期刊菜单

A Class of Mathematical Model Concerning Impulsive Pest Control Strategies
DOI: 10.12677/AAM.2021.102060, PDF, HTML, XML, 下载: 103  浏览: 159

Abstract: In this paper, we first propose a mathematical model concerning an impulsive pest control strategies. Therefore, our models are the impulsive differential equations. And then we obtain some critical value of control variable. It is observed that there exists a globally asymptotically stable boundary periodic solution when the amount of infective pests released periodically is larger than this critical value. When the amount of infective pests released is less than this critical value, the system is shown to be permanent, which implies that the trivial boundary periodic solution loses its stability.

1. 引言

2. 问题的提出

(H1)害虫群被分为两类：一类是易受感染的；另一类是有传染性的。

(H2)易受感染的害虫满足logistic增长，而有传染性的疾病通过媒介传播，即传染速度可以用 $\beta {S}^{2n}I$ 表示。

(H3)害虫群的总数不是常数，是可以变化的。

$\left\{\begin{array}{l}\stackrel{˙}{S}=rS\left(1-\frac{S+\theta I}{k}\right)-\beta {S}^{2n}I,\\ \stackrel{˙}{I}=\beta {S}^{2n}I-\omega I+u.\end{array}$ (1)

$\left\{\begin{array}{l}\stackrel{˙}{S}=rS\left(1-\frac{S+\theta I}{k}\right)-\beta {S}^{2n}I,\text{}t\ne n\tau ,\\ \stackrel{˙}{I}=\beta {S}^{2n}I-\omega I,\text{}t\ne n\tau ,\\ \Delta I=I\left({t}^{+}\right)-I\left(t\right)=\tau u,t=n\tau ,\text{}n=1,2,\cdots .\end{array}$ (2)

(H4)被感染的害虫永远不能恢复，有传染性的害虫能再生，而且对作物无损害，而易受感染的害虫可以对作物造成损伤。

(H5)存在一个临界值SM，对于害虫损害来讲，SM代表一个最具经济意义的水平。

$\left\{\begin{array}{l}\stackrel{˙}{S}=rS\left(1-\frac{S+\theta I}{k}\right)-\beta {S}^{2n}I=P\left(S,I\right),\\ \stackrel{˙}{I}=\beta {S}^{2n}I-\omega I=Q\left(S,I\right).\end{array}$ (3)

3. 被感染害虫的脉冲释放

1) V在 $\left(n\tau ,\left(n+1\right)\tau \right]×{R}_{+}^{2}$ 上是连续的，而且对于任意的 $x\in {R}_{+}^{2},n\in {Z}_{+}$，极限

${\mathrm{lim}}_{\left(t,y\right)\to \left(n{\tau }^{+},x\right)}V\left(t,y\right)=V\left(n{\tau }^{+},x\right)$

2) V关于x满足局部Lipschitzian条件。

${D}^{+}V\left(t,x\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}\mathrm{sup}\frac{1}{h}\left[V\left(t+h,x+hf\left(t,x\right)\right)-V\left(t,x\right)\right]$.

$\left\{\begin{array}{l}{D}^{+}V\left(t,x\left(t\right)\right)\le g\left(t,V\left(t,x\left(t\right)\right)\right),\text{}t\ne n\tau ,\\ V\left(t,x\left({t}^{+}\right)\right)\le {\Psi }_{n}\left(V\left(t,x\left(t\right)\right)\right),\text{}t=n\tau ,\end{array}$ (4)

${\mathrm{lim}}_{\left(t,y\right)\to \left(n{\tau }^{+},z\right)}g\left(t,y\right)=g\left(n{\tau }^{+},z\right)$

$\left\{\begin{array}{l}\stackrel{˙}{v}\left(t\right)=g\left(t,v\left(t\right)\right),\text{}t\ne n\tau ,\\ v\left({t}^{+}\right)={\psi }_{n}\left(v\left(t\right)\right),\text{}t=n\tau ,\\ v\left({0}^{+}\right)=v\left(0\right),\end{array}$ (5)

$\left[0,+\infty \right)$ 上的最大解。那么只要 $V\left({0}^{+},{x}_{0}\right)\le {v}_{0}$ 就有 $V\left(t,x\left(t\right)\right)\le R\left(t\right)$$t\ge 0$，这里 $x\left(t\right)$ 是系统(2)的任一解。

