x-y}T(x-s)g(s,0)ds,\hfill & y\in [0,x],\hfill \\ l_3(x,y),\hfill & y\in (x,+\infty )\hfill \end{array}\right.$$

uniformly for $x\in {\mathbb{R}}^{+}$ and $y\in {\mathbb{R}}^{+}$.

This completes the proof. □

Theorem 2.

Let F be a $({\omega }_1,{\omega }_2)$-quasi-periodic limit function. Then, the mild solution of Problem (1) is asymptotically quasi-periodic.

Remark 2.

Theorem 4.4 in [21] implies that the mild solution of Problem (1) is asymptotically quasi-periodic when $F=F_1+F_2$, where $F_1\in P_{{\omega }_1}L({\mathbb{R}}^{+},X)$, $F_2\in P_{{\omega }_2}L({\mathbb{R}}^{+},X)$. However, the method in [21] is not available when $F=F_1{F}_2$, where $F_1\in P_{{\omega }_1}L({\mathbb{R}}^{+},\mathbb{C})$, $F_2\in P_{{\omega }_2}L({\mathbb{R}}^{+},X)$. Therefore, we propose the concept of the quasi-periodic limit function and develop the method to get a more general result.

Next, consider the following abstract Cauchy problem:

$$\left\{\begin{array}{c}{x}^{\prime }\left(t\right)=Ax\left(t\right)+F(t,x\left(t\right)),t\in {\mathbb{R}}^{+};\hfill \\ x\left(0\right)=x_0\in X,\hfill \end{array}\right.$$

(9)

where A is the infinitesimal generator of an exponentially stable $C_0$-semigroup ${\left(T\left(t\right)\right)}_{t\ge 0}$.

Let us introduce the following definition.

Definition 4.

A joint continuous function $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times X\to X$ is said to be the $({\omega }_1,{\omega }_2)$-periodic limit uniformly for x in bounded subsets of X if for every bounded subset K of X, $\{f(t,s,x):t,s\in {\mathbb{R}}^{+},x\in K\}$ is bounded, for each $t\in {\mathbb{R}}^{+}$ ${lim}_{n\to \infty }f(t+n{\omega }_1,s,x)=g(t,s,x)$ uniformly for $s\in {\mathbb{R}}^{+}$ and $x\in K$, for each $s\in {\mathbb{R}}^{+}$${lim}_{n\to \infty }f(t,s+n{\omega }_2,x)=h(t,s,x)$ uniformly for $t\in {\mathbb{R}}^{+}$ and $x\in K$. Denote by $P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times X,X)$ the set of all such functions. If we define $F(t,x)=f(t,t,x)$, then $F(t,x)$ is said to be the $({\omega }_1,{\omega }_2)$-quasi-periodic limit uniformly for x in bounded subsets of X.

The following is a composition theorem.

Theorem 3.

Let $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times X\to X$ be the $({\omega }_1,{\omega }_2)$-periodic limit uniformly for x in bounded subsets of X, and assume that f satisfies a Lipschitz condition in x uniformly in $t,s\in {\mathbb{R}}^{+}$:

$$\parallel f(t,s,x)-f(t,s,y)\parallel \le L\parallel x-y\parallel$$

for all $x,y\in X$ and $t,s\in {\mathbb{R}}^{+}$, where $L>0$. Let $\phi \in P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$. The function $F:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to X$ is defined by $F(t,s)=f(t,s,\phi (t,s\left)\right)$. Then, $F\in P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$.

Proof. 

Since $\phi \in P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$, we have for each $t\in {\mathbb{R}}^{+}$:

$$\underset{n\to \infty }{lim}\phi (t+n{\omega }_1,s)={\phi }_g(t,s)$$

(10)

uniformly for $s\in {\mathbb{R}}^{+}$.

Select a bounded subset K of X such that $\phi (t,s),{\phi }_g(t,s)\in K$ for $t,s\in {\mathbb{R}}^{+}$. Thus, $F(t,s)$ is bounded.

On the other side, one has for each $t\in {\mathbb{R}}^{+}$:

$$\underset{n\to \infty }{lim}f(t+n{\omega }_1,s,x)=g(t,s,x)$$

(11)

uniformly for $s\in {\mathbb{R}}^{+}$ and $x\in K$.

Let us consider the function $G:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to X$ defined by $G(t,s)=g(t,s,{\phi }_g(t,s))$. Note that for each $t\in {\mathbb{R}}^{+}$:

$$\begin{array}{cc}\hfill \parallel F(t+n{\omega }_1,s)-G(t,s)\parallel \le & \parallel f(t+n{\omega }_1,s,\phi (t+n{\omega }_1,s))-f(t+n{\omega }_1,s,{\phi }_g(t,s))\parallel \hfill \\ \hfill & +\parallel f(t+n{\omega }_1,s,{\phi }_g(t,s))-g(t,s,{\phi }_g(t,s))\parallel \hfill \\ \hfill \le & L\parallel \phi (t+n{\omega }_1,s)-{\phi }_g(t,s)\parallel +\parallel f(t+n{\omega }_1,s,{\phi }_g(t,s))-g(t,s,{\phi }_g(t,s))\parallel \hfill \end{array}$$

uniformly for $s\in {\mathbb{R}}^{+}$.

