最近, 文献[1]考虑了如下一类非局部时滞的SVIR反应扩散模型:

(1.1) $ \begin{eqnarray} \left\{ \begin{array}{ll} S_{t} = \nabla\cdot(d_{S}(x)\nabla S)+\lambda(x)-\beta_1(x)SI-(\mu(x)+\alpha(x))S, \ t>0, \ x\in\Omega, \\ V_{t} = \nabla\cdot(d_{V}(x)\nabla V)+\alpha(x)S-\beta_2(x)VI-(\mu(x)+\gamma(x))V, \ t>0, \ x\in\Omega, \\ { } I_{t} = \nabla\cdot(d_{I}(x)\nabla I)+\int_{\Omega}G(\tau, x, y)[\beta_1(y)S(t-\tau, y)+\beta_2(y)V(t-\tau, y)]I(t-\tau, y){\rm d}y \\ \qquad-(\mu(x)+\delta_{I}(x))I, \ t>0, \ x\in\Omega, \\ R_{t} = \nabla\cdot(d_{R}(x)\nabla R)+\gamma(x)V+\delta_{I}(x)I-\mu(x)R, \ t>0, \ x\in\Omega, \\ {}[d_{S}(x)\nabla S]\cdot{\bf \nu} = [d_{V}(x)\nabla V]\cdot{\bf \nu} = [d_{I}(x)\nabla I]\cdot{\bf \nu} = [d_{R}(x)\nabla R]\cdot{\bf \nu} = 0, \ t>0, \ x\in\partial\Omega, \end{array} \right. \end{eqnarray} $

其中$ S(t, x) $, $ V(t, x) $, $ I(t, x) $和$ R(t, x) $分别表示时刻$ t $和空间位置$ x $处易感个体、疫苗接种个体、染病个体和康复个体的密度. $ \nabla $是梯度算子. $ G $是与$ \nabla\cdot(d_{E}(\cdot)\nabla)-(\mu(\cdot)+\delta_{E}(\cdot)) $和Neumann边值条件相关的格林函数. $ \Omega\in{{\Bbb R}} ^{n} $是具有光滑边界$ \partial \Omega $的有界区域.所有的参数都是连续和严格正的.

由于系统(1.1)的第四个方程与前三个方程是解耦的, 故只需考虑如下系统:

(1.2) $ \begin{eqnarray} \left\{ \begin{array}{ll} S_{t} = \nabla\cdot(d_{S}(x)\nabla S)+\lambda(x)-\beta_1(x)SI-(\mu(x)+\alpha(x))S, \ t>0, \ x\in\Omega, \\ V_{t} = \nabla\cdot(d_{V}(x)\nabla V)+\alpha(x)S-\beta_2(x)VI-(\mu(x)+\gamma(x))V, \ t>0, \ x\in\Omega, \\ { } I_{t} = \nabla\cdot(d_{I}(x)\nabla I)+\int_{\Omega}G(\tau, x, y)[\beta_1(y)S(t-\tau, y)+\beta_2(y)V(t-\tau, y)]I(t-\tau, y){\rm d}y \\ \qquad-(\mu(x)+\delta_{I}(x))I, \ t>0, \ x\in\Omega, \\ {}[d_{S}(x)\nabla S]\cdot{\bf \nu} = [d_{V}(x)\nabla V]\cdot{\bf \nu} = [d_{I}(x)\nabla I]\cdot{\bf \nu} = 0, \ t>0, \ x\in\partial\Omega. \end{array} \right. \end{eqnarray} $

由文献[2]知, 当基本再生数$ {\cal R}_0<1 $时, 系统(1.2)的唯一的无病平衡态是全局渐近稳定的; 当$ {\cal R}_0>1 $时, 系统(1.2)至少存在一个正平衡态并且系统是一致持续生存的.由于空间异质的影响, 研究系统(1.2)的正平衡态的全局吸引性是非常困难的.注意到, 文献[2]利用波动引理的方法讨论了一类非局部时滞的捕食-食饵反应扩散模型的正平衡点的全局吸引性.随后, 很多学者利用该方法讨论了具有非局部时滞的不同类型传染病模型的正平衡点的全局吸引性, 见文献[3-6].

