In this section, it will be shown that, under certain conditions, the system (1.4) has a bounded solution and an asymptotically periodic solution.

**Theorem 3.1** *Suppose that conditions* ({C}_{1}), ({C}_{2}) *are satisfied*. *Let* *r* *be defined as*

r=\mathrm{\Gamma}({r}_{0}+N),

(3.1)

*where*

\mathrm{\Gamma}=max\{1,\underset{{T}_{0}\le s\le t}{sup}{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma},\underset{{T}_{0}\le t}{sup}{\int}_{{T}_{0}}^{t}{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\phantom{\rule{0.2em}{0ex}}ds\},

(3.2)

*and*

{r}_{0}=\frac{N}{\delta}.

(3.3)

*Then Eq*. (1.4) *has a bounded solution* u(t) *on* {\mathbb{R}}^{+} *such that* \parallel u(t)\parallel \le r *for* t\in {\mathbb{R}}^{+}. *Furthermore*, *if* v(t) *is any solution of Eq*. (1.4), *then* \parallel u(t)-v(t)\parallel \to 0 *as* t\to +\mathrm{\infty}.

*Proof* If A(t,0)\not\equiv 0 for t\in {\mathbb{R}}^{+}, we replace A(t,x) and f(t) by A(t,x)-A(t,0) and f(t)+A(t,0), respectively. We assume, henceforth, that A(t,0)\equiv 0 and \parallel f(t)\parallel \le N for all t\in {\mathbb{R}}^{+} and fix a vector {u}_{0}\in {\mathbb{R}}^{n} with \parallel {u}_{0}\parallel ={r}_{0}. For each positive integer *n* with n>{T}_{0}>\frac{1}{n}, we consider the following Cauchy problem:

{x}^{\prime}=A(t,x)+f(t),\phantom{\rule{1em}{0ex}}x\left(\frac{1}{n}\right)={u}_{0}.

(c.p)

We find that the conditions ({V}_{1})-({V}_{3}) and ({K}_{1})-({K}_{4}) in [5] can be satisfied by ({C}_{1}), ({C}_{2}) in the present paper, then Corollary 5.1 in [5] can now be applied to guarantee the (c.p) has a unique solution {u}_{n} on [\frac{1}{n},n]. We first prove that

\parallel {u}_{n}(t)\parallel \le {r}_{0}\phantom{\rule{1em}{0ex}}\text{for all}t\in [\frac{1}{n},{T}_{0}].

In fact, otherwise there exists some {t}_{1} such that \parallel {u}_{n}({t}_{1})\parallel ={r}_{1}, where {r}_{1} is an arbitrary number such that {r}_{1}>{r}_{0}. Let \tau =sup\{t\in [\frac{1}{n},{t}_{1}];\parallel {u}_{n}(t)\parallel \le {r}_{1}\}, by the continuity of {u}_{n}(t), it follows easily that \frac{1}{n}<\tau \le {T}_{0}. Then \tau <{T}_{0} implies \parallel {u}_{n}(\tau )\parallel ={r}_{1}, and by Lemma 2.1 and ({C}_{2}), we have

\begin{array}{rcl}{D}_{+}\parallel {u}_{n}(\tau )\parallel & =& [{u}_{n}(\tau ),{u}_{n}^{\prime}(\tau )]\\ =& [{u}_{n}(\tau ),A(\tau ,{u}_{n}(\tau ))+f(\tau )]\\ \le & [{u}_{n}(\tau ),A(\tau ,{u}_{n}(\tau ))]+[{u}_{n}(\tau ),f(\tau )]\\ \le & p(\tau )\parallel {u}_{n}(\tau )\parallel +\parallel f(\tau )\parallel \\ \le & p(\tau ){r}_{1}+N.\end{array}

