In order to obtain our existence theorem, we make the following assumptions:
-
(H0)
\(a\in\operatorname{AAP}(\mathbb{R},\mathbb{R})\) with \(M(a)>0\), and \(b_{i},\tau_{i}\in\operatorname{AAP}(\mathbb{R} ,\mathbb{R}^{+})\) with \(b_{i}^{-}>0\) for all \(i=1,2,\ldots,k\).
-
(H1)
There exists \(\alpha\in(0,M(a))\) such that
$$M_{2}:=\frac{\sum_{i=1}^{k}\frac{b_{i}^{-}}{1+M_{1}^{n}}}{a^{+}}\leq\frac {e^{\alpha T_{0}}\sum_{i=1}^{k} b_{i}^{+}}{\alpha}:=M_{1}, $$
where \(T_{0}\) is defined in Lemma 2.1.
-
(H2)
For the case of \(n\in(0,1]\), we have
$$\frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha}< \frac {(1+M_{2}^{n})^{2} M_{2}^{1-n}}{n}; $$
for the case of \(n>1\), we have
$$\frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha}< \frac {4n}{n^{2}-1}\sqrt[n]{\frac{n-1}{n+1}}. $$
Theorem 3.1
Under the assumptions (H0)-(H2), equation (1.1) has a unique asymptotically almost periodic solution in
$$\Omega=\bigl\{ \varphi\in\operatorname{AAP}(\mathbb{R},\mathbb{R}): M_{2}\leq\varphi(t)\leq M_{1}, \forall t \in\mathbb{R}\bigr\} . $$
Proof
Fix \(\varphi\in\Omega\). Let us consider the following differential equation:
$$ x'(t)=-a(t)x(t)+\sum_{i=1}^{k} \frac{b_{i}(t)}{1+\varphi^{n}(t-\tau_{i}(t))}. $$
(3.1)
It follows from Lemma 2.3 that \(\varphi(\cdot-\tau_{i}(\cdot ))\in\operatorname{AAP}(\mathbb{R},\mathbb{R})\). Then, by Lemma 1.4, we can obtain
$$\sum_{i=1}^{k}\frac{b_{i}(\cdot)}{1+\varphi^{n}(\cdot-\tau_{i}(\cdot ))}\in \operatorname{AAP}(\mathbb{R},\mathbb{R}). $$
By Lemma 2.2, equation (3.1) has a unique asymptotically almost periodic solution given by
$$ x^{\varphi}(t)= \int_{-\infty}^{t}e^{-\int_{s}^{t}a(r)\,dr}\sum _{i=1}^{k}\frac{b_{i}(s)}{1+\varphi^{n}(s-\tau_{i}(s))}\,ds, \quad t\in \mathbb{R}. $$
(3.2)
Now we define a mapping T on Ω by
$$(T\varphi) (t)=x^{\varphi}(t) ,\quad \varphi\in\Omega, t\in\mathbb{R}. $$
Next, we show that \(T(\Omega) \subset\Omega\). It suffices to prove that
$$M_{2} \leq(T\varphi) (t) \leq M_{1} $$
for all \(t \in\mathbb{R}\) and \(\varphi\in\Omega\). For every \(t \in \mathbb{R}\) and \(\varphi\in\Omega\), by Lemma 2.1, we have