假定系统(1.2)中的参数都是常数.由$ G(\tau, x, y) $是与$ d_{E}\Delta-(\mu+\delta_{E}) $和Neumann边值条件相关的格林函数及Fourier变换可知

$ G(\tau, x, y) = {\rm e}^{-(\mu+\delta_{E})\tau}\left(\frac{1}{\sqrt{4\pi d_E\tau}}\right)^{n}\exp{\left(-\frac{|x-y|^{2}}{4d_E\tau}\right)}. $

定义

$ G(d_E\tau, x, y) = \left(\frac{1}{\sqrt{4\pi d_E\tau}}\right)^{n}\exp{\left(-\frac{|x-y|^{2}}{4d_E\tau}\right)}, $

故

$ G(\tau, x, y) = {\rm e}^{-(\mu+\delta_{E})\tau}G(d_E\tau, x, y). $

由

$ \frac{\partial G(\tau, x, y)}{\partial \tau} = d_{E}\Delta G(\tau, x, y)-(\mu+\delta_{E})G(\tau, x, y), $

得

$ \frac{\partial G(d_E\tau, x, y)}{\partial \tau} = d_{E}\Delta G(d_E\tau, x, y), $

即$ G(d_E\tau, x, y) $是与$ d_{E}\triangle $和Neumann边值条件相关的格林函数且$ \int_{\Omega}G(d_{E}\tau, x, y){\rm d}y = 1 $.因此, 系统(1.2)中的参数都是常数时可化为如下系统:

(1.3) $ \begin{eqnarray} \left\{ \begin{array}{ll} S_{t} = d_{S}\Delta S+\lambda-\beta_1SI-(\mu+\alpha)S, \ t>0, \ x\in\Omega, \\ V_{t} = d_{V}\Delta V+\alpha S-\beta_2VI-(\mu+\gamma)V, \ t>0, \ x\in\Omega, \\ { } I_{t} = d_{I}\Delta I+{\rm e}^{-(\mu+d_{E})\tau}\int_{\Omega}G(d_{E}\tau, x, y)[\beta_1S(t-\tau, y)+\beta_2V(t-\tau, y)]I(t-\tau, y){\rm d}y \\ \qquad-(\mu+\delta_{I})I, \ t>0, \ x\in\Omega, \\ { } \frac{\partial S}{\partial{\bf \nu}} = \frac{\partial V}{\partial{\bf \nu}} = \frac{\partial I}{\partial{\bf \nu}} = 0, \ t>0, \ x\in\partial\Omega. \end{array} \right. \end{eqnarray} $

目前, 有关系统(1.3)的正平衡点的全局吸引性方面的结果还没有.最近, 文献[7]通过构造合适的Lyapunov泛函讨论一类非局部时滞的蓝舌病反应扩散模型的正平衡点的全局吸引性.受文献[7]的启发, 本文通过构造Lyapunov泛函研究系统(1.3)的正平衡点的全局吸引性.关于Lyapunov泛函的构造方法, 见文献[8-10].

本文结构安排如下:第二节给出一些预备知识.第三节讨论当基本再生数$ {\cal R}_0>1 $时系统(1.3)的正平衡点的全局吸引性.最后, 给出简要的结论.

对$ \tau\geq0 $, 定义$ C_{\tau}: = C([-\tau, 0], {\Bbb X}) $, 范数为$ ||\psi|| = \max\limits_{\theta\in[-\tau, 0]}||\psi(\theta)||_{{\Bbb X}}, $其中$ {\Bbb X}: = C(\bar{\Omega}, {{\Bbb R}} ^{3}) $是具有范数$ ||\psi(\cdot)||_{{\Bbb X}} $的Banach空间.进一步, 定义$ {\Bbb X}^{+}: = C(\bar{\Omega}, {{\Bbb R}} _{+}^{3}) $和$ C_{\tau}^{+}: = C([-\tau, 0], {\Bbb X}^{+}) $.

由文献[1]的定理6.1有

定理2.1  对$ \forall\ \psi\in C_{\tau}^{+} $, 系统(1.3)在$ [0, \infty) $上存在唯一的解$ {\bf u}(t, \cdot, \psi) $, 且解半流$ \Psi(t) = {\bf u}_{t}(\cdot): C_{\tau}^{+}\rightarrow C_{\tau}^{+} $, $ t\geq0 $在$ C_{\tau}^{+} $上有一个全局紧吸引子.

