关于时间慢变的广义Logistic模型的多尺度分析Multiscale Analysis of a Generalized Logistic Model with Slowly Varying Parameters

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In the generalized logistic model, there are three parameters, R (intrinsic rate of growth), K (maxi-mum tolerance of population) and β (the degree of population using environmental resources), which are constants in the model. In fact, they vary slowly with time in the natural environment. It is hard to obtain the exact solution. In this paper we get the asymptotic solution of the model using the method of multi-scale. Numerical simulation illustrates the validity of the asymptotic solution obtained.

1. 引言

$\frac{\text{d}M}{M\text{d}T}=R\left(\frac{K-M}{K+\left(R/C\right)M}\right),$ (1)

$\frac{\text{d}X}{\text{d}T}=\frac{R\left(T\right)X\left(1-\frac{X}{K\left(T\right)}\right)}{1+\epsilon \stackrel{¯}{\beta }\left(T\right)X},\text{\hspace{0.17em}}\text{\hspace{0.17em}}X\left(0\right)={X}_{0}.$ (2)

$K\left(\epsilon t\right)={K}_{0}k\left(\epsilon t\right),\text{\hspace{0.17em}}\stackrel{¯}{\beta }\left(\epsilon t\right)={\beta }_{0}\beta \left(\epsilon t\right),\text{\hspace{0.17em}}R\left(\epsilon t\right)={R}_{0}r\left(\epsilon t\right),$ (3)

(2)式变换为

$\frac{\text{d}x\left(t,\epsilon \right)}{\text{d}t}=\frac{r\left(\epsilon t\right)x\left(1-\frac{x\left(t,\epsilon \right)}{k\left(\epsilon t\right)}\right)}{1+\epsilon \sigma \beta \left(\epsilon t\right)x\left(t,\epsilon \right)},\text{\hspace{0.17em}}x\left(0,\epsilon \right)=\mu .$ (4)

2. 多尺度分析

${t}_{0}=\frac{1}{\epsilon }g\left({t}_{1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{t}_{1}=\epsilon t,$ (5)

${t}_{1}$ 相对于 ${t}_{0}$ 为慢时间尺度，且需满足 $g\left({t}_{1}\right)>0\left({t}_{1}>0\right),\text{\hspace{0.17em}}g\left(0\right)=0$ ，这里 $r,k,\stackrel{¯}{\beta }$ 都变为关于 ${t}_{\text{1}}$ 的函数 $r\left({t}_{1}\right),k\left({t}_{1}\right),\stackrel{¯}{\beta }\left({t}_{1}\right)$ 。此外，为了让 ${t}_{0}$${t}_{1}$ 一一对应，我们需要假定 ${g}^{\prime }\left({t}_{1}\right)>0$

$x\left(t,\epsilon \right)$ 看作两个时间尺度的函数 $x\left({t}_{\text{0}},{t}_{1},\epsilon \right)$ ，再运用链式法则得到的多尺度等式

${g}^{\prime }\left({t}_{1}\right){D}_{\text{0}}x\left(t,\epsilon \right)+\epsilon {D}_{\text{1}}x\left(t,\epsilon \right)=\frac{r\left(\epsilon t\right)x\left(1-\frac{x\left(t,\epsilon \right)}{k\left(\epsilon t\right)}\right)}{1+\epsilon \sigma \beta \left(\epsilon t\right)x\left(t,\epsilon \right)},$ (6)

$x\left({t}_{\text{0}},{t}_{1},\epsilon \right)={x}_{0}\left({t}_{0},{t}_{1}\right)+\epsilon {x}_{1}\left({t}_{0},{t}_{1}\right)+{\epsilon }^{2}{x}_{2}\left({t}_{0},{t}_{1}\right)+\cdots ,$ (7)

