具有Holling-III型食饵自食性的Leslie-Gower时滞捕食模型的周期解

Periodic Solution for a Delayed Leslie-Gower and Holling-Type III Predator-Prey Model Incorporating Prey Cannibalism

摘要: 本文研究具有Holling-III型食饵自食性的Leslie-Gower时滞捕食模型的周期解，了解食饵与捕食者之间的动态关系。利用重合度理论的连续性定理以及比较定理，如果满足条件，则系统存在一个ω-周期正解。

Abstract: In this paper, we study the periodic solution of a delayed Leslie-Gower and Holling-Type III predator-prey model incorporating prey cannibalism to understand the dynamic relationship between prey and predator. Using the continuity theorem of coincidence degree theory and comparison theorem, if the condition is met, then the system has a ω-positive periodic solution.

1. 引言

生态系统中存在多种捕食行为，捕食关系是生物数学界研究的一个重要课题。近几十年来，传统捕食模型由于其非常丰富的动态和在可再生资源管理中的重要作用受到了广泛的关注和研究 [1] - [6]。然而，同类相食(种内捕食)的行为在很多种群也会发生，使得简单的食物网变得更为复杂。同类相食是指同一类型的动物或者植物为了生存繁殖需要或者某种目的互相厮杀竞争的现象，也是一种种群内部的食饵物种–捕食物种之间的相互作用。它是一种消耗相同物种并有助于提供食物来源的行为，这种行为特征已经在各种各样的动物中被发现，包括鱼类，蜘蛛，浮游生物和昆虫等种群，因此研究具有食饵自食性的种群模型动力学行为尤为重要 [7]。近年来，具有Holling-II型的Leslie-Gower捕食模型已经得到了广泛的研究 [8] [9] [10]，因为Holling-III型可以清楚地描述食饵物种与捕食者之间的关系，且时滞微分方程比常微分方程更能展现出生物界复杂的动态关系，研究具有时滞的周期系统的周期解的存在性是有十分重要的意义的。所以本文考虑具有Holling-III型食饵自食性的Leslie-Gower时滞捕食模型：

$\{\begin{array}{l}{H}^{\prime }\left(t\right)=H\left(t\right)\left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right)P\left(t-{\tau }_1\left(t\right)\right)-b_1\left(t\right)H\left(t\right)\right]-\frac{f\left(t\right)H\left(t\right)H\left(t-\sigma \left(t\right)\right)}{d^2\left(t\right)+H^2\left(t-\sigma \left(t\right)\right)},\\ P^{\prime }\left(t\right)=P\left(t\right)\left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{P\left(t-{\tau }_2\left(t\right)\right)}{H\left(t-{\tau }_2\left(t\right)\right)}\right],\end{array}$ (1)

其中 $H\left(t\right)$， $P\left(t\right)$ 分别表示在时间t时食饵和捕食者的密度， ${\gamma }_1\left(t\right)$， $i=1,2$ 分别是食饵和捕食者的内在

增长率， $\frac{{\gamma }_1\left(t\right)}{b_1\left(t\right)}$ 是食饵的环境承载力， $f\left(t\right)$ 表示食饵的同类相食率， $C\left(H\right)\left(t\right)=f\left(t\right)\times H\left(t\right)\times \frac{H\left(t\right)}{H^2\left(t\right)+d^2(\; t\; )}$

是同类相食行为， $c_1\left(t\right)\times H\left(t\right)$ 是由于自相残杀而产生的新后代。因为食饵会掠夺许多食饵以产生一个新后代，所以 $c_1\left(t\right)<f\left(t\right)$。 ${\alpha }_1\left(t\right)$， ${\alpha }_2\left(t\right)$， $b_1\left(t\right)$， $d\left(t\right)$， $\sigma \left(t\right)$， ${\tau }_i\left(t\right)$， $i=1,2$ 均为严格正的 $\omega$ -周期连

续函数 $\left(\omega >0\right)$， ${\gamma }_1\left(t\right)$， ${\gamma }_2\left(t\right)$， $c_1\left(t\right)$ 是 $\omega$ -周期连续函数，假设 ${\displaystyle {\int }_0^{\omega }{c}_1\left(t\right)\text{d}t}>0$， ${\displaystyle {\int }_0^{\omega }{\gamma }_i\left(t\right)\text{d}t}>0$， $i=1,2$。

