登革热是一种由五种不同血清型病毒(DENV 1-5) 引起的经蚊子传播的虫媒疾病, 其主要传播媒介为埃及伊蚊和白纹伊蚊^[1–4]. 登革热及各类虫媒疾病的严重性促使众多研究者借助数学模型来研究疾病的传播与控制^[5–9]. 众所周知, 包括登革热在内的多数传染病均具有潜伏期, 而潜伏期内的个体具有随机移动性. 因此, 考虑个体扩散和潜伏期对疾病传播的影响是必要的^[10]. 特别地, 虽然易感者和染病者具有不同扩散性, 但目前的研究工作更偏向于考虑相同扩散系数^[9]. 受上述工作启发, 为研究不同扩散和潜伏期对疾病传播的影响, 本文研究如下具有离散时滞的反应扩散登革热模型

(1.1) $ \begin{equation} \left\{\begin{array}{ll} &\partial_tS_{h}(x, t)=d_1\Delta S_{h}(x, t)-k_1S_{h}(x, t)I_{v}(x, t-\tau_1), \\ &\partial_tI_{h}(x, t)=d_2\Delta I_{h}(x, t)+k_1S_{h}(x, t)I_{v}(x, t-\tau_1) -(\mu_h+\alpha_h)I_{h}(x, t), \\ &\partial_tR_{h}(x, t)=d_1\Delta R_{h}(x, t)+\alpha_hI_{h}(x, t), \\ &\partial_tS_{v}(x, t)=d_3\Delta S_{v}(x, t)-k_2S_{v}(x, t)I_{h}(x, t-\tau_2), \\ &\partial_tI_{v}(x, t)=d_4\Delta I_{v}(x, t)+k_2S_{v}(x, t)I_{h}(x, t-\tau_2) -\mu_vI_{v}(x, t), \end{array} \right. \end{equation} $

其中$ \Delta $表示Laplace算子, $ (x, t)\in {{\Bbb R}} \times {{\Bbb R}} ^+ $. $ S_{h}(x, t) $, $ I_{h}(x, t) $和$ R_{h}(x, t) $表示$ t $时刻$ x $位置易感、染病和恢复人群的密度；$ S_{v}(x, t) $和$ I_{v}(x, t) $表示$ t $时刻$ x $位置易感和染病蚊子的密度. $ \tau_1 $和$ \tau_2 $分别表示登革病毒的外部和内部潜伏期. 模型(1.1) 中的所有系数均为正常数, 生物解释见表 1. 记$ \gamma_{2}:=\mu_h+\alpha_h $, $ \gamma_{4}:=\mu_{v} $.

众所周知, 行波在流行病学研究中扮演重要角色, 其重要性使许多学者致力于行波研究(参见文献[11–20]). 然而, 对于时滞蚊媒疾病模型的行波问题研究较少, 原因之一在于方程维数较高处理起来较为复杂; 其次, 模型对应的特征方程为超越方程, 使得现有方法不能直接用来讨论最小正特征根的存在性. 基于上述分析, 本文旨在研究模型(1.1) 行波解的存在性和不存在性. 由于(1.1)式中$ R_h $相对于其他方程解耦, 因此只需研究如下系统

(1.2) $ \begin{equation} \left\{\begin{array}{ll} &\partial_tS_{h}(x, t)=d_1\Delta S_{h}(x, t)-k_1S_{h}(x, t)I_{v}(x, t-\tau_1), \\ &\partial_tI_{h}(x, t)=d_2\Delta I_{h}(x, t)+k_1S_{h}(x, t)I_{v}(x, t-\tau_1) -\gamma_2I_{h}(x, t), \\ &\partial_tS_{v}(x, t)=d_3\Delta S_{v}(x, t)-k_2S_{v}(x, t)I_{h}(x, t-\tau_2), \\ &\partial_tI_{v}(x, t)=d_4\Delta I_{v}(x, t)+k_2S_{v}(x, t)I_{h}(x, t-\tau_2) -\gamma_4I_{v}(x, t). \end{array} \right. \end{equation} $

设$ S_{h}(x, 0)=S_{h}^0 $和$ S_{v}(x, 0)=S_{v}^0 $分别为疾病开始时人和蚊子的密度, 其中$ S_{h}^0>0 $和$ S_{v}^0>0 $为常数. 因此, 系统(1.2) 存在一无病平衡点$ E_0=(S_h^0, 0, S_v^0, 0) $.

