关于非自治生物动力系统, 现已有了大量的研究工作^[1-7].自1993年Gopalsamy^[8]提出带有反馈控制的动力学模型以来, 许多学者利用重合度理论研究了具有反馈控制的生物动力系统^[9-15].基于反馈原理所建立的非自治系统的正周期解相当于对应地自治系统的正平衡点, 通过反馈控制变量来改变系统平衡态的大小, 借以达到人们所期望的控制目标.在自然界中, 影响种群数量变化的因素既有密度制约因素, 又有气候、污染物等非生物因素构成的非密度制约因素, 当发生流行性传染病或火灾乃至寒潮时将会导致种群有一定比例的个体死亡.因此, 研究基于生态环境和反馈控制的生物动力系统是有实际意义和应用价值的.本文建立基于生态环境和反馈控制的时滞非自治$n$种群竞争系统如下

$ \begin{equation}\label{eq:1.1} \left\{ \begin{array}{ll} \frac{{\rm d}N_{i}(t)}{{\rm d}t}=N_{i}(t)[b_{i}(t)-h_{i}(t)N_{i}(t)-\sum^n\limits_{j=1, j\neq1}a_{ij}(t)N_{j}(t)-c_{i}(t)N_{i}(t-\tau_{i})-d_{i}(t)P_{i}(t)], \\ \frac{{\rm d}P_{i}(t)}{{\rm d}t}=P_{i}(t)[r_{i}(t)N_{i}(t)-D_{i}(t)P_{i}(t)], (i=1, 2, \cdots, n), \end{array} \right. \end{equation} $

(1.1)

其中, $N_{i}(t) (i=1, 2, \cdots, n)$代表时刻$t$有竞争关系的第$i$个种群的密度, $P_{i}(t)$ $(i=1, 2, \cdots, n)$代表时刻$t$的第$i$个控制变量的密度, 假设条件如下.

(A1) $b_{i}(t), h_{i}(t), a_{ij}(t), c_{iv}(t), d_{i}(t), r_{i}(t), D_{i}(t)$ $(i, j=1, 2, \cdots, n) $都是正的以$\omega>0$为周期的连续函数;

(A2) 假设$n$个竞争种群在生态环境中因天敌、气候、有毒物、食物短缺等因素影响所导致的死亡率为$c_i(t)$, 且时滞$\tau_i$为发生该因素影响的时间间隔;

(A3) 第$i$个种群$N_i$: $b_{i}(t)$为种群的内禀增长率, $h_{i}(t)$为密度制约系数, $a_{ij}(t)(i\neq j)$为其它$n-1$个种群与之竞争公共资源的比例系数, $d_{i}(t)$为控制变量所捕获的比例系数;

(A4) 第$i$个控制变量$P_i$: $r_{i}(t)$为控制变量捕食后的转化率, $D_{i}(t)$为其密度制约系数.

记${\Bbb R}_{+}^{2n}=\{(N(t), P(t))\in {\Bbb R}^{2n}:N_{i}(t)\geq0, P_{i}\geq0, i=1, 2, \cdots, n\}$, $C^{+}=C([-\tau, 0], {\Bbb R}_{+}^{2n})$表示由$[-\tau, 0]$到${\Bbb R}_{+}^{2n}$的非负连续向量函数的全体, 其中

$ \tau=\max\{\tau_{i}:i=1, 2, \cdots, n\}, ~~ N=(N_{1}, N_{2}, \cdots, N_{n})^T, ~~P(t)=(P_{1}, P_{2}, \cdots, P_{n})^{T}. $

系统(1.1) 的初始条件为

$ \begin{equation} \begin{array} {l} \varphi_{i}, \psi_{i}\in C^+, ~~N_{i}(s)=\varphi_{i}(s)\geq0, ~~ P_{i}(s)=\psi_{i}(s)\geq0, \\ s\in[-\tau, 0], ~~ \varphi_{i}(0)>0, ~~\psi_{i}(0)>0, ~~(i=1, 2, \cdots, n). \end{array} \end{equation} $

(1.2)

对于正的$\omega$ -周期连续函数, 引入如下记号

$ \bar{g}=\frac{1}{\omega}\int_0^\omega g(t){\rm d}t, ~~g^L=\min\limits_{t\in[0, \omega]}\{g(t)\}, ~~g^M=\max\limits_{t\in[0, \omega]}\{g(t)\}. $

且令

$ M_{i}=\ln{\frac{\bar{b}_{i}}{h_{i}^{L}}}+2\bar{b}_{i}\omega, ~~L_{i}=\ln{\frac{r_{i}^{M}\bar{b}_{i}}{h_{i}^{L}D_{i}^{L}}}+\frac{2r_{i}^{M}\bar{b}_{i}\omega}{h_{i}^{L}}, $

本文主要在${\Bbb R}_{+}^{2n}$内研究系统$(1.1)$满足正初值条件$(1.2)$正周期解存在性和全局吸引性.

