一类集合生态系统正周期解的存在性

doi:10.12677/PM.2021.112028

期刊菜单

一类集合生态系统正周期解的存在性
Existence of Positive Periodic Solutions for a Class of Meta-Ecosystems

DOI: 10.12677/PM.2021.112028, PDF, HTML, XML, 下载: 487 浏览: 1,752
作者: 刘菁, 杨喜陶：湖南科技大学数学与计算科学学院，湖南湘潭
关键词: 集合生态系统；正周期解；时滞；连续理论；Meta-Ecosystems； Positive Periodic Solutions； Delay； Continuation Theorem

摘要: 本文聚焦于自然界中一类集合生态系统的动力学行为，在微分方程理论及连续理论的基础上，研究其正周期解的存在性。

Abstract: The dynamic behaviors for a class of meta-ecosystems were studied based on the differential equations theory and the continuation theorem, and we will study the existence of positive periodic solutions for the model.

文章引用：刘菁, 杨喜陶. 一类集合生态系统正周期解的存在性[J]. 理论数学, 2021, 11(2): 201-207. https://doi.org/10.12677/PM.2021.112028

1. 引言

最近，Messan和Kopp等人为了研究元生态系统中两个生态系统之间的双向资源交换的动态产出，在 [1] 中建立了如下生态模型：

${\begin{cases} \frac{d P}{d t} = r_{p} P [1 - \frac{P}{K_{p} + a_{p} Q}] - b_{p} Q P \\ \frac{d Q}{d t} = r_{q} Q [1 - \frac{Q}{K_{q} + a_{q} P}] - b_{q} Q P \end{cases}$ ，

其中， $P (t)$ 、 $Q (t)$ 是在时间t进行资源交换的两个相邻生态系统的产出， $r_{i}, i = p, q$ 是生态系统中生物量的内在增长率， $K_{i}$ 是在没有资源交换的情况下生态系统i的承载能力； $a_{i}$ 表示从贡献者生态系统转移到接收系统i的资源数量； $b_{i}$ 表示生态系统层面上供体生态系统i的资源枯竭率。 [1] 指出上述系统当

$β_{i} = \frac{b_{i}}{r_{i}} K_{j} < 1 (i \neq j)$

时为一致持久的。

事实上，动力系统不仅仅依赖与现在和过去的状态，还与时滞的衍生物有关 [2] [3] [4] [5]。正如Kuang在 [6] 中所指出的，任何没有时滞的物种动态模型在最好情况下也只能是近似值。时滞模型在物理学 [7]、神经网络 [8] [9]、种群动力学等领域有着重要的应用，更多关于时滞的重要性的详细论述可以参见Kuang [6] 和Gopalsamy [10] 的经典书籍。是以在本文建立了以下带有时滞的模型：

${\begin{cases} \frac{d x}{d t} = x (t) [r_{1} (t) - \frac{r_{1} (t) x (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) y (t)} - b_{1} (t) y (t)] \\ \frac{d y}{d t} = y (t) [r_{2} (t) - \frac{r_{2} (t) y (t - τ_{2} (t))}{K_{2} (t) + a_{2} (t) x (t)} - b_{2} (t) x (t)] \end{cases}$

$r_{i} (t), K_{i} (t), a_{i} (t), b_{i} (t), τ_{i} (t), i = 1, 2$ 是定义在R上的非负连续有界函数。我们得到了该系统一致持久的充分条件。

在现实生活中，因为环境的变化具有周期性，是以研究周期系统及其周期解的存在性是非常重要的。近年来，研究周期解的存在性已经有了非常多的研究结果。受上述的启发，在本文我们将主要研究以下集合生态系统：

其中 $r_{i} (t), K_{i} (t), a_{i} (t), b_{i} (t), τ_{i} (t), i = 1, 2$ 是定义在R上的非负连续有界的 $ω$ 周期函数。据我们所知，系统(1)周期解的存在性未有相应成果,本文将致力于解决这一问题。

2. 正周期解的存在性

引理1 (连续理论 [11] [12] ) 假设X和Z是两个Banach空间，映射 $L : D o m L \subset X \to Z$ 是一个指标为0的Fredholm算子，映射 $N : X \to Z$ 在 $\bar{Ω}$ 上L-紧的。 $Ω$ 是X的一个有界开子集，另外，假设下列所有条件满足：