2. 预备知识

设 $X$， $Z$ 为实Banach空间， $L$ ： $\text{Dom}L\subset X\to Z$ 为线性映射， $N$ ： $X\to Z$ 为连续映射，且记 $K_P$ 为 $L_P$ 的倒数。如果满足条件 $\dim \text{Ker}L=\text{codimIm}L<+\infty$， $\mathrm{Im}L$ 在 $Z$ 中为闭，则称 $L$ 为指标为零的Fredholm映射。如果映射 $L$ 是指标为零的Fredholm映射，则存在连续的投影 $P$ ： $X\to X$ 、 $Q$ ： $Z\to Z$ 有 $\text{Ker}Q=\mathrm{Im}L=\text{Im}\left(I-Q\right)$， $\mathrm{Im}P=\text{Ker}L$。可见 $L$ 对 $\text{Dom}L\cap \text{Ker}P$ 的限制性 $L_P$ ： $\left(I-P\right)X\to \mathrm{Im}L$ 是可逆的。如果 $\text{Ω}$ 是 $X$ 的一个开放边界子集， $QN\left(\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}\right)$ 有界， $K_P\left(I-Q\right)N$ ： $\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}\to X$ 紧，那么可知 $Z$ 在 $\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}$ 上是L-紧。根据 $\mathrm{Im}Q$ 与 $\text{Ker}L$ 是同构性，可知存在同构映射 $J$ ： $\mathrm{Im}L\to \text{Ker}L$。

引理2.1 [11] (连续性定理) 若 $\text{Ω}\subset X$ 是一个有界开集， $L$ 是指标为零的Fredholm映射，且 $N$ 在 $\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}$ 上是L-紧的，假设

(i) 对于任意的 $\lambda \in \left(0,1\right)$， $x\in \partial \text{Ω}\cap \text{Dom}L$， $Lx\ne \lambda Nx$ ；

(ii) 对于任意的 $x\in \partial \text{Ω}\cap \text{Ker}L$， $QNx\ne 0$ ；

(iii) $\text{deg}\{JQN,\text{Ω}\cap \text{Ker}L,0\}\ne 0$。

那么 $Lx=Nx$ 在 $\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}\cap \text{Dom}L$ 内至少有一个解。

引理2.2 [12] 假设 $y\in P{C}_{\omega }^1=\{x:x\in C^1\left(R,R\right),x\left(t+\omega \right)\equiv x\left(t\right)\}$，则

$0\le \underset{s\in \left[0,\omega \right]}{\max }y\left(s\right)-\underset{s\in \left[0,\omega \right]}{\min }y\left(s\right)\le \frac{1}{2}{\displaystyle {\int }_0^{\omega }\left|y^{\prime }\left(s\right)\right|\text{d}s}.$

3. 周期解的存在性

方便起见，记 $\bar{f}=\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }f\left(t\right)\text{d}t}$，其中 $f\left(t\right)$ 是 $\omega$ -周期连续函数。

定理3.1 如果条件 $\overline{{\gamma }_1+c_1}>\overline{{\alpha }_1}{\text{e}}^{{\beta }_2}+\overline{\left(\frac{f}{d^2}\right)}{\text{e}}^{{\beta }_1}$ 成立，那么系统(1)至少存在一个 $\omega$ -周期正解。

证明 令 $H\left(t\right)={\text{e}}^{x_1\left(t\right)}$， $P\left(t\right)={\text{e}}^{x_2\left(t\right)}$，则系统(1)变为

$\{\begin{array}{l}{x^{\prime }}_1\left(t\right)={\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}},\\ {x^{\prime }}_2\left(t\right)={\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}},\end{array}$ (2)

从上可知系统(2)具有一个以 $\omega$ 为周期的解 ${\left(x_1^{\ast }\left(t\right),x_2^{\ast }\left(t\right)\right)}^{\text{T}}$，则 ${\left(H^{\ast }\left(t\right),P^{\ast }\left(t\right)\right)}^{\text{T}}={\left({\text{e}}^{x_1^{\ast }\left(t\right)},{\text{e}}^{x_2^{\ast }\left(t\right)}\right)}^{\text{T}}$

是系统(1)的一个以 $\omega$ 为周期的周期正解。所以，在这里只要证明系统(2)至少有一个 $\omega$ -周期解即可。

令 $X=Z=\left\{x\left(t\right)={\left(x_1\left(t\right),x_2\left(t\right)\right)}^{\text{T}}\in C\left(R,R^2\right):x\left(t+\omega \right)=x\left(t\right)\right\}$，且 $\Vert x\Vert =\underset{s\in \left[0,\omega \right]}{\max }\left\{\left|x_1\left(t\right)+x_2\left(t\right)\right|\right\}$，那么在 $\Vert \cdot \Vert$ 范数下， $X$ 和 $Z$ 为Banach空间。