先引入如下定理作为本文引理.

引理2.1(Mawhin's延拓定理^[16]) 设$X, Y$是Banach空间$L$是指标为零的Fredholm映射, $N:\bar{\Omega}\rightarrow Z$在$\bar{\Omega}$上是$L$ -紧的, 其中$\Omega$是$X$中的有界开集, 如果

(a) $\forall x\in \partial\Omega\cap {\rm Dom}L, \lambda\in(0, 1), Lx\neq \lambda Nx;$

(b) $\forall x\in \partial\Omega\cap {\rm Ker}L, QNx\neq 0, $ deg$\{JQN, \Omega\cap{\rm Ker}L, 0\}\neq 0$.则方程$Lx=Nx$在$\bar{\Omega}\cap {\rm Dom}L $内至少存在一个解.

设参数$\mu\in[0, 1], x=(x_1, x_2, \cdots, x_n, y_1, y_2, \cdots, y_n)^T\in {\Bbb R}^{2n}$, 满足方程组

$ \begin{eqnarray} \left\{ \begin{array}{ll} \bar{b}_{i}-\bar{h}_{i}{\rm e}^{x_{i}}-\mu(\bar{d}_{i}{\rm e}^{y_{i}}+\sum^n\limits_{j=1, j\neq i}\bar{a}_{ij}{\rm e}^{x_{j}}+\bar{c}_{i}{\rm e}^{x_{i}}+\bar{d}_{i}{\rm e}^{y_{i}})=0, \\ \bar{r}_{i}{\rm e}^{x_{i}}-\bar{D}_{i}{\rm e}^{y_{i}}=0, (i=1, 2, \cdots, n). \end{array} \right. \end{eqnarray} $

(2.1)

由方程组(2.1) 可得

$ \frac{1}{\bar{h}_{i}+\bar{c}_{i}} \bigg[\bar{b}_{i}-\sum^n\limits_{j=1, j\neq i}\frac{\bar{a}_{ij}\bar{b}_{j}}{\bar{h}_{j}}-\frac{\bar{d}_{i}\bar{r}_{i}\bar{b}_{i}}{\bar{D}_{i}\bar{h}_{i}} \bigg]>0, $

$ M_{0i}=\frac{\bar{b}_{i}}{\bar{h}_{i}}\geq {\rm e}^{x_{i}}\geq \frac{1}{\bar{h}_{i}+\bar{c}_{i}}\bigg[\bar{b}_{i}-\sum^n\limits_{j=1, j\neq i}\frac{\bar{a}_{ij}\bar{b}_{j}}{\bar{h}_{j}}-\frac{\bar{d}_{i}\bar{r}_{i}\bar{b}_{i}}{\bar{D}_{i}\bar{h}_{i}} \bigg]=m_{0i}, $

$ N_{0i}=\frac{\bar{r}_{i}\bar{b}_{i}}{\bar{D}_{i}\bar{h}_{i}}\geq {\rm e}^{y_{i}}\geq \frac{\bar{r}_{i}}{\bar{D}_{i}(\bar{h}_{i}+\bar{c}_{i})}\bigg[\bar{b}_{i}-\sum^n\limits_{j=1, j\neq i}\frac{\bar{a}_{ij}\bar{b}_{j}}{\bar{h}_{j}}-\frac{\bar{d}_{i}\bar{r}_{i}\bar{b}_{i}}{\bar{D}_{i}\bar{h}_{i}} \bigg] =n_{0i}. $

于是

$ l_{0i}=\ln m_{0i}\leq x_{i}\leq\ln M_{0i}=L_{0i}, l^*_{0i}=\ln n_{0i}\leq y_i\leq\ln N_{0i}=L^*_{0i}. $

令$A_{0i}=\max\limits_{1\leq i\leq n}\{|l_{0i}|, |L_{0i}|\}, A^*_{0i}=\max\limits_{1\leq i\leq n}\{|l^*_{0i}|, |L^*_{0i}|\}$, 显然$A_{0i}, A^*_{0i}(i=1, 2, \cdots, n)$与参数$\mu$无关且

$ |x_i|\leq A_{0i}, |y_i|\leq A^*_{0i}, ||x||=\sum^n\limits_{i=1}|x_i|+\sum^n\limits_{i=1}|y_i|\leq\sum^n\limits_{i=1}A_{0i}+\sum^n\limits_{i=1}A^*_{0i}:=A_0. $

综上讨论, 有如下结论.