$\begin{array}{l}{g}^{\prime }\left({t}_{1}\right){D}_{0}\left[{x}_{0}\left({t}_{0},{t}_{1}\right)+\epsilon {x}_{1}\left({t}_{0},{t}_{1}\right)+{\epsilon }^{2}{x}_{2}\left({t}_{0},{t}_{1}\right)+\cdots \right]+\epsilon {D}_{1}\left[{x}_{0}\left({t}_{0},{t}_{1}\right)+\epsilon {x}_{1}\left({t}_{0},{t}_{1}\right)+{\epsilon }^{2}{x}_{2}\left({t}_{0},{t}_{1}\right)+\cdots \right]\\ =\frac{r\left(\epsilon t\right)\left({x}_{0}\left({t}_{0},{t}_{1}\right)+\epsilon {x}_{1}\left({t}_{0},{t}_{1}\right)+{\epsilon }^{2}{x}_{2}\left({t}_{0},{t}_{1}\right)+\cdots \right)\left(1-\frac{{x}_{0}\left({t}_{0},{t}_{1}\right)+\epsilon {x}_{1}\left({t}_{0},{t}_{1}\right)+{\epsilon }^{2}{x}_{2}\left({t}_{0},{t}_{1}\right)+\cdots }{k\left(\epsilon t\right)}\right)}{1+\epsilon \sigma \beta \left(\epsilon t\right)\left({x}_{0}\left({t}_{0},{t}_{1}\right)+\epsilon {x}_{1}\left({t}_{0},{t}_{1}\right)+{\epsilon }^{2}{x}_{2}\left({t}_{0},{t}_{1}\right)+\cdots \right)}.\end{array}$ (8)

${g}^{\prime }\left({t}_{1}\right){D}_{0}{x}_{0}=r\left({t}_{1}\right){x}_{0}\left(1-\frac{{x}_{0}}{k\left({t}_{1}\right)}\right);$ (9)

${g}^{\prime }\left({t}_{1}\right){D}_{0}{x}_{1}+{D}_{1}{x}_{0}=\frac{r\left({t}_{1}\right)}{k\left({t}_{1}\right)}\left[{x}_{1}\left(k\left({t}_{1}\right)-2{x}_{0}\right)\right]-\frac{r\left({t}_{1}\right)}{k\left({t}_{1}\right)}\sigma \beta \left({t}_{1}\right)\left[{x}_{0}^{2}\left(k\left({t}_{1}\right)-{x}_{0}\right)\right],$ (10)

${x}_{0}=\frac{k\left({t}_{1}\right)}{1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}},$ (11)

$\begin{array}{c}{x}_{1}=\frac{-\theta \left({t}_{1}\right)\sigma \beta \left({t}_{1}\right)k{\left({t}_{1}\right)}^{3}c\left({t}_{1}\right)\left[{t}_{0}+\frac{1}{\theta \left({t}_{1}\right)}\mathrm{ln}\left(1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}\right)\right]}{{\text{e}}^{\theta \left({t}_{1}\right){t}_{0}}{\left(1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{-\frac{{k}^{\prime }\left({t}_{1}\right)}{\theta \left({t}_{1}\right)}+k{\left({t}_{1}\right)}^{2}{c}^{\prime }\left({t}_{1}\right){t}_{0}{\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}-\frac{1}{2}c\left({t}_{1}\right)k{\left({t}_{1}\right)}^{2}{\theta }^{\prime }\left({t}_{1}\right){t}_{0}^{2}{\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}}{{g}^{\prime }\left({t}_{1}\right){\left(1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}\right)}^{2}}.\end{array}$ (12)

$\begin{array}{c}x\left({t}_{0},{t}_{1},\epsilon \right)=\frac{k\left({t}_{1}\right)}{1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\epsilon \frac{-\frac{{k}^{\prime }\left({t}_{1}\right)}{\theta \left({t}_{1}\right)}}{{g}^{\prime }\left({t}_{1}\right){\left(1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\epsilon \frac{-\sigma \beta \left({t}_{1}\right)k{\left({t}_{1}\right)}^{3}c\left({t}_{1}\right)\mathrm{ln}\left(1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}\right)}{{\text{e}}^{\theta \left({t}_{1}\right){t}_{0}}{\left(1+c\left({t}_{1}\right)k\left({t}_{1}\right){\text{e}}^{-\theta \left({t}_{1}\right){t}_{0}}\right)}^{2}}.\end{array}$ (13)

${\theta }^{\prime }\left({t}_{1}\right)=0,\text{\hspace{0.17em}}\frac{{c}^{\prime }\left({t}_{1}\right)}{{g}^{\prime }\left({t}_{1}\right)}-\sigma \theta \left({t}_{1}\right)\beta \left({t}_{1}\right)k\left({t}_{1}\right)c\left({t}_{1}\right)=\text{0}\text{.}$ (14)

${t}_{0}=\frac{1}{\epsilon }{\int }_{0}^{\epsilon t}r\left(s\right)\text{d}s.$ (15)