易知, 系统(1.3)总是存在一个无病平衡点$ E_0 = (S^0, V^0, 0) $, 其中

$ S^0 = \frac{\lambda}{\mu+\alpha}, \quad V^0 = \frac{\alpha\lambda}{(\mu+\alpha)(\mu+\gamma)}. $

而且系统(1.3)的基本再生数

$ {\cal R}_0 = {\rm e}^{-(\mu+\delta_{E})\tau}\frac{\beta_1S^0+\beta_2V^0}{\mu+\delta_{I}}. $

由文献[11]知, 当$ {\cal R}_0>1 $时, 系统(1.3)有唯一的正平衡点$ E^* = (S^*, V^*, I^*) $, 其中

$ S^* = \frac{\lambda}{\beta_1I^*+\mu+\alpha}, \quad V^* = \frac{\alpha S^*}{\beta_2I^*+\mu+\gamma}, $

$ a(I^*)^{2}+bI^*+c = 0, $

这里

$ a = \beta_1\beta_2, \quad b = \beta_1(\mu+\gamma)+\beta_2(\mu+\alpha)-{\rm e}^{-(\mu+\delta_{E})\tau}\frac{\beta_1\beta_2\lambda}{\mu+\delta_{I}}, $

$ c = (\mu+\alpha)(\mu+\gamma)(1-{\cal R}_0). $

再由文献[1]的推论6.7有

定理2.2   (ⅰ)若$ {\cal R}_0<1 $, 则无病平衡点$ E_{0} $在$ C_{\tau}^{+} $内是全局吸引的.

(ⅱ) 若$ {\cal R}_0>1 $, 则存在一个常数$ \varrho>0 $使得对$ \forall\ \psi\in C_{\tau}^{+} $满足$ \psi_3(0, \cdot)\not\equiv0 $, 关于$ x\in \overline{\Omega} $一致成立

$ \liminf \limits_{t\rightarrow +\infty}S(t, x)\geq\varrho, \ \; \; \liminf \limits_{t\rightarrow +\infty}V(t, x)\geq\varrho, \ \; \; \liminf \limits_{t\rightarrow +\infty}I(t, x)\geq\varrho. $

下面给出本文的主要结果.

定理3.1  若$ {\cal R}_0>1 $, 则对$ \forall\ \psi\in C_{\tau}^{+} $满足$ \psi_3(0, \cdot)\not\equiv0 $, 关于$ x\in \overline{\Omega} $一致成立

$ \lim\limits_{t\rightarrow \infty}{\bf u}(t, x;\psi) = {\bf u}^{*}, $

其中$ {\bf u}^{*} = (S^*, V^*, I^*) $.

证  定义$ \omega(\psi) $是半流$ \Psi(t) $的轨道$ \gamma^{+}(\psi) $的$ \omega $ -极限集.定义集合

$ {\Bbb H}: = \left\{\psi\in C_{\tau}^{+}:\psi_{i}(0, x)>0, \ \forall x\in\bar{\Omega}, \ i = 1, 2, 3\right\}. $

由定理2.1和2.2(ⅱ)可知, $ \omega(\psi)\subset{\Bbb H} $.为方便讨论, 记$ {\cal X} = {\cal X}(t, x), \ \forall {\cal X}\in\{S, V, I\}. $

构造Lyapunov泛函$ {\Bbb V}:{\Bbb H}\rightarrow {{\Bbb R}} $如下:

$ \begin{eqnarray*} {\Bbb V}(t) = \int_{\Omega}({\Bbb V}_1(t, x)+{\Bbb V}_2(t, x)){\rm d}x, \end{eqnarray*} $

其中

$ \begin{eqnarray*} {\Bbb V}_1(t, x)& = & S^* g\left(\frac{S}{S^*}\right)+V^* g\left(\frac{V}{V^*}\right) +{\rm e}^{(\mu+\delta_{E})\tau}I^* g\left(\frac{I}{I^*}\right), \\ {\Bbb V}_2(t, x)& = & \beta_1S^* I^*\int_{-\tau}^{0}\int_{\Omega}G(d_{E}(-\theta), x, y)g\left(\frac{S(t+\theta, y)I(t+\theta, y)}{S^* I^*}\right){\rm d}y{\rm d}\theta \\ &&+\beta_2V^* I^*\int_{-\tau}^{0}\int_{\Omega}G(d_{E}(-\theta), x, y)g\left(\frac{V(t+\theta, y)I(t+\theta, y)}{V^* I^*}\right){\rm d}y{\rm d}\theta, \end{eqnarray*} $

$ g(x) = x-1-\ln x $, $ x>0 $.显然, $ x>0 $时$ g(x)\geq 0 $且$ \min\limits_{x>0}g(x) = g(1) $.