1) $\forall λ \in (0, 1), \forall x \in \partial Ω \cap D o m L, L x \neq λ N x$ ，

2) $Q N x \neq 0$ ，对 $\forall x \in \partial Ω \cap \ker L$ ，

3) $\deg {J Q N, Ω \cap \ker L, 0} \neq 0$ ，

那么方程 $L x = N x$ 在 $D o m L \cap \bar{Ω}$ 上至少存在一个解。

定理1 假设 $r_{i} (t), K_{i} (t), a_{i} (t), i = 1, 2$ 是定义在R上的非负连续有界的 $ω$ 周期函数，

$\exp M_{1} < K_{1} (t), \exp M_{2} < K_{2} (t), t \in [0, ω],$

则系统(1)至少有一个 $ω$ 正周期解。其中

$M_{1} = A_{2} + 2 \int_{0}^{ω} r_{2} (t) d t, M_{2} = A_{1} + 2 \int_{0}^{ω} r_{1} (t) d t, A_{1} = \ln \frac{\int_{0}^{ω} r_{2} (t) d t}{\int_{0}^{ω} b_{2} (t) d t}, A_{2} = \ln \frac{\int_{0}^{ω} r_{1} (t) d t}{\int_{0}^{ω} b_{1} (t) d t} .$

证明：考虑到(1)的现实生物意义，在此本文仅聚焦于该系统正周期解的存在性。因此，令

$x (t) = e^{u_{1} (t)}, y (t) = e^{u_{2} (t)},$

方程组(1)变为：

${\begin{cases} {u^{'}}_{1} (t) = r_{1} (t) - \frac{r_{1} (t) \exp u_{1} (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) \exp u_{2} (t)} - b_{1} (t) \exp u_{2} (t) \\ {u^{'}}_{2} (t) = r_{2} (t) - \frac{r_{2} (t) \exp u_{2} (t - τ_{2} (t))}{K_{2} (t) + a_{2} (t) \exp u_{1} (t)} - b_{2} (t) \exp u_{1} (t) \end{cases}$ (2)

其中所有函数定义同(1)。显然若(2)有一个周期解 ${(u_{1}^{*} (t), u_{2}^{*} (t))}^{T}$ ，那么

${(x^{*} (t), y^{*} (t))}^{T} = {(\exp {u_{1}^{*} (t)}, \exp {u_{2}^{*} (t)})}^{T}$

是(1)的一个 $ω$ 周期解。因此，我们只需证明定理1的条件能保证系统(2)有一个 $ω$ 周期解即可。令 $X = {(x_{1}, x_{2}) | x_{i} \in C (R), x_{i} (t + w) = x_{i} (t), \forall t \in R, i = 1, 2}, \forall (x_{1}, x_{2}) \in X$ ，定义

$‖ (x_{1}, x_{2}) ‖ = \max_{0 \leq t \leq ω} (| x_{1} (t) | + | x_{2} (t) |) .$

则X按照上述范数构成Banach空间。令 $Z = X, \forall (u_{1}, u_{2}) \in X$ ，定义 $L (u_{1}, u_{2}) = ({u^{'}}_{1}, {u^{'}}_{2})$ ，则

$D o m L = {(u_{1}, u_{2}) \in X | ({u^{'}}_{1}, {u^{'}}_{2}) \in X}, K e r L = R^{2},$

$Im L = {(u_{1}, u_{2}) \in D o m L | \int_{0}^{ω} u_{1} (t) d t = \int_{0}^{ω} u_{2} (t) d t = 0}, Z / Im L = R^{2} .$

分别定义P、Q算子如下形式：

$P : X \cap D o m L \to K e r L, P (u_{1}, u_{2}) = (\frac{1}{ω} \int_{0}^{ω} u_{1} (t) d t, \frac{1}{ω} \int_{0}^{ω} u_{2} (t) d t) .$

$Q : Z \to Z / Im L, Q (u_{1}, u_{2}) = (\frac{1}{ω} \int_{0}^{ω} u_{1} (t) d t, \frac{1}{ω} \int_{0}^{ω} u_{2} (t) d t) .$