定义映射 $L$， $P$， $Q$ 分别为

$L:\text{Dom}L\cap X\to Z,Lx=x^{\prime };$

$P\left(x\right)=\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }x\left(t\right)\text{d}t},x\in X;$

$Q\left(x\right)=\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }z\left(t\right)\text{d}t},z\in Z.$

其中 $DomL=\left\{x\in X:x\left(t\right)\in c^1\left(R,R^2\right)\right\}$。

定义变换 $N:X\to Z$

$Nx=\left[\begin{array}{c}{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\\ {\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\end{array}\right],$

显然， $\text{Ker}L=R^2$， $\mathrm{Im}L=\left\{z|z\in Z,{\displaystyle {\int }_0^{\omega }z\left(t\right)\text{d}t=0}\right\}$ 为 $Z$ 中的闭集， $\text{Ker}L=\text{codimIm}I=2$， $P$， $Q$ 是满足 $\mathrm{Im}P=\text{Ker}L$， $\text{Ker}Q=\mathrm{Im}L=\mathrm{Im}\left(I-Q\right)$ 的连续映射，故L是指标为零的Fredholm算子，而且 $K_P$ 为 $K_P\left(z\right)={\displaystyle {\int }_0^{t}z\left(s\right)\text{d}s}-\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }{\displaystyle {\int }_0^{t}z\left(s\right)\text{d}s\text{d}t}}$，于是

$QNx=\left[\begin{array}{c}\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }\left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right]}\\ \frac{1}{\omega }{\displaystyle {\int }_0^{\omega }\left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]}\end{array}\right],$

$\begin{array}{c}{K}_P\left(I-Q\right)Nx=\left(\begin{array}{c}{\displaystyle {\int }_0^t\left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right]}\\ {\displaystyle {\int }_0^t\left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]}\end{array}\right)\\ -\left(\begin{array}{c}\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }{\displaystyle {\int }_0^t\left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right]}}\\ \frac{1}{\omega }{\displaystyle {\int }_0^{\omega }{\displaystyle {\int }_0^t\left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]}}\end{array}\right)\\ -\left(\begin{array}{c}\left(\frac{t}{\omega }-\frac{1}{2}\right){\displaystyle {\int }_0^{\omega }\left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right]}\\ \left(\frac{t}{\omega }-\frac{1}{2}\right){\displaystyle {\int }_0^{\omega }\left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]}\end{array}\right).\end{array}$

易证， $QN$ 和 $K_P\left(I-Q\right)N$ 都是连续的。对于任意的有界开集 $\text{Ω}\subset X$， $QN\left(\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}\right)$， $K_P\left(I-Q\right)N\left(\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}\right)$ 紧，所以有 $N$ 在 $\text{<mover accent="true"> <mo> Ω <mo> ¯ </mo> </mo> </mover>}$ 上是L-紧的。

对应方程 $Lx=\lambda Nx$， $\lambda \in \left(0,1\right)$，有

$\{\begin{array}{l}{x^{\prime }}_1\left(t\right)=\lambda \left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right],\\ {x^{\prime }}_2\left(t\right)=\lambda \left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right],\end{array}$ (3)

将(3)式从 $\left[0,\omega \right]$ 积分有

$\{\begin{array}{l}{\displaystyle {\int }_0^{\omega }\left[{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right]\text{d}t}=0,\\ {\displaystyle {\int }_0^{\omega }\left[{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]\text{d}t}=0,\end{array}$ (4)

由(4)得

${\displaystyle {\int }_0^{\omega }\left[{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}+b_1\left(t\right){\text{e}}^{x_1\left(t\right)}+\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right]\text{d}t}=\overline{{\gamma }_1+c_1}\omega ,$ (5)

${\displaystyle {\int }_0^{\omega }\left[{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]\text{d}t}=\overline{{\gamma }_2}\omega ,$ (6)

(3)结合(5)有

$\begin{array}{c}{\displaystyle {\int }_0^{\omega }\left|x_1\left(t\right)\right|\text{d}t}=\lambda {\displaystyle {\int }_0^{\omega }\left|{\gamma }_1\left(t\right)+c_1\left(t\right)-{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}-b_1\left(t\right){\text{e}}^{x_1\left(t\right)}-\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\right|\text{d}t}\\ \le \left(\overline{\left|{\gamma }_{１}+c_1\right|}+\overline{{\gamma }_{１}+c_1}\right)\omega .\end{array}$ (7)