$\left\{\begin{array}{l}{D}^{+}V\left(t\right)+\omega V\left(t\right)=\left(r+\omega \right)S-\frac{r{S}^{2}+r\theta SI}{k},\text{}t\ne n\tau ,\\ V\left(n{\tau }^{+}\right)=V\left(n\tau \right)+\tau u,\text{}n=1,2,\cdots .\end{array}$ (6)

$\left\{\begin{array}{l}{D}^{+}V\left(t\right)+\omega V\left(t\right)

$V\left(t\right)\le \left(V\left({0}^{+}\right)-\frac{L}{\omega }\right){\text{e}}^{-\omega t}+\frac{\tau u\left(1-{\text{e}}^{-n\omega \tau }\right)}{1-{\text{e}}^{-\omega \tau }}{\text{e}}^{-\omega \left(t-n\tau \right)}+\frac{L}{\omega }$

$\left\{\begin{array}{l}{I}^{\prime }\left(t\right)=-\omega I\left(t\right),\text{}t\ne n\tau ,\\ \Delta I\left(t\right)=I\left({t}^{+}\right)-I\left(t\right)=\tau u,\text{}t=n\tau ,\\ I\left({0}^{+}\right)={I}_{0}\ge 0.\end{array}$ (7)

$n\tau 时，系统(7)的解是

$I\left(t\right)=\left(I\left({0}^{+}\right)-\frac{\tau u}{1-{\text{e}}^{-\omega \tau }}\right){\text{e}}^{-\omega t}+\frac{\tau u{\text{e}}^{-\omega \left(t-n\tau \right)}}{1-{\text{e}}^{-\omega \tau }}.$

$\left(0,{I}^{*}\left(t\right)\right)=\left(0,\frac{\tau u{\text{e}}^{-\omega \left(t-n\tau \right)}}{1-{\text{e}}^{-\omega \tau }}\right)$, $n\tau .

$u>\frac{k\omega }{\theta }$.(8)

$\left(S\left(t\right),I\left(t\right)\right)$ 是系统(2)的任一解，而且 $S\left(t\right)={u}_{1}\left(t\right)$, $I\left(t\right)={u}_{2}\left(t\right)+{I}^{*}\left(t\right)$。系统(2)在 $\left(0,{I}^{*}\left(t\right)\right)$ 的相应线性系统是

$\left\{\begin{array}{l}{\stackrel{˙}{u}}_{1}\left(t\right)=r\text{\hspace{0.17em}}{u}_{1}-\frac{r\theta {I}^{*}\left(t\right){u}_{1}}{k},\text{\hspace{0.17em}}\text{}t\ne n\tau ,\\ {\stackrel{˙}{u}}_{2}\left(t\right)=-\omega \text{\hspace{0.17em}}{u}_{2},\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }\text{}t\ne n\tau ,\\ {u}_{1}\left({t}^{+}\right)={u}_{1}\left(t\right),\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }t=n\tau ,\\ {u}_{2}\left({t}^{+}\right)={u}_{2}\left(t\right),\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{}t=n\tau .\end{array}$ (9)

$\varphi \left(t\right)$ 是(9)的基解矩阵，则 $\varphi \left(t\right)$ 满足

$\frac{\text{d}\varphi \left(t\right)}{\text{d}t}=\left(\begin{array}{cc}r-\frac{r\theta {I}^{*}\left(t\right)}{k}& 0\\ 0& -\omega \end{array}\right)\varphi \left(t\right)=A\varphi \left(t\right)$, (10)

$\varphi \left(t\right)=\left(\begin{array}{cc}\mathrm{exp}\left({\int }_{0}^{t}\left(r-\frac{r\theta }{k}{I}^{*}\left(s\right)\right)\text{d}s\right)& 0\\ \Delta & \mathrm{exp}\left(-\omega t\right)\end{array}\right)$.

$\left(\begin{array}{c}{u}_{1}\left(n{\tau }^{+}\right)\\ {u}_{2}\left(n{\tau }^{+}\right)\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{u}_{1}\left(n\tau \right)\\ {u}_{2}\left(n\tau \right)\end{array}\right)$.

$M=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\varphi \left(\tau \right)=\varphi \left(\tau \right)$.