引理2.2 如果

$ \rho_{i}=\bar{b}_{i}-\sum^n\limits_{j=1, j\neq i}\frac{\bar{a}_{ij}\bar{b}_{j}}{\bar{h}_{j}}-\frac{\bar{d}_{i}\bar{r}_{i}\bar{b}_{i}}{\bar{D}_{i}\bar{h}_{i}}>0, ~~i=1, 2, \cdots, n, $

则方程组(2.1) 存在一个与参数$\mu$无关的常数$A_0>0$使得$||x||\leq A_0$.

由文献[17]的类似方法易得下面的引理.

引理2.3 设$D=\{(N(t), P(t))^T\in {\Bbb R}_+^{2n}:k_i\leq N_i(t)\leq K_i, \tilde{k}_i\leq P_i(t)\leq \tilde{K}_i, $ $i=1, 2, \cdots, n\}$.若$S_i=b_i^L-\sum^n\limits_{j=1, j\neq i}a_{ij}^MK_j-c_{i}^MK_i-d_i^M\tilde{K}_i>0, i=1, 2, \cdots, n, $则对任何解$(N(t), P(t))=(N_1(t), N_2(t), \cdots, N_n(t), P_1(t), P_2(t), \cdots, P_n(t)), $存在$ T>0, $当$t>T$时有$(N(t), P(t))\in D.$其中

$ K_{i}=\frac{b_{i}^M}{h_{i}^L}+\varepsilon, ~~k_i= \frac{S_{i}}{h_{i}^M}-\varepsilon, ~~ \tilde{K}_{i}=\frac{r_{i}^MK_i}{D_{i}^L}+\varepsilon, ~~ \tilde{k}_i=\frac{r_{i}^Lk_i}{D_{i}^M}-\varepsilon, $

这里$\varepsilon$为任意小的正数.

引理2.4  (Barbalat引理^[18])设$f$是定义在$[0, +\infty)$上的非负函数, 在$[0, +\infty)$上可积, 而且在$[0, +\infty)$上一致连续, 则$\lim\limits_{t\rightarrow+\infty}f(t)=0.$

定理3.1 如果除初值(1.2) 条件外, 还满足条件

$ S_i^*=\bar{b}_i-\sum^n\limits_{j=1, j\neq i}a_{ij}^M{\rm e}^{M_j}-d_i^M{\rm e}^{L_i}>0, i=1, 2, \cdots, n. $

则系统(1.1) 至少存在一个$\omega$-周期正解.

证  作变换$N_{i}(t)={\rm e}^{x_i(t)}, P_i(t)={\rm e}^{y_i(t)}, (i=1, 2, \cdots, n), $将周期系统(1.1) 化为如下等价系统

$ \begin{eqnarray} \left\{ \begin{array}{ll} x'_i(t)=b_i(t)-h_i(t){\rm e}^{x_i(t)}-\sum^n\limits_{j=1, j\neq i}a_{ij}(t){\rm e}^{x_j(t)}-c_{i}(t){\rm e}^{x_i(t-\tau_i)}-d_i{\rm e}^{y_i(t)}:=\Lambda_i(x, t), \\ y'_i(t)=r_i(t){\rm e}^{x_i(t)}-D_i(t){\rm e}^{y_i(t)}:=\Delta_i(x, t), (i=1, 2, \cdots, n). \end{array} \right. \end{eqnarray} $

(3.1)

记$x(t)=(x_1(t), x_2(t), \cdots, x_n(t), y_1(t), y_2(t), \cdots, y_n(t))^T, X=Z=\{x\in C(R, {\Bbb R}^{2n}):x(t+\omega)=x(t)\}, $范数为

$ \|x\|=\sum^n\limits_{i=1}\max\limits_{t\in[0, \omega]}|x_i(t)|+\sum^n\limits_{i=1}\max\limits_{t\in[0, \omega]}|y_i(t)|, $

其中$|\cdot|$表示Euelid范数, 则在范数$\|\cdot\|$下$X, Z$都是Banach空间.