$$\begin{aligned} (T\varphi) (t) =&x^{\varphi}(t) \\ =& \int_{-\infty}^{t}e^{-\int_{s}^{t}a(r)\,dr}\sum _{i=1}^{k}\frac {b_{i}(s)}{1+\varphi^{n}(s-\tau_{i}(s))}\,ds \\ \leq& \int_{-\infty}^{t}e^{-\int_{s}^{t}a(r)\,dr}\sum _{i=1}^{k}b_{i}(s)\,ds \\ \leq& \sum_{i=1}^{k}b_{i}^{+} \int_{-\infty}^{t}e^{-\int _{s}^{t}a(r)\,dr}\,ds \\ =& \sum_{i=1}^{k}b_{i}^{+} \int_{0}^{+\infty}e^{-\int _{t-s}^{t}a(r)\,dr}\,ds \\ \leq& \sum_{i=1}^{k}b_{i}^{+} \int_{0}^{+\infty}e^{\alpha T_{0}}e^{-\alpha s}\,ds \\ =& \frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha}=M_{1}. \end{aligned}$$
Moreover, for every \(t \in\mathbb{R}\) and \(\varphi\in\Omega\), we have
$$\begin{aligned} (T\varphi) (t) =&x^{\varphi}(t) \\ =& \int_{-\infty}^{t}e^{-\int_{s}^{t}a(r)\,dr}\sum _{i=1}^{k}\frac {b_{i}(s)}{1+\varphi^{n}(s-\tau_{i}(s))}\,ds \\ \geq& \int_{-\infty}^{t}e^{-\int_{s}^{t}a(r)\,dr}\sum _{i=1}^{k}\frac {b_{i}(s)}{1+M_{1}^{n}}\,ds \\ \geq& \sum_{i=1}^{k}\frac{b_{i}^{-}}{1+M_{1}^{n}} \int_{0}^{+\infty }e^{-\int_{t-s}^{t}a(r)\,dr}\,ds \\ \geq& \sum_{i=1}^{k}\frac{b_{i}^{-}}{1+M_{1}^{n}} \int_{0}^{+\infty }e^{-a^{+}s}\,ds \\ =& \frac{\sum_{i=1}^{k}\frac{b_{i}^{-}}{1+M_{1}^{n}}}{a^{+}}=M_{2}. \end{aligned}$$
Thus, T is a self-mapping from Ω to Ω.
Next, let us show that T is a contraction mapping. We consider two cases.
Case I. \(n\in(0,1]\).
By mean value theorem and direct calculations, one can obtain
$$ \biggl\vert \frac{1}{1+x^{n}} - \frac{1}{1+y^{n}}\biggr\vert \leq\frac {n}{(1+M_{2}^{n})^{2} M_{2}^{1-n}}\cdot|x-y| $$
(3.3)
for all \(x,y\geq M_{2}\). By using (3.3) and Lemma 2.1, we conclude, for every \(\varphi,\psi\in\Omega\),
$$\begin{aligned}& \|T\varphi- T\psi\| \\& \quad = \sup_{t\in\mathbb{R}}\bigl\vert (T\varphi) (t) - (T\psi) (t) \bigr\vert \\& \quad = \sup_{t\in\mathbb{R}}\Biggl\vert \int_{-\infty}^{t}e^{-\int _{s}^{t}a(r)\,dr}\sum _{i=1}^{k}b_{i}(s)\biggl[\frac{1}{1+\varphi^{n}(s-\tau_{i}(s))} - \frac{1}{1+\psi^{n}(s-\tau_{i}(s))}\biggr]\,ds\Biggr\vert \\& \quad \leq \sup_{t\in\mathbb{R}} \int_{-\infty}^{t}e^{-\int _{s}^{t}a(r)\,dr}\sum _{i=1}^{k}b_{i}(s)\cdot\frac{n}{(1+M_{2}^{n})^{2} M_{2}^{1-n}} \cdot\bigl|\varphi\bigl(s-\tau_{i}(s)\bigr) - \psi\bigl(s- \tau_{i}(s)\bigr)\bigr|\,ds \\& \quad \leq \frac{n\sum_{i=1}^{k}b_{i}^{+}}{(1+M_{2}^{n})^{2} M_{2}^{1-n}}\cdot \|\varphi- \psi\|\cdot\sup _{t\in\mathbb{R}} \int_{-\infty }^{t}e^{-\int _{s}^{t}a(r)\,dr}\,ds \\& \quad \leq \frac{ne^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha (1+M_{2}^{n})^{2} M_{2}^{1-n}}\cdot\|\varphi- \psi\|. \end{aligned}$$
Case II. \(n>1\).