${\lambda }_{1}={\text{e}}^{-\omega \tau }<1$, ${\lambda }_{2}=\mathrm{exp}\left({\int }_{0}^{\tau }\left(r-\frac{r\theta }{k}{I}^{*}\left(s\right)\right)\text{d}s\right)$.

$I\left(t\right)>{I}^{*}\left(t\right)-\epsilon$.(11)

$\stackrel{˙}{S}\left(t\right)\le rS\left(t\right)\left(1-\frac{\theta }{k}\left({I}^{*}-\epsilon \right)\right)$ (12)

$S\left(\left(n+1\right)\tau \right)\le S\left(n\tau \right)\mathrm{exp}\left({\int }_{n\tau }^{\left(n+1\right)\tau }\left(r-\frac{r\theta }{k}\left({I}^{*}\left(t\right)-\epsilon \right)\right)\text{d}t\right)=S\left(n\tau \right)\rho .$

$-\omega I\left(t\right)\le \stackrel{˙}{I}\left(t\right)=\beta {S}^{2n}I\left(t\right)-\omega I\left(t\right)\le \left(\epsilon -\omega \right)I\left(t\right)$ (13)

$\left\{\begin{array}{l}{{y}^{\prime }}_{1}\left(t\right)=-\omega {y}_{1}\left(t\right),\text{}t\ne n\tau ,\\ \Delta {y}_{1}\left(t\right)={y}_{1}\left({t}^{+}\right)-{y}_{1}\left(t\right)=\tau u,\text{}t=n\tau ,\\ {y}_{1}\left({0}^{+}\right)={I}_{0}\ge 0,\end{array}$ (14)

$\left\{\begin{array}{l}{{y}^{\prime }}_{2}\left(t\right)=\left(\epsilon -\omega \right){y}_{2}\left(t\right),\text{}t\ne n\tau ,\\ \Delta {y}_{2}\left(t\right)={y}_{2}\left({t}^{+}\right)-{y}_{2}\left(t\right)=\tau u,\text{}t=n\tau ,\\ {y}_{2}\left({0}^{+}\right)={I}_{0}\ge 0,\end{array}$ (15)

${y}_{2}^{*}\left(t\right)=\frac{\tau u{\text{e}}^{\left(\epsilon -\omega \right)\left(t-n\tau \right)}}{1-{\text{e}}^{\left(\epsilon -\omega \right)\tau }},n\tau .

$u<\frac{k\omega }{\theta }$. (16)

1) 首先令 ${m}_{3}>0,\text{}{\epsilon }_{1}>0$ 充分小，使

${m}_{3}<\frac{\omega }{\beta },\text{}\delta =\beta {m}_{3}^{2n}-\omega <0$,

$\sigma =r\tau -\frac{r{m}_{3}\tau }{k}+\left(\frac{r\theta }{k}+{m}_{3}\beta \right)\frac{\tau u}{\delta }-{m}_{3}^{2n-1}\beta {\epsilon }_{1}\tau >0$,

$\eta =r-\frac{r{m}_{3}}{k}-\frac{r\theta M}{k}-{m}_{3}^{2n-1}\beta M<0$,

$\stackrel{˙}{I}\left(t\right)=\beta {S}^{2n}\left(t\right)I\left(t\right)-\omega I\left(t\right)\le I\left(t\right)\left(\beta {m}_{3}^{2n}-\omega \right)=\delta I\left( t \right)$

$\left\{\begin{array}{l}{{y}^{\prime }}_{3}\left(t\right)=\left(\beta {m}_{3}^{2n}-\omega \right){y}_{3}\left(t\right),\text{}t\ne n\tau ,\\ \Delta {y}_{3}\left(t\right)={y}_{3}\left({t}^{+}\right)-{y}_{3}\left(t\right)=\tau u,\text{}t=n\tau ,\\ {y}_{3}\left({0}^{+}\right)={I}_{0}\ge 0,\end{array}$ (17)

${y}_{3}^{*}\left(t\right)=\frac{\tau u{\text{e}}^{\delta \left(t-n\tau \right)}}{1-{\text{e}}^{\delta \tau }},\text{}n\tau .