定义

$ Lx=x', Nx=(\Lambda_1(x, t), \Lambda_2(x, t), \cdots, \Lambda_n(x, t), \Delta_1(x, t), \Delta_2(x, t), \cdots, \Delta_n(x, t))^T, $

$ Px=Qx=\frac{1}{\omega}\int_0^\omega x(t){\rm d}t, x\in X. $

令

$ {\rm Dom}L=\{x\in X:x\in C^1(R, {\Bbb R}^{2n})\}\subset X\rightarrow X. $

易知$P$和$Q$是连续投影算子且${\rm Im}P={\rm Ker}L, {\rm Im}L={\rm Ker}Q={\rm Im}(I-Q).$进而可知Ker$L={\rm Im}P={\Bbb R}^{2n}, {\rm Im}L={\rm Ker}Q=\{x\in X:\bar{x}_i=0, \bar{y}_i=0, i=1, 2, \cdots, n\}$是$X$的闭子集, dim${\rm Ker}L=2n=\dim (Z/{\rm Im}L), $故$L$是指标为零Fredholm算子.定义$L$的广义逆为$K_P:{\rm Im}L\rightarrow {\rm Dom}L\bigcap {\rm Ker}P$如下

$ K_P(x)=\int_0^tx(s){\rm d}s-\frac{1}{\omega}\int_0^\omega\int_0^tx(s){\rm d}s{\rm d}t. $

于是

$ QNx=\left(\frac{1}{\omega}\int_0^t\Lambda_1(x, s){\rm d}s, \cdots, \frac{1}{\omega}\int_0^t\Lambda_n(x, s){\rm d}s, \frac{1}{\omega}\int_0^t\Delta_1(x, s){\rm d}s, \cdots, \frac{1}{\omega}\int_0^t\Delta_n(x, s){\rm d}s \right)^T, $

$ K_P(I-Q)Nx=\left(\varphi_1(x, t), \cdots, \varphi_2(x, t), \psi_1(x, t), \cdots, \psi_n(x, t)\right)^T, $

其中

$ \varphi_i(x, t)=\int_0^t\Lambda_i(x, s){\rm d}s-\frac{1}{\omega}\int_0^\omega\int_0^t\Lambda_i(x, s){\rm d}s{\rm d}t- \Big(\frac{t}{\omega}-\frac{1}{2}\Big)\int_0^t\Lambda_i(x, s){\rm d}s, $

$ \psi_i(x, t)=\int_0^t\Delta_i(x, s){\rm d}s-\frac{1}{\omega}\int_0^\omega \int_0^t\Delta_i(x, s){\rm d}s{\rm d}t- \Big(\frac{t}{\omega}-\frac{1}{2}\Big)\int_0^t\Delta_i(x, s){\rm d}s, $

利用Lebesgue收敛定理可以证明$QN$及$K_P(I-Q)N$是连续的.不难证明, 对$X$中的任何开的有界的子集$\Omega$, $QN(\bar{\Omega})$及$\bar{K_P(I-Q)N(\bar{\Omega})}$是相对紧的.因此, 对$X$的任何开的有界子集$\Omega$, $N$在$\bar{\Omega}$上是$L$紧的.对应算子方程$Lx=\lambda Nx, \lambda\in(0, 1)$有

$ \begin{equation}\label{eq:3.2} x'_i(t)=\lambda \Lambda_i(x, t), y'_i(t)=\lambda \Delta_i(x, t). \end{equation} $

(3.2)

对系统$(3.2)$在$[0, \omega]$上取积分得

$ \begin{equation} \int_0^\omega \bigg[h_i(t){\rm e}^{x_i(t)}+\sum^n\limits_{j=1, j\neq i}a_{ij}(t){\rm e}^{x_j(t)}+c_{i}(t){\rm e}^{x_i(t-\tau_i)}+d_i{\rm e}^{y_i(t)} \bigg]{\rm d}t=\bar{b}_i\omega. \end{equation} $

(3.3)

$ \begin{equation} \int_0^\omega r_i(t){\rm e}^{x_i(t)}{\rm d}t=\int_0^\omega D_i(t){\rm e}^{y_i(t)}{\rm d}t. \end{equation} $

(3.4)