We deduce from (10) and (11) that for each $t\in {\mathbb{R}}^{+}$:

$$\underset{n\to \infty }{lim}F(t+n{\omega }_1,s)=G(t,s)$$

uniformly for $s\in {\mathbb{R}}^{+}$.

In a similar way, we can show there exists a function H such that for each $s\in {\mathbb{R}}^{+}$:

$$\underset{n\to \infty }{lim}F(t,s+n{\omega }_2)=H(t,s)$$

uniformly for $t\in {\mathbb{R}}^{+}$. □

Definition 5.

A function $x\in C_b({\mathbb{R}}^{+},X)$ is said to be a mild solution of Problem (9) if:

$$x\left(t\right)=T\left(t\right)x_0+{\int }_0^{t}T(t-s)F(s,x\left(s\right))ds,t\in {\mathbb{R}}^{+},$$

where ${\left(T\left(t\right)\right)}_{t\ge 0}$ is an exponentially stable $C_0$-semigroup.

Now, we can establish the following theorem.

Theorem 4.

Let $F:{\mathbb{R}}^{+}\times X\to X$ be the $({\omega }_1,{\omega }_2)$-quasi-periodic limit uniformly for x in bounded subsets of X, and assume $F(t,x)=f(t,t,x)$, where $f\in P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times X,X)$. Assume that f satisfies a Lipschitz condition in x uniformly in $t,s\in {\mathbb{R}}^{+}$:

$$\parallel f(t,s,x)-f(t,s,y)\parallel \le L\parallel x-y\parallel$$

for all $x,y\in X$ and $t,s\in {\mathbb{R}}^{+}$, where $L>0$. If $ML<r$, then there exists an asymptotically quasi-periodic mild solution of Problem (9).

Proof. 

Define the function $T:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to X$ by:

$$T(x,y)=\left\{\begin{array}{c}T\left(y\right)x_0,y\ge x,\hfill \\ T\left(x\right)x_0,x\ge y.\hfill \end{array}\right.$$

Then, we can define the operator $\Gamma$ on the space $AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ by:

$$\Gamma \phi (x,y)=T(x,y)+\left\{\begin{array}{c}{\int }_0^{x}T(x-s)f(s,s+y-x,\phi (s,s+y-x))ds,y\ge x,\hfill \\ {\int }_0^{y}T(y-s)f(s+x-y,s,\phi (s+x-y,s))ds,x\ge y,\hfill \end{array}\right.$$

where $\phi \in AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)\subset P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$.

Clearly, $T(x,y)\in AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$. By Lemma 1 and Theorem 3, one has $\Gamma \phi \in AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$. Note that the space $AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ is a Banach space by Theorem 1. Finally, for ${\phi }_1,{\phi }_2\in AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$, one has:

$$\parallel \Gamma {\phi }_1-\Gamma {\phi }_2{\parallel }_{\infty }\le \frac{ML}{r}{\parallel {\phi }_1-{\phi }_2\parallel }_{\infty },$$

which shows that $\Gamma$ is a contraction. By the contraction mapping principle, there exists a unique $\overline{\phi }(x,y)\in AQ{P}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},X)$ such that $\Gamma \overline{\phi }(x,y)=\overline{\phi }(x,y)$, that is:

$$\overline{\phi }(x,y)=T(x,y)+\left\{\begin{array}{c}{\int }_0^{x}T(x-s)f(s,s+y-x,\overline{\phi }(s,s+y-x))ds,y\ge x,\hfill \\ {\int }_0^{y}T(y-s)f(s+x-y,s,\overline{\phi }(s+x-y,s))ds,x\ge y.\hfill \end{array}\right.$$

If we denote $\overline{\Phi }\left(t\right)=\overline{\phi }(t,t)$, then we have $\Gamma \overline{\Phi }\left(t\right)=\overline{\Phi }\left(t\right)$, that is:

$$\overline{\Phi }\left(t\right)=T\left(t\right)x_0+{\int }_0^{t}T(t-s)F(s,\overline{\Phi }\left(s\right))ds.$$

Therefore, $\overline{\Phi }$ is a solution of Problem (9). Moreover, $\overline{\Phi }$ is asymptotically quasi-periodic by Proposition 8. □

Remark 3.

The operator Γ in Theorem 4 may be constructed in a different way, so we cannot show the uniqueness of the solution of Problem (9).

Example 4.