By the mean value theorem and direct calculations, one can obtain
$$ \biggl\vert \frac{1}{1+x^{n}} - \frac{1}{1+y^{n}}\biggr\vert \leq\frac {n^{2}-1}{4n}\sqrt[n]{\frac{n+1}{n-1}}\cdot|x-y| $$
(3.4)
for all \(x,y\geq0\). By using (3.4) and Lemma 2.1, we conclude that, for every \(\varphi,\psi\in\Omega\),
$$\begin{aligned}& \|T\varphi- T\psi\| \\& \quad = \sup_{t\in\mathbb{R}}\Biggl\vert \int_{-\infty}^{t}e^{-\int _{s}^{t}a(r)\,dr}\sum _{i=1}^{k}b_{i}(s)\biggl[\frac{1}{1+\varphi^{n}(s-\tau_{i}(s))} - \frac{1}{1+\psi^{n}(s-\tau_{i}(s))}\biggr]\,ds\Biggr\vert \\& \quad \leq \sup_{t\in\mathbb{R}} \int_{-\infty}^{t}e^{-\int _{s}^{t}a(r)\,dr}\sum _{i=1}^{k}b_{i}(s)\cdot\frac{n^{2}-1}{4n} \sqrt [n]{\frac {n+1}{n-1}}\cdot\bigl\vert \varphi\bigl(s- \tau_{i}(s)\bigr) - \psi\bigl(s-\tau_{i}(s)\bigr)\bigr\vert \,ds \\& \quad \leq \frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha}\cdot\frac {n^{2}-1}{4n}\sqrt[n]{\frac{n+1}{n-1}} \cdot\|\varphi- \psi\|. \end{aligned}$$
In all cases, by (H2), T is a contraction. Thus, T has a unique fixed point in Ω, i.e., equation (1.1) has a unique asymptotically almost periodic solution in Ω. □
Remark 3.2
Compared with most earlier results concerning almost periodic solutions to equation (1.1), in Theorem 3.1, we do not assume that \(a^{-} > 0\) (see also Remark 3.6).
Next, let us study exponential stability of asymptotically almost periodic solution of (1.1). For convenience, we only discuss the case of \(n>1\).
Theorem 3.3
Let
\(n>1\). Suppose that (H0)-(H2) are satisfied, \(x(t)\)
is the unique asymptotically almost periodic solution of (1.1) in Ω, and
\(y(t)\)
is an arbitrary nonnegative global solution of (1.1). Then there exists a constant
\(\lambda> 0\)
such that
$$ \bigl\vert x(t)-y(t)\bigr\vert \leq Me^{\alpha T_{0}}e^{-\lambda t} $$
(3.5)
for all
\(t \in[-\tau, +\infty)\), where
\(\tau=\max_{1\leq i \leq k}\tau_{i}^{+}\)
and
\(M=\sup_{t\in[-\tau,0] }|x(t)-y(t)|\).
Proof
Let \(N=\frac{n^{2}-1}{4n}\sqrt[n]{\frac{n+1}{n-1}}\). By (H2), we have
$$\frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha}< \frac{1}{N}. $$
Thus, there exists \(\lambda\in(0, \alpha)\) such that
$$ \lambda- \alpha+ e^{\alpha T_{0}}\sum _{i=1}^{k}b_{i}^{+}\cdot Ne^{\lambda\tau} < 0. $$
(3.6)
Now, setting \(z(t)=x(t)-y(t)\), it is not difficult to see that
$$z(t) = e^{-\int_{0}^{t}a(u)\,du}z(0)+ \int_{0}^{t}e^{-\int _{s}^{t}a(u)\,du}\sum _{i=1}^{k}b_{i}(s)\cdot \biggl[ \frac{1}{1+x^{n}(s-\tau _{i}(s))} - \frac{1}{1+y^{n}(s-\tau_{i}(s))} \biggr]\,ds. $$
We claim that, for every \(\varepsilon>0\), we have