从而L是指标为0的Fredholm算子。令 $N : X \to X$ ：

$N (\begin{matrix} u_{1} \\ u_{2} \end{matrix}) = (\begin{matrix} r_{1} (t) - \frac{r_{1} (t) \exp u_{1} (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) \exp u_{2} (t)} - b_{1} (t) \exp u_{2} (t) \\ r_{2} (t) - \frac{r_{2} (t) \exp u_{2} (t - τ_{2} (t))}{K_{2} (t) + a_{2} (t) \exp u_{1} (t)} - b_{2} (t) \exp u_{1} (t) \end{matrix}) .$

$Ω \subset X$ 为有界开集，则

$Q N (\begin{matrix} u_{1} \\ u_{2} \end{matrix}) = \frac{1}{ω} (\begin{matrix} \int_{0}^{ω} (r_{1} (t) - \frac{r_{1} (t) \exp u_{1} (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) \exp u_{2} (t)} - b_{1} (t) \exp u_{2} (t)) d t \\ \int_{0}^{ω} (r_{2} (t) - \frac{r_{2} (t) \exp u_{2} (t - τ_{2} (t))}{K_{2} (t) + a_{2} (t) \exp u_{1} (t)} - b_{2} (t) \exp u_{1} (t)) d t \end{matrix}) .$

从而 $Q N$ 连续，且 $Q N (\bar{Ω})$ 有界。

L的广义逆 $K_{p} : Im L \to K e r P \cap D o m L$ ：

$K_{p} (\begin{matrix} u_{1} \\ u_{2} \end{matrix}) = (\begin{matrix} \int_{0}^{t} u_{1} (s) d s - \frac{1}{ω} \int_{0}^{ω} \int_{0}^{t} u_{1} (s) d s d t \\ \int_{0}^{t} u_{2} (s) d s - \frac{1}{ω} \int_{0}^{ω} \int_{0}^{t} u_{2} (s) d s d t \end{matrix}) .$

从而 $K_{P} (I - Q) N (\begin{matrix} u_{1} \\ u_{2} \end{matrix}) = (\begin{matrix} g (t) \\ h (t) \end{matrix})$ ，其中

$\begin{matrix} g (t) = \int_{0}^{t} (r_{1} (s) - \frac{r_{1} (s) \exp u_{1} (s - τ_{1} (s))}{K_{1} (s) + a_{1} (s) \exp u_{2} (s)} - b_{1} (s) \exp u_{2} (s)) d s \\ - \frac{1}{ω} \int_{0}^{ω} \int_{0}^{t} (r_{1} (s) - \frac{r_{1} (s) \exp u_{1} (s - τ_{1} (s))}{K_{1} (s) + a_{1} (s) \exp u_{2} (s)} - b_{1} (s) \exp u_{2} (s)) d s \\ + (\frac{t}{ω} + \frac{1}{2}) \int_{0}^{ω} (r_{1} (s) - \frac{r_{1} (s) \exp u_{1} (s - τ_{1} (s))}{K_{1} (s) + a_{1} (s) \exp u_{2} (s)} - b_{1} (s) \exp u_{2} (s)) d s . \end{matrix}$

$\begin{matrix} h (t) = \int_{0}^{t} (r_{2} (s) - \frac{r_{2} (s) \exp u_{2} (s - τ_{2} (s))}{K_{2} (s) + a_{2} (s) \exp u_{1} (s)} - b_{2} (s) \exp u_{1} (s)) d s \\ - \frac{1}{ω} \int_{0}^{ω} \int_{0}^{t} (r_{2} (s) - \frac{r_{2} (s) \exp u_{2} (s - τ_{2} (s))}{K_{2} (s) + a_{2} (s) \exp u_{1} (s)} - b_{2} (s) \exp u_{1} (s)) d s \\ + (\frac{t}{ω} + \frac{1}{2}) \int_{0}^{ω} (r_{2} (s) - \frac{r_{2} (s) \exp u_{2} (s - τ_{2} (s))}{K_{2} (s) + a_{2} (s) \exp u_{1} (s)} - b_{2} (s) \exp u_{1} (s)) d s . \end{matrix}$

从而 $Q N (Ω)$ 及 $K_{P} (I - Q) N (Ω)$ 为相对紧，从而N在 $\bar{Ω}$ 上为L-紧。

设 $η \in (0, 1)$ ， ${(\begin{matrix} u_{1} \\ u_{2} \end{matrix})}^{T} \in X$ ，满足 $L {(u_{1}, u_{2})}^{T} = η N {(u_{1}, u_{2})}^{T}$ ，则