Consider the following problem:

$$\left\{\begin{array}{c}\frac{\partial }{\partial t}u(t,x)=\frac{{\partial }^2}{\partial x^2}u(t,x)+A\left(t\right)f\left(u(t,x)\right),t\in {\mathbb{R}}^{+},x\in [0,\pi ];\hfill \\ u(t,0)=u(t,\pi )=0,t\in {\mathbb{R}}^{+};\hfill \\ u(0,x)=g\left(x\right),x\in [0,\pi ].\hfill \end{array}\right.$$

(12)

where the functions $g:[0,\pi ]\to \mathbb{R}$ and $f:\mathbb{R}\to \mathbb{R}$ are appropriate bounded continuous functions and A defined as $A\left(t\right)=a(t,t)$, where $a\in P{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+}\times {\mathbb{R}}^{+},\mathbb{R})$. Besides, f satisfies:

$$\left|f\right(x)-f(y\left)\right|\le L|x-y|,x,y\in \mathbb{R},$$

where $L>0$. Let $X=L^2\left([0,\pi ]\right)$, and let A be the operator given by $Au=u^{\prime \prime }$ with domain $D\left(A\right)=\{u\in X:u^{\prime \prime }\in X,u\left(0\right)=u\left(\pi \right)=0\}$. Clearly, A is the infinitesimal generator of an analytic semigroup ${\left(T\left(t\right)\right)}_{t\ge 0}$ on X. Moreover, A has a discrete spectrum with eigenvalues $-n^2$, $n\in \mathbb{N}$, and corresponding normalized eigenfunctions given by $z_n\left(\xi \right)={\left(\frac{2}{\pi }\right)}^{\frac{1}{2}}sin\left(n\xi \right)$. Furthermore, $\{z_n:n\in \mathbb{N}\}$ is an orthonormal basis of X and $T\left(t\right)x={\sum }_{n=1}^{\infty }{e}^{-n^{2}t}\langle x,z_n\rangle z_n$ for $x\in X$. Therefore, one has $\parallel T\left(t\right)\parallel \le e^{-t},t\in {\mathbb{R}}^{+}$. Therefore, if ${\parallel a\parallel }_{\infty }L<1$, (12) has an asymptotically quasi-periodic mild solution by Theorem 4.

4. Some Remarks and Questions

Proposition 6 and Proposition 8 make sure that these quasi-periodic limit functions can be regarded as a generalization of asymptotically quasi-periodic functions. Here, a question arises: Could we generalize quasi-periodic limit functions again?

It seems more general if we define $F\left(t\right)=f(t,t)$ without H1 and H2 in Definition 2. However, we have the following example.

Example 5.

Define:

$$f(x,y)=\left\{\begin{array}{cc}{f}_1(x,y),\hfill & y\ge k_{1}x,\hfill \\ f_2(x,y),\hfill & k_{2}x<y<k_{1}x,\hfill \\ f_3(x,y),\hfill & y\le k_{2}x,\hfill \end{array}\right.$$

where $x,y\in {\mathbb{R}}^{+}$, $k_1>1,0<k_2<1$. $f_1$ is a continuous function such that ${lim}_{n\to \infty }{f}_1(x,y+n{\omega }_2)=g_1(x,y)$. $f_3$ is a continuous function such that ${lim}_{n\to \infty }{f}_3(x+n{\omega }_1,y)=h_3(x,y)$. $f_1(0,0)=f_3(0,0)$. $f_2$ is an arbitrary continuous function such that f is continuous on ${\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$. Then, f is a $({\omega }_1,{\omega }_2)$-periodic limit function. However, $F\left(t\right)=f(t,t)=f_2(t,t)(t\ge t_0,t_0>0)$ can be an arbitrary continuous function.

It is interesting to ask whether there is another way to generalize asymptotically quasi-periodic functions without using a function with several variables.

Let $x\in C_b({\mathbb{R}}^{+},X)$. If for every $\epsilon >0$, there exist $f_i\in P_{{\omega }_1}L({\mathbb{R}}^{+},\mathbb{C})$, $g_i\in P_{{\omega }_2}L({\mathbb{R}}^{+},\mathbb{C})$, $F_i\in P_{{\omega }_2}L({\mathbb{R}}^{+},X)$ and $G_i\in P_{{\omega }_1}L({\mathbb{R}}^{+},X)(i=1,2,\cdots ,N)$ such that:

$$\parallel x-\sum _{i=1}^N(f_i{F}_i+g_i{G}_i)\parallel <\epsilon ,$$

then denote the set of x by ${\overline{QPL}}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+},X)$.

Note that ${\sum }_{i=1}^N(f_i{F}_i+g_i{G}_i)\in QP{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+},X)$ for each $N\in \mathbb{N}$. Therefore, there is a natural question as follows.

Question 4.2: Does ${\overline{QPL}}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+},X)$ equal $QP{L}_{({\omega }_1,{\omega }_2)}({\mathbb{R}}^{+},X)$?