注意到

$ \lambda = \beta_1S^* I^*+(\mu+\alpha)S^*, \ \alpha S^* = \beta_2V^* I^*+(\mu+\gamma)V^* $

和

$ {\rm e}^{-(\mu+d_{E})\tau}(\beta_1S^*+\beta_2V^*) = \mu+\delta_{I}. $

通过计算得

$ \begin{eqnarray*} \frac{\partial{\Bbb V}_1(t, x)}{\partial t} & = &\left(1-\frac{S^*}{S}\right)(d_{S}\triangle S+(\mu+\alpha)S^*-(\mu+\alpha)S+\beta_1S^* I^*-\beta_1SI)\\ && +\left(1-\frac{V^*}{V}\right)(d_{V}\triangle V+\alpha(S-S^*)+(\mu+\gamma)(V^*-V)+\beta_2V^* I^*-\beta_2VI)\\ & &+{\rm e}^{(\mu+\delta_{E})\tau}\left(1-\frac{I^*}{I}\right)\left(d_{I}\Delta I-{\rm e}^{-(\mu+\delta_{E})\tau} (\beta_1S^*+\beta_2V^*)I\right.\\ &&\left.+{\rm e}^{-(\mu+\delta_{E})\tau}\int_{\Omega}G(d_{E}\tau, x, y)[\beta_1S (t-\tau, y)+\beta_2V(t-\tau, y)]I(t-\tau, y){\rm d}y\right)\\ & = & A-\mu S^*\left(g\left(\frac{S^*}{S}\right)+g\left(\frac{S}{S^*}\right)\right)+\alpha S^*\left(1-\frac{S^*}{S} -\frac{SV^*}{S^* V}+\frac{V^*}{V}\right)\\ && -(\mu+\gamma)V^*\left(g\left(\frac{V^*}{V}\right)+g\left(\frac{V}{V^*}\right)\right)\\ && +\beta_1S^* I^*\left(2-\frac{SI}{S^*I^*}-\frac{S^*}{S}-\int_{\Omega}G(d_{E}\tau, x, y)\frac{S(t-\tau, y)I(t-\tau, y)} {S^* I}{\rm d}y\right)\\ &&+\beta_2V^* I^*\left(2-\frac{VI}{V^*I^*}-\frac{V^*}{V}-\int_{\Omega}G(d_{E}\tau, x, y)\frac{V(t-\tau, y)I(t-\tau, y)} {V^* I}{\rm d}y\right)\\ &&+\int_{\Omega}G(d_{E}\tau, x, y)[\beta_1S(t-\tau, y)+\beta_2V(t-\tau, y)]I(t-\tau, y){\rm d}y, \end{eqnarray*} $

其中

$ A = \frac{d_{S}(S-S^*)}{S}\triangle S+\frac{d_{V}(V-V^*)}{V}\triangle V+{\rm e}^{(\mu+\delta_{E})\tau}\frac{d_{I}(I-I^*)}{I}\triangle I. $

$ \begin{eqnarray*} \frac{\partial{\Bbb V}_2(t, x)}{\partial t} & = &\beta_1S^* I^*\left[g\left(\frac{SI}{S^* I^*}\right)-\int_{\Omega}G(d_{E}\tau, x, y)g\left(\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I^*}\right){\rm d}y\right]\\ &&+\beta_2V^* I^*\left[g\left(\frac{VI}{V^* I^*}\right)-\int_{\Omega}G(d_{E}\tau, x, y)g\left(\frac{V(t-\tau, y)I(t-\tau, y)}{V^* I^*}\right){\rm d}y\right]\\ & = &\beta_1S^* I^*\left[\frac{SI}{S^* I^*}-\ln\frac{SI}{S^* I^*}-\int_{\Omega}G(d_{E}\tau, x, y)\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I^*}{\rm d}y\right.\\ &&\left.+\int_{\Omega}G(d_{E}\tau, x, y)\ln\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I^*}{\rm d}y\right]\\ &&+\beta_2V^* I^*\left[\frac{VI}{V^* I^*}-\ln\frac{VI}{V^* I^*}-\int_{\Omega}G(d_{E}\tau, x, y)\frac{V(t-\tau, y)I(t-\tau, y)}{V^* I^*}{\rm d}y\right.\\ &&\left.+\int_{\Omega}G(d_{E}\tau, x, y)\ln\frac{V(t-\tau, y)I(t-\tau, y)}{V^* I^*}{\rm d}y\right]. \end{eqnarray*} $