$$ \bigl\vert z(t)\bigr\vert =\bigl\vert x(t)-y(t)\bigr\vert \leq(M+\varepsilon)e^{\alpha T_{0}}e^{-\lambda t}, \quad t\in[-\tau,+\infty). $$
(3.7)
Otherwise, for some \(\varepsilon>0\),
$$\bigl\{ t>0:\bigl\vert z(t)\bigr\vert >(M+\varepsilon)e^{\alpha T_{0}}e^{-\lambda t} \bigr\} \neq \emptyset. $$
Let
$$t_{0}=\inf\bigl\{ t>0:\bigl\vert z(t)\bigr\vert >(M+ \varepsilon)e^{\alpha T_{0}}e^{-\lambda t}\bigr\} . $$
Then \(t_{0}>0\) and
$$\bigl\vert z(t_{0})\bigr\vert =(M+\varepsilon)e^{\alpha T_{0}}e^{-\lambda t_{0}}, \qquad \bigl\vert z(t)\bigr\vert \leq (M+\varepsilon)e^{\alpha T_{0}}e^{-\lambda t}, \quad t\in[-\tau, t_{0}). $$
Combining this with (3.4), (3.6), and Lemma 2.1, we conclude
$$\begin{aligned} \bigl\vert z(t_{0})\bigr\vert =& \Biggl\vert e^{-\int_{0}^{t_{0}}a(u)\,du}z(0)+ \int_{0}^{t_{0}}e^{-\int _{s}^{t_{0}}a(u)\,du}\cdot\sum _{i=1}^{k}b_{i}(s) \\ &{}\cdot\biggl[ \frac {1}{1+x^{n}(s-\tau_{i}(s))} - \frac{1}{1+y^{n}(s-\tau_{i}(s))} \biggr]\,ds\Biggr\vert \\ \leq& e^{\alpha T_{0}}e^{-\alpha t_{0}}\bigl\vert z(0)\bigr\vert + \int_{0}^{t_{0}}e^{\alpha T_{0}}e^{-\alpha(t_{0}-s)}\cdot \sum_{i=1}^{k}b_{i}^{+}N\bigl\vert z\bigl(s-\tau_{i}(s)\bigr)\bigr\vert \,ds \\ \leq& e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+\varepsilon)+Ne^{\alpha T_{0}} \int _{0}^{t_{0}}e^{-\alpha(t_{0}-s)}\sum _{i=1}^{k}b_{i}^{+}\cdot(M+\varepsilon )e^{\alpha T_{0}}e^{-\lambda(s-\tau_{i}(s))}\,ds \\ \leq& e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+\varepsilon)+Ne^{\alpha T_{0}}e^{\lambda\tau} \int_{0}^{t_{0}}e^{-\alpha(t_{0}-s)}\sum _{i=1}^{k}b_{i}^{+}\cdot(M+ \varepsilon)e^{\alpha T_{0}}e^{-\lambda s}\,ds \\ =& e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+\varepsilon)+e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+ \varepsilon) \int_{0}^{t_{0}} Ne^{\lambda\tau }e^{\alpha T_{0}}\sum _{i=1}^{k}b_{i}^{+} \cdot e^{(\alpha-\lambda )s}\,ds \\ < & e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+\varepsilon)+e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+ \varepsilon) \int_{0}^{t_{0}} (\alpha-\lambda) e^{(\alpha-\lambda)s}\,ds \\ =& e^{\alpha T_{0}}e^{-\alpha t_{0}}(M+\varepsilon) e^{(\alpha-\lambda )t_{0}}=(M+ \varepsilon)e^{\alpha T_{0}}e^{-\lambda t_{0}}, \end{aligned}$$
which is a contradiction. Thus, for every \(\varepsilon>0\), (3.7) holds. By the arbitrariness of ε, we conclude that
$$\bigl\vert z(t)\bigr\vert \leq Me^{\alpha T_{0}}e^{-\lambda t},\quad t \in[-\tau,+\infty). $$
This completes the proof. □
Next, we give two examples to illustrate our main results.