${\begin{cases} {u^{'}}_{1} (t) = η (r_{1} (t) - \frac{r_{1} (t) \exp u_{1} (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) \exp u_{2} (t)} - b_{1} (t) \exp u_{2} (t)) \\ {u^{'}}_{2} (t) = η (r_{2} (t) - \frac{r_{2} (t) \exp u_{2} (t - τ_{2} (t))}{K_{2} (t) + a_{2} (t) \exp u_{1} (t)} - b_{2} (t) \exp u_{1} (t)) \end{cases}$ (3)

从0到 $ω$ 对(3)积分得：

${\begin{cases} \int_{0}^{ω} (r_{1} (t) - \frac{r_{1} (t) \exp u_{1} (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) \exp u_{2} (t)} - b_{1} (t) \exp u_{2} (t)) d t = 0 \\ \int_{0}^{ω} (r_{2} (t) - \frac{r_{2} (t) \exp u_{2} (t - τ_{2} (t))}{K_{2} (t) + a_{2} (t) \exp u_{1} (t)} - b_{2} (t) \exp u_{1} (t)) d t = 0 \end{cases}$ (4)

由(3)及(4)知：

$\begin{matrix} \int_{0}^{ω} | {u^{'}}_{1} (t) | d t \leq η (\int_{0}^{ω} r_{1} (t) d t + \int_{0}^{ω} (\frac{r_{1} (t) \exp u_{1} (t - τ_{1} (t))}{K_{1} (t) + a_{1} (t) \exp u_{2} (t)} + b {}_{1}{(t)} \exp u_{2} (t)) d t) \\ = 2 η \int_{0}^{ω} r_{1} (t) d t \leq 2 \int_{0}^{ω} r_{1} (t) d t . \end{matrix}$ (5)

同理可得

$\int_{0}^{ω} | {u^{'}}_{2} (t) | d t \leq 2 \int_{0}^{ω} r_{2} (t) d t .$ (6)

由(4)知：

$\int_{0}^{ω} b_{2} (t) \exp u_{1} (t) d t \leq \int_{0}^{ω} r_{2} (t) d t .$ (7)

记 $A_{1} = \ln \frac{\int_{0}^{ω} r_{2} (t) d t}{\int_{0}^{ω} b_{2} (t) d t}$ ， $A_{2} = \ln \frac{\int_{0}^{ω} r_{1} (t) d t}{\int_{0}^{ω} b_{1} (t) d t}$ 。若 $\forall t \in [0, ω]$ ， $u_{1} (t) > A_{1}$ ，由(7)推出矛盾，从而存在 $t_{0} \in [0, ω]$ ，使 $u_{1} (t_{0}) \leq A_{1}$ 。故由(5)， $\forall t \in [0, ω]$ ，有

$u_{1} (t) = u_{1} (t_{0}) + \int_{t_{0}}^{t} {u^{'}}_{1} (s) d s \leq u_{1} (t_{0}) + | \int_{t_{0}}^{t} {u^{'}}_{1} (s) d s | \leq A_{1} + \int_{t_{0}}^{t} | {u^{'}}_{1} (s) | d s \leq A_{1} + 2 \int_{0}^{ω} r_{1} (t) d t ≜ M_{2} .$

类似可证 $\forall t \in [0, ω]$ , $u_{2} (t) \leq A_{2} + 2 \int_{0}^{ω} r_{2} (t) d t ≜ M_{1}$

由(4)知 $0 < \int_{0}^{ω} r_{1} (t) d t - \int_{0}^{ω} \frac{r_{1} (t)}{K_{1} (t)} \exp M_{1} d t \leq \int_{0}^{ω} b_{1} (t) \exp u_{2} (t) d t$ 。从而存在 $t_{0} \in [0, ω]$ 使

$u_{2} (t_{0}) > - \ln \frac{\int_{0}^{ω} (r_{1} (t) - \frac{r_{1} (t)}{K_{1} (t)} \exp M_{1}) d t}{\int_{0}^{ω} b_{1} (t) d t} ≜ - B_{2} .$

由(6)知， $\forall t \in [0, ω]$ ，有

$u_{2} (t) = u_{2} (t_{0}) + \int_{t_{0}}^{t} {u^{'}}_{2} (s) d s > - B_{2} - | \int_{t_{0}}^{t} {u^{'}}_{2} (s) d s | > - B_{2} - \int_{0}^{ω} | {u^{'}}_{2} (s) | d s \geq - B_{2} - 2 \int_{0}^{ω} r_{2} (t) d t = m_{2} .$