${x}_{\text{0}}$${x}_{\text{1}}$ 代入 $x\left({t}_{0},{t}_{1},\epsilon \right)$ ，并将初值条件代入可得

$d=\frac{k\left(0\right)-\mu }{\mu k\left(0\right)},\text{\hspace{0.17em}}c=\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}}.$ (16)

$\theta =1,\text{\hspace{0.17em}}c=\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}},$ (17)

$\begin{array}{c}x\left({t}_{0},{t}_{1},\epsilon \right)=\frac{k\left({t}_{1}\right)}{1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}}k\left({t}_{1}\right){\text{e}}^{-{t}_{0}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\epsilon \frac{-\sigma \beta \left({t}_{1}\right)k{\left({t}_{1}\right)}^{3}\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}}\mathrm{ln}\left(1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}}k\left({t}_{1}\right){\text{e}}^{-{t}_{0}}\right)}{{\text{e}}^{\theta \left({t}_{1}\right){t}_{0}}{\left(1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}}k\left({t}_{1}\right){\text{e}}^{-{t}_{0}}\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\epsilon \frac{-{k}^{\prime }\left({t}_{1}\right)}{r\left({t}_{1}\right){\left(1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \int \beta \left({t}_{1}\right)k\left({t}_{1}\right)r\left({t}_{1}\right)\text{d}{t}_{1}}k\left({t}_{1}\right){\text{e}}^{-{t}_{0}}\right)}^{2}}.\end{array}$ (18)

$\begin{array}{c}x\left(t,\epsilon \right)=\frac{k\left(\epsilon t\right)}{1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \epsilon \int \beta \left(\epsilon t\right)k\left(\epsilon t\right)r\left(\epsilon t\right)\text{d}t}k\left(\epsilon t\right){\text{e}}^{-\frac{1}{\epsilon }{\int }_{0}^{\epsilon t}r\left(s\right)\text{d}s}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\epsilon \frac{-\sigma \beta \left(\epsilon t\right)k{\left(\epsilon t\right)}^{3}\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \epsilon \int \beta \left(\epsilon t\right)k\left(\epsilon t\right)r\left(\epsilon t\right)\text{d}t}\mathrm{ln}\left(1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \epsilon \int \beta \left(\epsilon t\right)k\left(\epsilon t\right)r\left(\epsilon t\right)\text{d}t}k\left(\epsilon t\right){\text{e}}^{-\frac{1}{\epsilon }{\int }_{0}^{\epsilon t}r\left(s\right)\text{d}s}\right)}{{\text{e}}^{{t}_{0}}{\left(1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \epsilon \int \beta \left(\epsilon t\right)k\left(\epsilon t\right)r\left(\epsilon t\right)\text{d}t}k\left(\epsilon t\right){\text{e}}^{-\frac{1}{\epsilon }{\int }_{0}^{\epsilon t}r\left(s\right)\text{d}s}\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\epsilon \frac{-{k}^{\prime }\left(\epsilon t\right)}{r\left(\epsilon t\right){\left(1+\frac{k\left(0\right)-\mu }{\mu k\left(0\right)}{\text{e}}^{\sigma \epsilon \int \beta \left(\epsilon t\right)k\left(\epsilon t\right)r\left(\epsilon t\right)\text{d}t}k\left(\epsilon t\right){\text{e}}^{-\frac{1}{\epsilon }{\int }_{0}^{\epsilon t}r\left(s\right)\text{d}s}\right)}^{2}}.\end{array}$ (19)

3. 数值模拟

$\begin{array}{l}r\left(\epsilon t\right)=\frac{9}{10}+\frac{2}{10}\mathrm{sin}\epsilon t，\\ k\left(\epsilon t\right)=1+\frac{2}{10}\mathrm{sin}\epsilon t，\\ \beta \left(\epsilon t\right)=\frac{1}{2}+\frac{2}{10}\mathrm{sin}\epsilon t,\end{array}$

Figure 1. Curve: Comparison between the numerical solution and the asymptotic solution

$\epsilon =\frac{1}{100},\mu =\frac{2}{10}$ 。那么，我们可以得到模型(4)的渐近解与数值解，由图1所示，蓝色曲线(数值解)与红色的点组成的曲线(形式渐近解)在 $t\ge 0$ 时，是完全吻合的。

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