存在 ${\xi }_i,{\eta }_i,i=1,2$ 满足

$x_i\left({\xi }_i\right)=\underset{s\in \left[0,\omega \right]}{\text{sup}}{x}_i\left(t\right),x_i\left({\eta }_i\right)=\underset{s\in \left[0,\omega \right]}{\inf }{x}_i\left(t\right),$ (8)

由(5)和(8)得

$\overline{{\gamma }_1+c_1}\omega \ge {\displaystyle {\int }_0^{\omega }{b}_1\left(t\right){\text{e}}^{x_1\left(t\right)}}\text{d}t\ge \omega \overline{b_1}{\text{e}}^{x_1\left({\eta }_1\right)},$ (9)

化简有

$x_1\left({\eta }_1\right)\le \ln \frac{\overline{{\gamma }_1+c_1}}{\overline{b_1}}.$ (10)

结合(7)，(10)和引理2.2得

$x_1\left(t\right)\le x_1\left({\eta }_1\right)+\frac{1}{2}{\displaystyle {\int }_0^{\omega }\left|x_1\left(t\right)\right|\text{d}t}\le \ln \frac{\overline{{\gamma }_1+c_1}}{\overline{b_1}}+\frac{1}{2}\left(\overline{\left|{\gamma }_{１}+c_1\right|}+\overline{{\gamma }_{１}+c_1}\right)\omega =:{\beta }_1.\text{}$ (11)

根据(3)和(6)有

${\displaystyle {\int }_0^{\omega }\left|x_2\left(t\right)\right|\text{d}t}=\lambda {\displaystyle {\int }_0^{\omega }\left|{\gamma }_2\left(t\right)-{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right|\text{d}t}\le \left(\overline{\left|{\gamma }_2\right|}+\overline{{\gamma }_2}\right)\omega ,$ (12)

由(6)，(8)和(11)得

$\frac{\omega \overline{{\alpha }_2}{\text{e}}^{x_2\left({\eta }_2\right)}}{{\text{e}}^{{\beta }_1}}\le {\displaystyle {\int }_0^{\omega }\left[{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]\text{d}t}=\overline{{\gamma }_2}\omega ,$ (13)

化简得

$x_2\left({\eta }_2\right)\le \ln \frac{\overline{{\gamma }_2}{\text{e}}^{{\beta }_1}}{\overline{{\alpha }_2}}=\ln \frac{\overline{{\gamma }_2}}{\overline{{\alpha }_2}}+{\beta }_1,$ (14)

结合(12)，(14)和引理2.2得

$x_2\left(t\right)\le x_2\left({\eta }_2\right)+\frac{1}{2}{\displaystyle {\int }_0^{\omega }\left|x_2\left(t\right)\right|\text{d}t}\le \ln \frac{\overline{{\gamma }_2}}{\overline{{\alpha }_2}}+{\beta }_1+\frac{1}{2}\left(\overline{\left|{\gamma }_2\right|}+\overline{{\gamma }_2}\right)\omega =:{\beta }_2.$ (15)

由(4)和(8)得

$\begin{array}{c}\omega \overline{b_1}{\text{e}}^{x_1\left({\xi }_1\right)}\ge \overline{{\gamma }_1+c_1}\omega -{\displaystyle {\int }_0^{\omega }{\alpha }_1\left(t\right){\text{e}}^{x_2\left(t-{\tau }_1\left(t\right)\right)}}\text{d}t-{\displaystyle {\int }_0^{\omega }\frac{f\left(t\right){\text{e}}^{x_1\left(t-\sigma \left(t\right)\right)}}{d^2\left(t\right)+{\text{e}}^{2x_1\left(t-\sigma \left(t\right)\right)}}\text{d}t}\\ \ge \overline{{\gamma }_1+c_1}\omega -\overline{{\alpha }_1}\omega {\text{e}}^{{\beta }_2}-\overline{\left(\frac{f}{d^2}\right)}\omega {\text{e}}^{{\beta }_{１}},\end{array}$ (16)

即

$x_{１}\left({\xi }_{１}\right)\ge \ln \frac{\overline{{\gamma }_1+c_1}-\overline{{\alpha }_1}{\text{e}}^{{\beta }_2}-\overline{\left(\frac{f}{d^2}\right)}{\text{e}}^{{\beta }_1}}{\overline{b_1}}.$ (17)