For each \u03f5>0, there exists an {h}_{0}>0 such that

\begin{array}{rcl}\parallel {u}_{n}(\tau +h)\parallel & \le & \parallel {u}_{n}(\tau )\parallel +h(p(\tau ){r}_{1}+N+\u03f5)\\ =& {r}_{1}+h(p(\tau ){r}_{1}+N+\u03f5)\end{array}

for 0<h\le {h}_{0}. Since {r}_{1}>{r}_{0}=\frac{N}{\delta}, p(\tau ){r}_{1}\le -\delta {r}_{1}<-N, p(\tau ){r}_{1}+N<0. Thus, for *ϵ* with 0<\u03f5<-(p(\tau ){r}_{1}+N), there exists a sufficiently small h>0 such that \parallel {u}_{n}(\tau +h)\parallel <{r}_{1}. This contradicts the definition of *τ*. Then \parallel {u}_{n}(t)\parallel \le {r}_{0} for all t\in [\frac{1}{n},{T}_{0}].

On the other hand, using the following differential inequality:

{D}_{+}\parallel {u}_{n}(t)\parallel \le p(t)\parallel {u}_{n}(t)\parallel +\parallel f(t)\parallel \phantom{\rule{1em}{0ex}}\text{for}t\in [{T}_{0},n],

we have

\begin{array}{rcl}\parallel {u}_{n}(t)\parallel & \le & \parallel {u}_{n}({T}_{0})\parallel {e}^{{\int}_{{T}_{0}}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}+{\int}_{{T}_{0}}^{t}N{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\phantom{\rule{0.2em}{0ex}}ds\\ \le & \mathrm{\Gamma}({r}_{0}+N)=r.\end{array}

It thus follows that \parallel {u}_{n}(t)\parallel \le r for all t\in [\frac{1}{n},n]. Lemma 8.1 in [8] can now be applied to guarantee the existence on (0,+\mathrm{\infty}) of a bounded solution u(t) of Eq. (1.4). By the continuity of u(t), u(t) is a bounded solution on {\mathbb{R}}^{+} which also satisfies \parallel u(t)\parallel \le r. If v(t) is any solution of Eq. (1.4) on {\mathbb{R}}^{+}, by Lemma 2.3, we have

\begin{array}{rcl}\parallel u(t)-v(t)\parallel & \le & \parallel u(0)-v(0)\parallel {e}^{{\int}_{0}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\\ =& \parallel u(0)-v(0)\parallel {e}^{{\int}_{0}^{{T}_{0}}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\cdot {e}^{{\int}_{{T}_{0}}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}.\end{array}

By Lemma 2.4, we obtain {e}^{{\int}_{{T}_{0}}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\to 0, when t\to +\mathrm{\infty}, {\int}_{{T}_{0}}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma \to -\mathrm{\infty}, and then \parallel u(t)-v(t)\parallel \to 0 as t\to +\mathrm{\infty}. This completes the proof. □

**Theorem 3.2** *Suppose that* A(t,x) *is asymptotically almost periodic in* *t* *uniformly for* x\in Sr(0), *where* *r* *is a positive number defined by Eq*. (3.1) *and* Sr(0)=\{x\in {\mathbb{R}}^{n};\parallel x\parallel \le r\}, *and* f(t) *is an asymptotically almost periodic function*. *Suppose*, *furthermore*, *that the condition* ({C}_{2}) *is satisfied*. *Then Eq*. (1.4) *has an asymptotically almost periodic solution on* {\mathbb{R}}^{+}.