由系统$(3.2)$和式(3.3)-(3.4) 有

$ \begin{equation} \int_0^\omega |x'_i(t)|{\rm d}t=\lambda\int_0^\omega|\Lambda_i(x, t)|{\rm d}t <2\int_0^\omega b_i(t){\rm d}t=2\bar{b}_i\omega. \end{equation} $

(3.5)

$ \begin{equation} \int_0^\omega |y'_i(t)|{\rm d}t=\lambda\int_0^\omega|\Delta_i(x, t)|{\rm d}t <2\int_0^\omega r_i(t){\rm e}^{x_i(t)}{\rm d}t. \end{equation} $

(3.6)

由式$(3.3)$解得

$ \begin{equation} \int_0^\omega {\rm e}^{x_i(t)}{\rm d}t <\frac{\bar{b}_i\omega}{h_{i}^L}. \end{equation} $

(3.7)

进而, 由式(3.6)-(3.7) 有

$ \begin{equation} \int_0^\omega |y'_i(t)|{\rm d}t < 2r_i^M\int_0^\omega {\rm e}^{x_i(t)}{\rm d}t <\frac{2r_i^M\bar{b}_i\omega}{h_{i}^L}. \end{equation} $

(3.8)

因为$x\in X$所以存在$\xi_i, \zeta_i, \eta_i, \sigma_i\in[0, \omega]$使得

$ \begin{equation} \begin{array}{l} x_i(\eta_i)=\max\limits_{t\in[0, \omega]}x_i(t), ~~x_i(\xi_i)=\min\limits_{t\in[0, \omega]}x_i(t), \\ y_i(\sigma_i)=\max\limits_{t\in[0, \omega]}y_i(t), ~~ y_i(\zeta_i)=\min\limits_{t\in[0, \omega]}y_i(t), ~~i=1, 2, \cdots, n. \end{array} \end{equation} $

(3.9)

一方面, 由式(3.7) 和(3.9) 解得

$ \begin{equation} x_i(\xi_i) <\ln{\frac{\bar{b}_i}{h_{i}^L}}. \end{equation} $

(3.10)

于是, $\forall t\in[0, \omega]$有

$ \begin{equation} x_i(t)\leq x_i(\xi_i)+\int_0^\omega |x'_i(t)|{\rm d}t<\ln{\frac{\bar{b}_i}{h_{i}^L}}+2\bar{b}_i\omega:=M_i. \end{equation} $

(3.11)

由式(3.4) 和(3.7) 解得

$ \begin{equation} y_i(\zeta_i)<\ln{\frac{r_i^M\bar{b}_i}{h_{i}^LD_i^L}}. \end{equation} $

(3.12)

因此, $\forall t\in[0, \omega]$有

$ \begin{equation} y_i(t)\leq y_i(\zeta_i)+\int_0^\omega |y'_i(t)|{\rm d}t<\ln{\frac{r_i^M\bar{b}_i}{h_{i}^LD_i^L}}+\frac{2r_i^M\bar{b}_i\omega}{h_{i}^L}:=L_i. \end{equation} $

(3.13)

另一方面, 利用式(3.11) 和(3.13), 由式(3.3) 解得

$ \begin{equation} x_i(\eta_i)>\ln{\frac{S^*_i}{h_{i}^M+c^M_i}}, \end{equation} $

(3.14)

其中

$ S_i^*=\bar{b}_i-\sum^n\limits_{j=1, j\neq i}a_{ij}^M{\rm e}^{M_j}-d_i^M{\rm e}^{L_i}>0, i=1, 2, \cdots, n. $

这样, $\forall t\in[0, \omega]$有

$ \begin{equation}x_i(t)\geq x_i(\eta_i)-\int_0^\omega |x'_i(t)|{\rm d}t>\ln{\frac{S_i^*}{h_{i}^M+c^M_i}}-2\bar{b}_i\omega:=m_i. \end{equation} $

(3.15)

再由式(3.4) 和(3.14), 可得

$ \begin{equation} y_i(\sigma_i)>\ln{\frac{r_i^LS^*_i}{h_{i}^MD_i^M}} \end{equation} $

(3.16)

进而, $\forall t\in[0, \omega]$有

$ \begin{equation} y_i(t)\geq y_i(\sigma_i)-\int_0^\omega |y'_i(t)|{\rm d}t>\ln{\frac{r_i^LS^*_i}{h_{i}^MD_i^M}}-\frac{2r_i^M\bar{b}_i\omega}{h_{i}^L}:=l_i. \end{equation} $