$I\left(t\right)\le {y}_{3}\left(t\right)\le {y}_{3}^{*}\left(t\right)+{\epsilon }_{1}$

$\stackrel{˙}{S}\left(t\right)\ge S\left(t\right)\left(r-\frac{r{m}_{3}}{k}-\frac{r\theta }{k}\left({y}_{3}^{*}\left(t\right)+{\epsilon }_{1}\right)-{m}_{3}^{2n-1}\beta \left({y}_{3}^{*}\left(t\right)+{\epsilon }_{1}\right)\right).$ (18)

${N}_{1}\in {Z}_{+}$，满足 ${N}_{1}\tau \ge {T}_{1}$，在区间 $\left(n\tau ,\left(n+1\right)\tau \right]$, $n>{N}_{1}$ 上积分(18)得

$S\left(\left(n+1\right)\tau \right)\ge S\left(n{\tau }^{+}\right)\mathrm{exp}\left({\int }_{n\tau }^{\left(n+1\right)\tau }\Delta \text{d}t\right)=S\left(n\tau \right){\text{e}}^{\sigma }$.

2) 其次，如果 $S\left(t\right)\ge {m}_{3}$ 对所有的 $t\ge {t}_{1}$ 成立，结论自然成立。因此，我们只需考虑离开区域 $\Gamma =\left\{\left(S\left(t\right),I\left(t\right)\right)\in {R}_{+}^{2}:S\left(t\right)<{m}_{3}\right\}$ 后又进入该区域的那些解。记 ${t}^{*}={\mathrm{inf}}_{t\ge {t}_{1}}\left\{t:S\left(t\right)<{m}_{3}\right\}$，则 $S\left(t\right)\ge {m}_{3}$$t\in \left[{t}_{1},{t}^{*}\right)$，且由于 $S\left(t\right)$ 连续，所以 $S\left({t}^{*}\right)={m}_{3}$。选取 ${n}_{2},{n}_{3}\in {Z}_{+}$，使

${n}_{2}\tau >\frac{1}{\delta }\mathrm{ln}\frac{{\epsilon }_{1}}{M+\tau u},{\text{e}}^{{n}_{2}\eta \tau }{\text{e}}^{{n}_{3}\sigma }>1$.

$T={n}_{2}\tau +{n}_{3}\tau$，则我们可以断言，存在 ${t}_{2}\in \left({t}^{*},{t}^{*}+T\right]$ 使得 $S\left({t}_{2}\right)>{m}_{3}$。否则，我们在 ${y}_{3}^{*}\left({t}^{{*}^{+}}\right)=I\left({t}^{{*}^{+}}\right)$ 下考虑(17)得

${y}_{3}\left(t\right)=\left({y}_{3}\left(\left({n}_{1}+1\right){\tau }^{+}\right)-\frac{\tau u}{1-{\text{e}}^{\delta \tau }}\right){\text{e}}^{\left(t-\left({n}_{1}+1\right)\tau \right)\delta }+{y}_{3}^{*}\left(t\right)$, $n\tau .

$|{y}_{3}\left(t\right)-{y}_{3}^{*}\left(t\right)|<\left(M+\tau u\right){\text{e}}^{\left(t-\left({n}_{1}+1\right)\tau \right)\delta }<{\epsilon }_{1}$, $I\left(t\right)\le {y}_{3}\left(t\right)\le {y}_{3}^{*}\left(t\right)+{\epsilon }_{1}$, ${t}^{*}+{n}_{2}\tau \le t\le {t}^{*}+T$,

$\stackrel{˙}{S}\left(t\right)\ge S\left(t\right)\left(r-\frac{r{m}_{3}}{k}-\frac{r\theta M}{k}-{m}_{3}^{2n-1}\beta M\right)=\eta S\left(t\right)$.

$\stackrel{¯}{t}={\mathrm{inf}}_{t\ge {t}^{\ast }}\left\{t:S\left(t\right)\ge {m}_{3}\right\}$，则 $S\left(\stackrel{¯}{t}\right)\ge {m}_{3}$，于是对于任意的 $t\in \left[{t}^{*},\stackrel{¯}{t}\right]$ 总有

$S\left(t\right)\ge S\left({t}^{*}\right){m}_{3}{\text{e}}^{\left(t-{t}^{*}\right)\eta }\ge {m}_{3}{\text{e}}^{\left({n}_{2}+{n}_{3}\right)\eta \tau }\triangleq {m}_{1}$.

NOTES

*山东省本科高校教学改革研究项目，项目批准号：2015M139。

#通讯作者。

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