*Proof* First, we prove that f(t) is bounded. f(t)=g(t)+\alpha (t) in {\mathbb{R}}^{+}, and g(t) is an almost periodic function in ℝ. For any \epsilon \le 1, there is an l(\epsilon )>0, when t\in [0,l(\epsilon )], there is an M>0, \parallel g(t)\parallel \le M. For any t\in \mathbb{R}, choose \tau \in [-t,-t+l(\epsilon )], then t+\tau \in [0,l(\epsilon )], \parallel g(t+\tau )\parallel <M and \parallel g(t+\tau )-g(t)\parallel <1, so for any *t*, \parallel g(t)\parallel <M+1. While \alpha (t)\to 0 (t\to \mathrm{\infty}), we have a positive N>0 such that \parallel f(t)\parallel <N, then the condition ({C}_{1}) is satisfied. Conditions ({C}_{1}) and ({C}_{2}) are satisfied, let u(t) be a bounded solution of (1.4) on {\mathbb{R}}^{+} obtained in Theorem 3.1. Note that \parallel u(t)\parallel \le r for all t\in {\mathbb{R}}^{+}, where *r* is a number defined by Eq. (3.1).

Notice that B(t,x)+g(t) is also an almost periodic function in *t* uniformly for x\in Sr(0). For each \u03f5>0, there exist a positive number {t}_{1}(\u03f5,Sr(0)) and a positive number L(\u03f5,Sr(0)) such that any interval of length L(\u03f5,Sr(0)) contains an *ω*,

\begin{array}{l}\parallel A(t+\omega ,x)-A(t,x)+f(t+\omega )-f(t)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel B(t+\omega ,x)+g(t+\omega )-B(t,x)-g(t)\parallel \\ \phantom{\rule{2em}{0ex}}+\parallel \beta (t+\omega ,x)-\beta (t,x)\parallel +\parallel \alpha (t+\omega )-\alpha (t)\parallel \\ \phantom{\rule{1em}{0ex}}\le \u03f5\phantom{\rule{1em}{0ex}}\text{for all}t{t}_{1}(\u03f5,Sr(0))\text{and}x\in Sr(0).\end{array}

(3.4)

By Lemma 2.1, ({C}_{2}) and Eq. (3.4), we have

\begin{array}{l}{D}_{+}\parallel u(t+\omega )-u(t)\parallel \\ \phantom{\rule{1em}{0ex}}=[u(t+\omega )-u(t),A(t+\omega ,u(t+\omega ))+f(t+\omega )-(A(t,u(t))+f(t))]\\ \phantom{\rule{1em}{0ex}}\le [u(t+\omega )-u(t),A(t,u(t+\omega ))-A(t,u(t))]\\ \phantom{\rule{2em}{0ex}}+\parallel A(t+\omega ,u(t+\omega ))-A(t,u(t+\omega ))+f(t+\omega )-f(t)\parallel \\ \phantom{\rule{1em}{0ex}}\le p(t)\parallel u(t+\omega )-u(t)\parallel +\u03f5\end{array}

(3.5)

for all t>{t}_{1}. Solving this differential inequality, we have

\begin{array}{rcl}\parallel u(t+\omega )-u(t)\parallel & \le & \parallel u(t-\eta +\omega )-u(t-\eta )\parallel {e}^{{\int}_{t-\eta}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}+\u03f5{\int}_{t-\eta}^{t}{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\phantom{\rule{0.2em}{0ex}}ds\\ \le & 2r{e}^{{\int}_{t-\eta}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}+\u03f5{\int}_{t-\eta}^{t}{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\phantom{\rule{0.2em}{0ex}}ds,\end{array}

(3.6)

where *η* is a positive number to be chosen later appropriately, and t-\eta \ge 0. We show that

K=sup\{{\int}_{t-\eta}^{t}{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\phantom{\rule{0.2em}{0ex}}ds;t\ge {t}_{1},t-\eta \ge 0\}

(3.7)

is finite. In fact, this follows from the following estimates:

{\int}_{t-\eta}^{t}{e}^{{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\phantom{\rule{0.2em}{0ex}}ds\le \{\begin{array}{ll}(1+\frac{1}{\delta})\mathrm{\Gamma},& t\ge {T}_{0},t-\eta \le {T}_{0};\\ \mathrm{\Gamma},& t\ge {T}_{0},t-\eta \ge {T}_{0}.\end{array}

(3.8)

Let {T}_{1}>{T}_{0} be a number such that

{\int}_{s}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma \le -\frac{\gamma}{2}(t-s)\phantom{\rule{1em}{0ex}}(t\ge {T}_{1},s\ge {T}_{0})

(3.9)

and

{e}^{-\gamma {T}_{1}}<\frac{\u03f5}{2\mathrm{\Gamma}}.