Example 3.4
Let \(n=k=1\), \(a(t)=1+\ln2\cdot(\sin t+\sin\pi t)-\frac {1}{2}e^{-t^{2}}\), and
$$b_{1}(t)=\frac{1+\sin^{2}t+\sin^{2} \sqrt{2}t}{48},\qquad \tau _{1}(t)= \cos^{2}t+\cos^{2} \sqrt{2}t+\frac{1}{1+t^{2}}. $$
It is easy to see that (H0) holds. For all \(t,s\in\mathbb{R}\) with \(s\leq t\), we have
$$\int_{t}^{s}a(u)\,du \leq(s-t)+\ln2\cdot\biggl(2+ \frac{2}{\pi}\biggr)+\frac {1}{2}(t-s)\leq\frac{1}{2} \bigl[(s-t)+6\ln2\bigr]. $$
So, we can choose \(\alpha=\frac{1}{2}\) and \(T_{0}=6\ln2\). By a direct calculation, we can obtain
$$e^{\alpha T_{0}}=8,\qquad b_{1}^{+}=\frac{1}{16}, \qquad b_{1}^{-}=\frac {1}{48},\qquad a^{+}=1+\ln4. $$
Thus, we have
$$M_{1}=\frac{e^{\alpha T_{0}}\sum_{i=1}^{k} b_{i}^{+}}{\alpha}=1,\qquad M_{2}= \frac{\sum_{i=1}^{k}\frac{b_{i}^{-}}{1+M_{1}^{n}}}{a^{+}}=\frac {1}{96(1+\ln4)}\leq M_{1}, $$
which shows that (H1) holds. Moreover,
$$\frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha }=1< (1+M_{2})^{2}=\frac{(1+M_{2}^{n})^{2} M_{2}^{1-n}}{n}, $$
which means that (H2) holds. Thus, by Theorem 3.1, equation (1.1) has a unique asymptotically almost periodic solution in Ω. A numerical simulation is given in Figure 1.
Example 3.5
Let \(n=k=2\), \(a(t)=10+5 (\sin\frac{15}{\ln2} t+\sin\frac {15\pi}{\ln 2} t )+e^{-t^{2}}\),
$$b_{1}(t)=1+\sin^{2}t+\sin^{2} \sqrt{2}t,\qquad b_{2}(t)=1+\cos^{2}t+\cos^{2} \sqrt{2}t, $$
and
$$\tau_{1}(t)=\cos^{2}t+\cos^{2} \sqrt{2}t+ \frac{1}{1+t^{2}},\qquad \tau _{2}(t)=\sin^{2}t+ \sin^{2} \sqrt{2}t+\frac{1}{1+t^{4}}. $$
It is easy to see that (H0) holds. For all \(t,s\in\mathbb{R}\) with \(s\leq t\), we have
$$\int_{t}^{s}a(u)\,du \leq10(s-t)+\frac{5\ln2}{15} \biggl(2+\frac{2}{\pi }\biggr)\leq 10(s-t)+\ln2\leq9\biggl[(s-t)+ \frac{1}{9}\ln2\biggr]. $$
So, we can choose \(\alpha=9\) and \(T_{0}=\frac{1}{9}\ln2\). By a direct calculation, we can obtain
$$e^{\alpha T_{0}}=2,\qquad b_{1}^{+}=b_{2}^{+}=3,\qquad b_{1}^{-}=b_{2}^{-}=1, \qquad a^{+}\in[20,21]. $$
Thus, we have
$$M_{1}=\frac{e^{\alpha T_{0}}\sum_{i=1}^{k} b_{i}^{+}}{\alpha}=\frac {4}{3},\qquad M_{2}= \frac{\sum_{i=1}^{k}\frac {b_{i}^{-}}{1+M_{1}^{n}}}{a^{+}}\leq\frac{\sum_{i=1}^{k}\frac {b_{i}^{-}}{1+M_{1}^{n}}}{20}=\frac{9}{250}\leq M_{1}, $$
which shows that (H1) holds. Moreover,
$$\frac{e^{\alpha T_{0}}\sum_{i=1}^{k}b_{i}^{+}}{\alpha}=\frac {4}{3}< \frac{8}{3\sqrt{3}}=\frac{4n}{n^{2}-1} \sqrt[n]{\frac{n-1}{n+1}}, $$
which means that (H2) holds. Thus, by Theorem 3.1 and Theorem 3.3, equation (1.1) has a unique asymptotically almost periodic solution \(x_{0}(t)\) in Ω, and every nonnegative global solution of (1.1) converges exponentially to \(x_{0}(t)\) as \(t\to+\infty\). A numerical simulation is given in Figure 2.
Remark 3.6
It is easy to see that \(a^{-}<0\) in Example 3.4 and \(a^{-}=0\) in Example 3.5. So, many earlier results, which requires \(a^{-}>0\), cannot be applied to the above two examples.