## Stability of a Class of Nonlinear Hierarchical Age-Dependent Population Model

He Zerong,, Zhang Zhiqiang, Wang Yang

 基金资助: 国家自然科学基金.  11871185浙江省自然科学基金.  LY18A010010

Abstract

The article is concerned with the existence of positive equilibria and stability of zero state in a nonlinear hierarchical species. Based on the assumption that young individuals are more competitive than older ones, an integro-partial differential equation is taken to model the revolution process of the population. The net reproductive number is defined and used to show that there are positive steady states in the system. Furthermore, stability results for zero equilibrium are derived via the characteristic equation and a Liapunov function. Finally, some numerical experiments are presented.

Keywords： Hierarchy of ages ; Positive equilibria ; Stability ; Non-zero fixed points

He Zerong, Zhang Zhiqiang, Wang Yang. Stability of a Class of Nonlinear Hierarchical Age-Dependent Population Model. Acta Mathematica Scientia[J], 2020, 40(6): 1712-1722 doi:

## 2 正平衡态的存在性

$$$\left\{\begin{array}{ll} { } \frac{\partial x}{\partial t} + \frac{\partial x}{\partial a} = - \mu(a, E(x)(a, t))x(a, t), & (a, t) \in Q, \\ { } x(0, t) = \int_0^A \beta(a, E(x)(a, t))x(a, t){\rm d}a, & t \in (0, T), \\ x(a, 0) = x_0(a), & a \in [0, A], \\ { } E(x)({a, t}) = \int_0^a x( r, t) {\rm d}r + \alpha \int_a^A x(r, t){\rm d}r, & (a, t)\in Q, \ 0 \le \alpha < 1, \end{array}\right.$$$

(H1)   对任意$(a, s)\in [0, A]\times [0, \infty)$, $\mu(a, s)>0$, 且$\int_0^A \mu(a, s){\rm d}a = +\infty$; $\mu$连续, 且关于$s$单调增加、满足局部Lipschitz条件;

(H2)   对任意$(a, s)\in [0, A]\times [0, \infty)$, $0\leq\beta(a, s)\leq M_1, M_1$为常数; $\beta$连续且关于$s$单调不增、满足局部Lipschitz条件;

(H3)   函数$0\leq x_0 (a)$非负有界.

$$$x^* (0) = x^* (0)\int_0^A \beta(a, E(x^*)(a))\exp \Big\{-\int_0^a\mu(r, E(x^*)(r)){\rm d}r\Big\}{\rm d}a.$$$

$x^* (0)>0$时, 上式即为$S(x^*) = 1$, 其中

$$$S(x^*) = \int_0^A\beta(a, E(x^*)(a))\exp\Big\{-\int_0^a\mu(r, E(x^*)(r)){\rm d}r\Big\}{\rm d}a,$$$

(1)   对任意满足$\|z\| = r$的元$z,$及数$\lambda >1,$必有$Fz\neq \lambda z;$

(2)   存在$\rho \in(0, r), e\in K\backslash \{0\},$使得:当$\|z\| = \rho, \lambda >0$都有$z-Fz\neq \lambda e.$

$$$\sup\limits_{f\in K}\|f\| = \sup\limits_{f\in K}\Big(\int_{S}|f(s)|^p{\rm d}s\Big)^{1/p}<\infty,$$$

$$$\lim\limits_{t\rightarrow 0}\int_{S}|f(t+s)-f(s)|^p {\rm d}s = 0, \quad \mbox{对f\in K一致成立},$$$

$$$\lim\limits_{\gamma\rightarrow\infty}\int_{|s|>\gamma}|f(s)|^p{\rm d}s = 0, \quad \mbox{对f\in K 一致成立}.$$$

为了应用引理2.3, 将$L^1(0, A)$中的函数$v(a)$的定义域拓展到$(-\infty, \infty)$上.当$a\notin[0, A]$时, 令$v(a) = 0$.先证$F(K_r)$满足$(2.6)$式.对任意$(v, c)\in K_r$, 令

$(2.6)$式得证.

$a\in (A-\epsilon , A)$时, 注意到$\int_0^A\mu(\theta, x){\rm d}\theta = +\infty$, 可导出

$F(K_r)$中的元素一致成立.应用引理2.3即得结论.

若$F(v, c) = \lambda (v, c)$, 由映射$F$的定义得

$$$\left\{\begin{array}{ll} { } \lambda v(a) = c\exp\Big\{-\int_0^a\mu(r, E(v)(r)){\rm d}r\Big\}, \\ { } \lambda c = c\int_0^A\beta(a, E(v)(a))\exp\Big\{-\int_0^a\mu(r, E(v)(r)){\rm d}r\Big\}{\rm d}a. \end{array}\right.$$$

$(2.9)$式中的第一式可知, 当$c = 0$时, $v(a) = 0$, 这不可能!因为$\|(v, c)\| = r>0$.

(1) $\forall \|(v, c)\| = r, \exists \lambda>1$使得$(2.9)$式成立.由(2.9)式中第一式得

$$$\|v\|\le \lambda \|v\| = c\Big\|\exp\Big\{-\int_0^\cdot \mu(r, E(v)(r)){\rm d}r\Big\}\Big\|\le cA.$$$

(2) $\exists \|(v, c)\| = r, \forall \lambda>1$, 使得$(2.9)$式成立.由$(2.9)$式和$0\le \beta(a, x)\le M_1$可知

用反证法, 分四种情形处理.