(3.17)

综上, 由式(3.11), (3.13), (3.15) 和(3.17) 有

$ \max\limits_{t\in[0, \omega]}|x_i(t)|\leq\max\{|m_i|, |M_i|\}:=A_{1i}, \max\limits_{t\in[0, \omega]}|y_i(t)|\leq\max\{|l_i|, |L_i|\}:= A_{2i}, i=1, 2, \cdots, n. $

显然$A_{1i}, A_{2i}, (i=1, 2, \cdots, n)$不依赖于参数$\lambda$.令$A=\sum^n\limits_{i=1}A_{1i}+\sum^n\limits_{i=1}A_{2i}+A_0$这里$A_0$由引理$2.2$所给出的.取$\Omega=\{x\in X:||x||<A\}, $对$\lambda\in(0, 1), x\in\partial \Omega\bigcap {\rm Ker}L=\partial\Omega\bigcap {\Bbb R}^{2n}$有$Lx\neq\lambda Nx$, 故$\Omega$满足引理$2.1$条件(a).当$x\in\partial\Omega\bigcap {\rm Ker}L$时, $x$是${\Bbb R}^{2n}$中的常值向量且$||x||=A, $由引理$2.1$知$QNx\neq0$构造映射$H_\mu(x)=\mu QNx+(1-\mu)G_x, \mu\in[0,1], $这里

$ G_x={(\bar{b}_i-\bar{h}_{i}{\rm e}^{x_{i}})_{n\times 1} \choose (\bar{r}_i{\rm e}^{x_i}-\bar{D}_i{\rm e}^{y_i})_{n\times 1}}. $

由引理2.2知, 对于$\mu\in[0,1], x\in\partial\Omega\bigcap {\rm Ker}L, H_\mu(x)\neq0$, 所以$H_\mu(x)$是一个同伦映射.根据同伦不变性, 我们取恒同映射$J=I$于是

$ {\rm deg}\{JQN, \Omega\cap {\rm Ker}L, 0\}={\rm deg}\{H_{1}(x), \Omega\cap {\rm Ker}L, 0\}= {\rm deg}\{H_{0}(x), \Omega\cap {\rm Ker}L, 0\}, $

由于代数方程组

$ \begin{eqnarray*} \left\{ \begin{array}{ll} \bar{b}_i-\bar{h}_{i}{\rm e}^{x_{i}}=0, \\ \bar{r}_i{\rm e}^{x_i}-\bar{D}_i{\rm e}^{y_i}=0, (i=1, 2, \cdots, n) \end{array} \right. \end{eqnarray*} $

存在唯一解$(x_1^*, x_2^*, \cdots, x_n^*, y_1^*, y_2^*, \cdots, y_n^*)^T, $其中

$ x_i^*=\ln{\frac{\bar{b}_i}{\bar{h}_{i}}}, ~~y_i^*=\ln{\frac{\bar{r}_i\bar{b}_i}{\bar{h}_{i}\bar{D}_i}}, ~~i=1, 2, \cdots, n. $

所以直接计算得

$ {\rm deg}\{JQN, \Omega\cap {\rm Ker}L, 0\}={\rm deg}\{G_{x}, \Omega\cap {\rm Ker}L, 0\}={\rm sgn}\bigg\{ \prod^n\limits_{i=1}\bar{a}_{ii}\bar{D}_{i}{\rm e}^{x_{i}^{*}+y_{i}^{*}} \bigg\}=1\neq0. $

这样$\Omega$满足引理$2.1$的条件(b), 因此方程$Lx=Nx$在${\rm Dom}L\bigcap\bar{\Omega}$至少有一个解, 进而由变换$N_{i}(t)={\rm e}^{x_i(t)}, P_i(t)={\rm e}^{y_i(t)}$ $(i=1, 2, \cdots, n), $知系统$(1.1)$在${\rm Dom}L\bigcap\bar{\Omega}$内至少存在一个$\omega$ -周期正解.证毕.

定理4.1 如果除初值(1.2) 和定理3.1的条件外, 还满足条件

$ \beta_i=h_{i}^L-r^M_i+c_{i}^M-\sum^n\limits_{j=1, j\neq i}a_{ij}^M>0, \eta_i=D_i^L-d_i^M>0, i=1, 2, \cdots, n, $

则系统(1.1) 存在唯一的$\omega$ -周期正解且为全局吸引的.