进一步, 有

$ \begin{eqnarray*} &&\frac{\partial({\Bbb V}_1(t, x)+{\Bbb V}_2(t, x))}{\partial t}\\ & = & A-\mu S^*\left(g\left(\frac{S^*}{S}\right)+g\left(\frac{S}{S^*}\right)\right)-(\mu+\gamma)V^* g\left(\frac{V^*}{V}\right)-(\mu+\gamma)V^* g\left(\frac{V}{V^*}\right)\\ &&+\alpha S^*\left(g\left(\frac{V^*}{V}\right)-g\left(\frac{S^*}{S}\right)-g\left(\frac{SV^*}{S^* V}\right)\right)\\ &&+\beta_1S^* I^*\left(2-\frac{S^*}{S}-\int_{\Omega}G(d_{E}\tau, x, y)\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I}{\rm d}y\right.\\ &&\left.-\ln\frac{SI}{S^* I^*}+\int_{\Omega}G(d_{E}\tau, x, y)\ln\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I^*}{\rm d}y\right)\\ &&+\beta_2V^* I^*\left(2-\frac{V^*}{V}-\int_{\Omega}G(d_{E}\tau, x, y)\frac{V(t-\tau, y)I(t-\tau, y)}{V^* I}{\rm d}y\right.\\ &&\left.-\ln\frac{VI}{V^* I^*}+\int_{\Omega}G(d_{E}\tau, x, y)\ln\frac{V(t-\tau, y)I(t-\tau, y)}{V^* I^*}{\rm d}y\right)\\ & = & A-\mu S^*\left(g\left(\frac{S^*}{S}\right)+g\left(\frac{S}{S^*}\right)\right)-(\mu+\gamma)V^* g\left(\frac{V}{V^*}\right)\\ &&-\alpha S^*\left(g\left(\frac{S^*}{S}\right)+g\left(\frac{SV^*}{S^* V}\right)\right)\\ &&-\beta_1S^* I^*\left(g\left(\frac{S^*}{S}\right)+\int_{\Omega}G(d_{E}\tau, x, y)g\left(\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I}\right){\rm d}y\right)\\ &&-\beta_2V^* I^*\int_{\Omega}G(d_{E}\tau, x, y)g\left(\frac{V(t-\tau, y)I(t-\tau, y)}{V^* I}\right){\rm d}y. \end{eqnarray*} $

由于$ \int_{\Omega}\Delta \kappa {\rm d}x = 0 $和$ \int_{\Omega}\frac{\Delta \kappa}{\kappa}{\rm d}x = \int_{\Omega}\frac{\|\nabla\kappa\|^2}{\kappa^2}{\rm d}x\geq 0 $, 故

(3.1) $ \begin{eqnarray} \frac{{\rm d}{\Bbb V}(t)}{{\rm d}t}& = &\int_{\Omega}\frac{\partial({\Bbb V}_1(t, x)+{\Bbb V}_2(t, x))}{\partial t}{\rm d}x {}\\ &\leq&-\int_{\Omega}\left[\mu S^*\left(g\left(\frac{S^*}{S}\right)+g\left(\frac{S}{S^*}\right)\right) +\alpha S^*\left(g\left(\frac{S^*}{S}\right)+g\left(\frac{SV^*}{S^*V}\right)\right)\right.{}\\ &&\left.+(\mu+\gamma)V^* g\left(\frac{V}{V^*}\right)+\beta_1S^* I^*g\left(\frac{S^*}{S}\right)\right]{{\rm d}x}{}\\ &&-\beta_1S^* I^*\int_{\Omega}\int_{\Omega}G(d_{E}\tau, x, y)g\left(\frac{S(t-\tau, y)I(t-\tau, y)}{S^* I}\right){\rm d}y{\rm d}x{}\\ &&-\beta_2V^* I^*\int_{\Omega}\int_{\Omega}G(d_{E}\tau, x, y)g\left(\frac{V(t-\tau, y)I(t-\tau, y)} {V^* I}\right){\rm d}y{\rm d}x: = {\cal U}_{\psi}(t). \end{eqnarray} $