We will show that \parallel u(t+\omega )-u(t)\parallel \le {K}_{0}\u03f5, where {K}_{0} is a positive constant independent of *ϵ* and *ω*.

We must estimate {e}^{{\int}_{t-\eta}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma} for *t* large enough.

Since {e}^{{\int}_{t-2{T}_{1}}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\le {e}^{-\frac{\gamma}{2}(t-t+2{T}_{1})}={e}^{-\gamma {T}_{1}} (when t-2{T}_{1}\ge {T}_{0}), if t-\eta \ge {T}_{0}, t-4{T}_{1}\ge t-\eta, t\ge {T}_{1}, t-4{T}_{1}\ge {T}_{0}, then

\begin{array}{rcl}{e}^{{\int}_{t-\eta}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}& =& {e}^{{\int}_{t-\eta}^{t-2{T}_{1}}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}{e}^{{\int}_{t-2{T}_{1}}^{t}p(\sigma )\phantom{\rule{0.2em}{0ex}}d\sigma}\\ \le & \mathrm{\Gamma}{e}^{-\gamma {T}_{1}}.\end{array}

When we choose \eta =4{T}_{1} and t-4{T}_{1}\ge {T}_{0}, that is, t\ge 4{T}_{1}+{T}_{0}, then, for any \u03f5>0, there exists a positive number L(\u03f5,Sr(0)) such that any interval of length L(\u03f5,Sr(0)) contains an *ω*, when t\ge T=max\{{t}_{1}(\u03f5,Sr(0)),4{T}_{1}+{T}_{0}\}, \eta =4{T}_{1}, {K}_{0}=\gamma +K,

\begin{array}{rcl}\parallel u(t+\omega )-u(t)\parallel & \le & 2r\cdot \mathrm{\Gamma}{e}^{-\gamma {T}_{1}}+\u03f5K\\ \le & r\u03f5+K\u03f5=(r+K)\u03f5={K}_{0}\u03f5.\end{array}

From Lemma 2.2, u(t) is an asymptotically almost periodic solution of Eq. (1.4). This completes the proof. □

**Remark 3.1** In [4], employing the dissipative-type condition for A(t,x), the authors gave some sufficient conditions to prove the existence of a bounded solution, a periodic or almost periodic solution of the equation {x}^{\prime}=A(t,x)+f(t). Extension of this result has been obtained in one direction: from periodic and almost periodic to asymptotically almost periodic forcing. The equation can be more widely used with asymptotically almost periodic functions.

**Remark 3.2** The condition ({C}_{2}) implies the following hypothesis.

({C}_{2}^{\prime}) Suppose that there exist p(t)\in ({\mathbb{R}}^{+},\mathbb{R}) and positive constants *δ*, {\delta}_{1}, {T}_{0} and {T}_{1} such that

\begin{array}{c}p(t)\le -\delta \phantom{\rule{1em}{0ex}}(t\in [0,{T}_{0}]),\hfill \\ p(t)\le -{\delta}_{1}\phantom{\rule{1em}{0ex}}(t\in [{T}_{1},+\mathrm{\infty}]).\hfill \end{array}

And for all (t,x),(t,y)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{n},

[x-y,A(t,x)-A(t,y)]\le p(t)\parallel x-y\parallel .

We know that ({C}_{2}^{\prime}) can also be used to prove the lemmas in Section 2 and the theorems in Section 3 leaving the conclusion unchanged. ({C}_{2}^{\prime}) as well as ({C}_{2}) yields the existence of a bounded solution, and the process of the proof is similar to the proof before, and we need not necessarily do it again.