记 $B_{1} = \ln \frac{\int_{0}^{ω} (r_{2} (t) - \frac{r_{2} (t)}{K_{2} (t)} \exp M_{2}) d t}{\int_{0}^{ω} b_{2} (t) d t}$ ，与上面类似可证 $\forall t \in [0, ω]$ ，有

$u_{1} (t) > - B_{1} - 2 \int_{0}^{ω} r_{1} (t) d t = m_{1} .$

令 $J = \max (2 | m_{i} |, 2 | M_{i} |, i = 1, 2)$ ， $Ω = {(u_{1}, u_{2}) \in X | ‖ (u_{1}, u_{2}) ‖ < J}$ ，则J为X的有界开子集,由J的取法知， $\forall (u_{1}, u_{2}) \in \partial Ω \cap K e r L$ ，有

令 $φ (λ, (u_{1}, u_{2})) = λ (u_{1}, u_{2}) + Q N ((u_{1}, u_{2}))$ ， $0 \leq λ \leq 1$ 。 $\forall (u_{1}, u_{2}) \in \partial Ω \cap K e r L$ ， $λ \in [0, 1]$ ，

$φ (λ, (u_{1}, u_{2})) \neq 0$ 。由同伦不变性知

$\deg (Q N (u_{1}, u_{2}), \partial Ω \cap K e r L, 0) \neq 0.$

从而 $Ω$ 满足引理的所有条件。故存在 $(u_{1}, u_{2}) \in Ω$ ，使 $L (u_{1}, u_{2}) = N (u_{1}, u_{2})$ 。即(2)存在 $ω$ 周期解 $(u_{1} (t), u_{2} (t))$ ，从而(1)存在正 $ω$ 周期解 $(x_{1} (t), x_{2} (t)) = (\exp u_{1} (t), \exp u_{2} (t))$ 。定理得证。

3. 应用

下面将举一个例子来支持前面所得出的结果。

例3.1 对于以下系统

${\begin{cases} \frac{d x}{d t} = x (t) [r_{1} (t) - \frac{r_{1} (t) x (t - (\exp (4 π + \sin t)))}{K_{1} (t) + (2 + \sin t) y (t)} - b_{1} (t) y (t)] \\ \frac{d y}{d t} = y (t) [r_{2} (t) - \frac{r_{2} (t) y (t - (\exp (4 π + \cos t)))}{K_{2} (t) + (2 + \sin t) x (t)} - b_{2} (t) x (t)] \end{cases}$ (8)

至少有一个 $\frac{1}{5}$ 周期解。其中

$r_{1} (t) = 3 + 0.3 \sin (10 π t), r_{2} (t) = 2 + 0.4 \sin (10 π t),$

$K_{1} (t) = 10 - \cos (10 π t), K_{2} (t) = 10 -0 .25 \cos (10 π t)$

$b_{1} (t) = 1 + 0.1 \sin (10 π t), b_{2} (t) = 1 + 0.2 \sin (10 π t) .$

证明：由(8)可得：

$A_{1} = \ln \frac{\int_{0}^{\frac{1}{5}} (2 + 0.4 \sin (10 π t)) d t}{\int_{0}^{\frac{1}{5}} (1 + 0.2 \sin (10 π t)) d t} = \ln 2, A_{2} = \ln \frac{\int_{0}^{\frac{1}{5}} (3 + 0.3 \sin (10 π t)) d t}{\int_{0}^{\frac{1}{5}} (1 + 0.1 \sin (10 π t)) d t} = \ln 3,$

$M_{1} = ln3 + 2 \int_{0}^{\frac{1}{5}} (2 + 0.4 \sin (10 π t)) d t \approx 1.8986122886681,$

$M_{2} = \ln 2 + 2 \int_{0}^{\frac{1}{5}} (3 + 0.3 \sin (10 π t)) d t \approx 1.8931471805599,$

$\exp M_{1} \approx 6.67662 < K_{1} (t), \exp M_{2} \approx 6.64023 < K_{2} (t) .$

由定理1，方程(8)至少有一个 $\frac{1}{5}$ 周期解。

参考文献

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