对$ t\geq0 $, $ {\Bbb V}({\bf u}_{t}(\psi)) $是非增和有下界的.因此, 存在一个常数$ {\cal G}>0 $使得$ \lim\limits_{t\rightarrow \infty}{\Bbb V}({\bf u}_{t}(\psi)) = {\cal G}. $进一步, 存在一个序列$ l_{n}\rightarrow \infty $使得在$ {\Bbb H} $内有$ \lim\limits_{n\rightarrow \infty}{\bf u}_{l_n}(\psi) = \varphi $.这意味着对$ \forall\varphi\in\omega(\psi) $有$ {\Bbb V}(\varphi) = {\cal G} $.故当$ t\geq0 $时, 由$ {\bf u}_t(\varphi)\in\omega(\psi) $知$ {\Bbb V}({\bf u}_t(\varphi)) = {\cal G} $.故有$ \frac{{\rm d}{\Bbb V}({\bf u}_t(\varphi))}{{\rm d}t} = 0 $.根据不等式(3.1)得

$ 0 = \frac{{\rm d}{\Bbb V}({\bf u}_t(\varphi))}{{\rm d}t}\leq{\cal U}_{\varphi}(t)\leq0. $

因此, $ t\geq0 $时有$ {\cal U}_{\varphi}(t) = 0 $.由系统(1.3)知, 对$ \forall\ t\geq\tau $有$ {\bf u}_t(\varphi) = {\bf u}^* $.故$ t\geq\tau $时, 由于$ \varphi\in\omega(\psi) $是任意的, 故得$ {\bf u}_t(\omega(\psi)) = {\bf u}^* $.利用$ \omega $ -极限集的不变性, 得$ \omega(\psi) = {\bf u}_{\tau}(\omega(\psi)) = {\bf u}^* $, 即$ \lim\limits_{t\rightarrow \infty}{\bf u}(t, x;\psi) = {\bf u}^{*} $.

当不考虑扩散和空间异质时, 系统(1.3)即为如下的时滞系统:

(4.1) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } \frac{{\rm d}S(t)}{{\rm d}t} = \lambda-\beta_1SI-(\mu+\alpha)S, \\ { } \frac{{\rm d}V(t)}{{\rm d}t} = \alpha S-\beta_2VI-(\mu+\gamma)V, \\ { } \frac{{\rm d}I(t)}{{\rm d}t} = {\rm e}^{-(\mu+d_{E})\tau}[\beta_1S(t-\tau)+\beta_2V(t-\tau)]I(t-\tau)-(\mu+\delta_{I})I. \end{array} \right. \end{eqnarray} $

有关系统(4.1)的正平衡点的全局吸引性的结果可参见文献[12-14].当时滞$ \tau = 0 $时, 参见文献[15].

当不考虑固定潜伏期时, 系统(1.3)即为如下的反应扩散系统:

(4.2) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } S_{t} = d_{S}\Delta S+\lambda-\beta_1SI-(\mu+\alpha)S, \ t>0, \ x\in\Omega, \\ { } V_{t} = d_{V}\Delta V+\alpha S-\beta_2VI-(\mu+\gamma)V, \ t>0, \ x\in\Omega, \\ { } I_{t} = d_{I}\Delta I+\beta_1SI+\beta_2VI-(\mu+\delta_{I})I, \ t>0, \ x\in\Omega, \\ { } \frac{\partial S}{\partial{\bf \nu}} = \frac{\partial V}{\partial{\bf \nu}} = \frac{\partial I}{\partial{\bf \nu}} = 0, \ t>0, \ x\in\partial\Omega. \end{array} \right. \end{eqnarray} $

系统(4.2)的正平衡点的全局吸引性的结果见文献[16].

当$ \beta_2(x) = 0 $, 系统(1.2)已被文献[17]讨论过.进一步, 文献[18]在文献[17]的基础上考虑时间异质并推广了文献[17]中的结果.注意到文献[17, 18]中并未讨论正平衡态的全局吸引性.对于空间异质的反应扩散传染病模型, 研究其正平衡态的全局吸引性是非常困难的.当$ {\cal R}_0>1 $时, 本文利用Lyapunov泛函的方法讨论了系统(1.3)的正平衡点的全局吸引性.