证 设$(N(t), P(t))$和$(\widetilde{N}(t), \widetilde{P}(t))$分别是系统(1.1) 的周期正解和任意正解, 其中

$ \widetilde{N}(t)=(\widetilde{N}_1(t), \cdots, \widetilde{N}_n(t))^T, \widetilde{P}(t)=(\widetilde{P}_1(t), \cdots, \widetilde{P}_n(t))^T. $

由引理2.3知存在$T>0$, 当$t>T$时, $(N(t), P(t))\in D, (\widetilde{N}(t), \widetilde{P}(t))\in D$构造Lyapunov函数

$ V_i(t)=|\ln{N_i(t)}-\ln{\widetilde{N}_i(t)}|+|\ln{P_i(t)}-\ln{\widetilde{P}_i(t)}| +c_i^M\int_{t-\tau_i}^t |N_i(s)-\tilde{N}_i(s)|{\rm d}s, $

直接计算系统(1.1) 的的右上导数, 得

$ \begin{eqnarray*} V'_i(t)&=&{\rm sgn}(N_i(t)-\widetilde{N}_i(t))\{-h_i(t)(N_i(t)-\widetilde{N}_i(t)) \\ && -\sum^n\limits_{j=1, j\neq i}a_{ij}(t)(N_j(t)-\widetilde{N}_j(t)) -c_{i}(t)(N_i(t-\tau_{i})-\widetilde{N}_i(t-\tau_{i})) \\ && -d_i(t)(P_i(t)-\widetilde{P}_i(t))\} +{\rm sgn}(P_i(t)-\widetilde{P}_i(t))\{r_i(t)(N_i(t)-\widetilde{N}_i(t)) \\ &&-D_i(t)(P_i(t)-\widetilde{P}_i(t))\} +c_i^M|N_i(t)-\widetilde{N}_i(t)|-c_i^M|N_i(t-\tau_i)-\widetilde{N}_i(t-\tau_i)| \\ &\leq& -(h_{i}^L-r_i^M-c_{i}^M)|N_i(t)-\widetilde{N}_i(t)| \\ && +\sum^n\limits_{j=1, j\neq i}a_{ij}^M|(N_j(t)-\widetilde{N}_j(t))|-(D_i^L-d_i^M)|P_i(t)-\widetilde{P}_i(t)|. \end{eqnarray*} $

定义Lyapunov函数

$ V(t)=V_1(t)+V_2(t)+\cdots+V_n(t), $

据此, 计算系统(1.1) 的右上导数为

$ V'(t)\leq-\sum^n\limits_{i=1}\bigg[h_{i}^L-r^M_i-c_{i}^M-\sum^n\limits_{j=1, j\neq i}a_{ij}^M\bigg]|N_i(t)-\widetilde{N}_i(t)| -\sum^n\limits_{i=1}(D_i^L-d_i^M)|P_i(t)-\widetilde{P}_i(t)|. $

取$\alpha=\min\limits_{1\leq i\leq n}\{\beta_i, \eta_i\}$, 从$T$到$t$积分得

$ V(t)+\alpha\int_T^t \bigg\{\sum^n\limits_{i=1}[|N_i(s)-\widetilde{N}_i(s)|+|P_i(s)-\widetilde{P}_i(s)|] \bigg\}{\rm d}s<V(T)<+\infty, t>T, $

$ \sum^n\limits_{i=1}[|N_i(t)-\widetilde{N}_i(t)|+|P_i(t)-\widetilde{P}_i(t)|]\in L^1[T, +\infty), $

$ \sup\int_T^t \bigg\{\sum^n\limits_{i=1}[|N_i(s)-\widetilde{N}_i(s)|+|P_i(s)-\widetilde{P}_i(s)|] \bigg \}{\rm d}s<\frac{V(T)}{\alpha}<+\infty. $

由系统(1.1) 的一致持久性及其解的导数有界性知

$ \sum^n\limits_{i=1}[|N_i(t)-\widetilde{N}_i(t)|+|P_i(t)-\widetilde{P}_i(t)|] $

在$[0.+\infty]$上是一致连续的, 根据引理2.4知

$ \lim\limits_{t\rightarrow+\infty}\sum^n\limits_{i=1}[|N_i(t)-\widetilde{N}_i(t)|+|P_i(t)-\widetilde{P}_i(t)|]=0. $

综上, 系统(1.1) 在区域$D$内存在唯一的$\omega$ -周期正解且为全局吸引的.证毕.