(1) $\forall \rho\in(0, r), \exists (\bar{v}, \bar{c})\in K \backslash \{0\}, $$\exists\|(v, c)\| = \rho, \forall \lambda>0 , 使得等式 (v, c)-F(v, c) = \lambda (\bar{v}, \bar{c}) 成立. 由映射 F 的定义知 $$\left\{\begin{array}{ll} { } v(a)-c\exp\Big\{-\int_0^a\mu(r, E(v)(r)){\rm d}r\Big\} = \lambda \bar{v}(a), \\ { } c-c\int_0^A\beta(a, E(v)(a))\exp\Big\{-\int_0^a\mu(r, E(v)(r)){\rm d}r\Big\}{\rm d}a = \lambda \bar{c}. \end{array}\right.$$ \rho \to 0 时可得 (\bar{v}, \bar{c}) = 0 , 推出矛盾. (2) \forall \rho\in(0, r), \exists (\bar{v}, \bar{c})\in K \backslash \{0\} , \exists\lambda_0>0, \forall \|(v, c)\| = \rho , 使得 \rho\to 0 可知 \lambda_0(\bar{v}, \bar{c}) = 0 , 这是矛盾, 因为 \lambda_0 (\bar{v}, \bar{c}) 为非零元. (3) \exists \rho\in(0, r), \forall (\bar{v}, \bar{c})\in K \backslash \{0\},$$ \exists \lambda_0>0, \forall \|(v, c)\| = \rho$, 使得

$(\bar{v}, \bar{c}) = \frac{2}{\lambda_0}(v, c)$, 则$F(v, c) = -(v, c)$, 这不可能, 因$F(v, c)$为非负向量.

(4) $\exists \rho\in(0, r), \forall (\bar{v}, \bar{c})\in K \backslash \{0\}, $$\exists \|(v, c)\| = \rho, \forall \lambda_0>0, 使得 \lambda_0 \to 0 时, 可知 (v, c) = F(v, c) .这意味着 (v(a), c) 就是正平衡态. 综合以上分析知:引理2.1中的所有条件都满足.因此, 映射 F 存在非零元不动点 x^* , 它就是系统 (2.1) 的正平衡态. ## 3 零平衡态的稳定性 系统(2.1)在零平衡态处的线性化结果如下 \begin{eqnarray} \left\{\begin{array}{ll} { } \frac{\partial x}{\partial a}+\frac{\partial x}{\partial t} = - \mu(a, 0)x(a, t), (a, t) \in Q, \\ { } x(0, t) = \int_0^A \beta(a, 0)x(a, t){\rm d}a, t\in (0, \infty), \\ x(a, 0) = x_0(a), a\in [0, A]. \end{array}\right. \end{eqnarray} 下面推导零平衡态的特征方程.考虑系统 (3.1) 形如 x(a, t) = e^{\lambda t}X(a) 的解, 其中 \lambda 可为复数.将其代入系统 (3.1) 中的第一式可得 解得 由此利用系统 (3.1) 中的第二式, 可以导出下列特征方程 \begin{eqnarray} 1 = K(\lambda)&\triangleq&\int_0^A\beta(a, 0)\exp\Big\{-\int_0^a [\mu(s, 0)+\lambda]{\rm d}s\Big\}{\rm d}a{}\\ & = &\int_0^A\beta(a, 0)\exp\Big\{-\int_0^a \mu(s, 0){\rm d}s\Big\}\exp\{-\lambda a\}{\rm d}a. \end{eqnarray} 易知 K(\lambda) 关于实数 \lambda 严格单调递减.当 K(0)>1 时, 因 \lim\limits_{\lambda\to +\infty}K(\lambda) = 0 , 故 K(\lambda) = 1 有唯一正实根.根据线性稳定性的一般理论可知:当 K(0)>1 时系统 (3.1) 的零平衡态不稳定. 另一方面, 当 K(0)<1 时, 特征方程有唯一负实根, 记为 \lambda_0 .若特征方程(3.2)存在复特征根 y+{\rm i}z , 则必有 y\le \lambda_0 .否则如果 y> \lambda_0 , 那么 推出矛盾.因此当 K(0)<1 时, 所有特征根的实部均为负, 系统 (3.1) 的零平衡态渐近稳定. 再假设 \mu(a, 0)\ge \beta(a, 0), a.e. a\in(0, A) .考虑Liapunov函数 V(x(t)) = \int_0^Ax(a, t){\rm d}a , 它沿着系统 (3.1) 的变化率为 其中应用了 x(A, t) = 0, 以及 \beta$$ \mu$关于$E(x)$的单调性.因此零平衡态全局渐近稳定.

(2) 当$S(0)<1$时, 系统(2.1)的零平衡态渐近稳定.进一步, 当$\mu(a, 0)\ge\beta(a, 0),$ a.e. $a \in(0, A)$时, 零平衡态全局渐近稳定.

## 4 数值模拟

$x_0^3(a) = 3(10-a)^2\cos^2(a)$; $x_0^4(a) = 5(10-a)^2(\cos(a)+1)$$S(0)$值为:$S(0) = 0.008; 0.0933; 0.0024$.

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