由(7)，(17)和引理2.2得

$\begin{array}{c}{x}_1\left(t\right)\ge x_1\left({\xi }_1\right)-\frac{1}{2}{\displaystyle {\int }_0^{\omega }\left|x_1\left(t\right)\right|\text{d}t}\\ \ge \ln \frac{\overline{{\gamma }_1+c_1}-\overline{{\alpha }_1}{\text{e}}^{{\beta }_2}-\overline{\left(\frac{f}{d^2}\right)}{\text{e}}^{{\beta }_1}}{\overline{b_1}}-\frac{1}{2}\left(\overline{\left|{\gamma }_{１}+c_1\right|}+\overline{{\gamma }_{１}+c_1}\right)\omega =:{\beta }_3.\end{array}$ (18)

结合(6)，(8)和(18)得

$\frac{\overline{{\alpha }_2}\omega {\text{e}}^{x_2\left({\xi }_2\right)}}{{\text{e}}^{{\beta }_3}}\ge {\displaystyle {\int }_0^{\omega }\left[{\alpha }_2\left(t\right)\frac{{\text{e}}^{x_2\left(t-{\tau }_2\left(t\right)\right)}}{{\text{e}}^{x_1\left(t-{\tau }_2\left(t\right)\right)}}\right]\text{d}t}=\overline{{\gamma }_2}\omega ,$ (19)

化简有

$x_2\left({\xi }_2\right)\ge \ln \frac{\overline{{\gamma }_2}{\text{e}}^{{\beta }_3}}{\overline{{\alpha }_2}}=\ln \frac{\overline{{\gamma }_2}}{\overline{{\alpha }_2}}+{\beta }_3.$ (20)

根据(12)，(20)和引理2.2得

$\begin{array}{c}{x}_2\left(t\right)\ge x_2\left({\xi }_2\right)-\frac{1}{2}{\displaystyle {\int }_0^{\omega }\left|x_2\left(t\right)\right|\text{d}t}\\ \ge \ln \frac{\overline{{\gamma }_2}}{\overline{{\alpha }_2}}+{\beta }_3-\frac{1}{2}\left(\overline{\left|{\gamma }_2\right|}+\overline{{\gamma }_2}\right)\omega =:{\beta }_4.\end{array}$ (21)

结合(11)，(15)，(18)，(21)有 $\Vert x\Vert \le \left|{\beta }_1\right|+\left|{\beta }_2\right|+\left|{\beta }_3\right|+\left|{\beta }_4\right|=:{\beta }_0$ ( ${\beta }_0$ 与 $\lambda$ 无关)。

考虑下面方程组

$\{\begin{array}{l}\overline{{\gamma }_1+c_1}-\overline{{\alpha }_1}{\text{e}}^{x_2}-\overline{b_1}{\text{e}}^{x_1}-\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }\frac{f\left(t\right){\text{e}}^{x_1}}{d^2\left(t\right)+{\text{e}}^{2x_1}}\text{d}t=0,}\\ \overline{{\gamma }_2}-\frac{1}{\omega }{\displaystyle {\int }_0^{\omega }{\alpha }_2\frac{{\text{e}}^{x_2}}{{\text{e}}^{x_1}}\text{d}t=0,}\end{array}$ (22)

若(22)有解 $x^{\ast }={\left(x_1^{\ast },x_2^{\ast }\right)}^{\text{T}}$，则满足 $\Vert {\left(x_1^{\ast },x_2^{\ast }\right)}^{\text{T}}\Vert =\max \left\{\left|x_1^{\ast }\right|+\left|x_2^{\ast }\right|\right\}<{\beta }_0$。当 $x\in \partial \text{Ω}$，且 $\lambda \in \left(0,1\right)$，满足引理2.1 (i)。假设当 $x\in \partial \text{Ω}\cap \text{Ker}L$ 且 $\Vert x\Vert ={\beta }_0$， $QNx\ne 0$，满足引理2.1 (ii)。定义 $J:\mathrm{Im}Q\to \ker L$， ${\left(x_1,x_2\right)}^{\text{T}}={\left(x_1,x_2\right)}^{\text{T}}$，可直接算得 $\text{deg}\{JQN,\text{Ω}\cap \text{Ker}L,0\}\ne 0$，满足引理2.1 (iii)。因此，(2)至少一个 $\omega$ -周期解。所以，系统(1)至少有一个 $\omega$ -周期正解。即可证定理3.1。

基金项目

国家自然科学基金青年项目(No. 11801398)。

NOTES

^*第一作者。

^#通讯作者。