本文安排如下: 第二节, 给出本文主要结果; 第三节, 利用不动点定理证明辅助系统行波解的存在性; 第四节, 通过Ascoli-Arzelà引理证明系统(1.2) 行波解的存在性；第五节, 借助双边Laplace变换和反证法研究行波的不存在性; 第六节, 探讨最小波速关于模型参数的连续依赖性.

若不考虑潜伏期和扩散率, 则系统(1.2)变为如下常微分系统

(2.1) $ \begin{equation} \left\{\begin{array}{ll} &{\rm d}S_{h}(t)/{\rm d}t=-k_1S_{h}(t)I_{v}(t), \\ &{\rm d}I_{h}(t)/{\rm d}t=k_1S_{h}(t)I_{v}(t) -\gamma_2I_{h}(t), \\ &{\rm d}S_{v}(t)/{\rm d}t=-k_2S_{v}(t)I_{h}(t), \\ &{\rm d}I_{v}(t)/{\rm d}t=k_2S_{v}(t)I_{h}(t) -\gamma_4I_{v}(t).\\ \end{array} \right. \end{equation} $

令

$ \begin{eqnarray*} F=\left(\begin{array}{ccc} 0 \; & k_1S_h^0 \\ k_2S_v^0 \; & 0 \end{array}\right), {\qquad} V=\left(\begin{array}{ccc} \gamma_2 \; & 0 \\ 0 \; & \gamma_4 \end{array}\right). \end{eqnarray*} $

根据文献[23]知(2.1)式的基本再生数为$ {\cal R}_0=(\tilde{k}/\tilde{\gamma})^{1/2} $, 其中$ \tilde{\gamma}=\gamma_2\gamma_4 $, $ \tilde{k}=\tilde{k}_1\tilde{k}_2 $, $ \tilde{k}_1=k_1S_{h}^0 $, $ \tilde{k}_2=k_2S_{v}^0 $. 称$ (S_{h}, I_{h}, S_{v}, I_{v}) $是(1.2)式的行波解, 若$ (S_{h}(\cdot, t), I_{h}(\cdot, t), S_{v}(\cdot, t), I_{v}(\cdot, t))= (U_1(z), U_2(z), U_3(z), U_4(z)):=U(z) $, 其中$ z=\cdot+ct $, $ (\cdot, t)\in {{\Bbb R}} \times {{\Bbb R}} ^+ $, $ c>0 $表示波速, $ U(z) $满足如下系统

(2.2) $ \begin{equation} \left\{\begin{array}{ll} &cU'_1(z)=d_1U''_1(z)-k_1U_1(z)U_4(z-c\tau_1), \\ &cU'_2(z)=d_2U''_2(z)+k_1U_1(z)U_4(z-c\tau_1)-\gamma_2 U_2(z), \\ &cU'_3(z)=d_3U''_3(z)-k_2U_3(z)U_2(z-c\tau_2), \\ &cU'_4(z)=d_4U''_4(z)+k_2U_3(z)U_2(z-c\tau_2)-\gamma_4U_4(z), \end{array} \right. \end{equation} $

及边界条件

(2.3) $ \begin{equation} (U_1(-\infty), U_2(-\infty), U_3(-\infty), U_4(-\infty)) =(S_{h}^{0}, 0, S_{v}^{0}, 0), \end{equation} $

(2.4) $ \begin{equation} (U_1(+\infty), U_2(+\infty), U_3(+\infty), U_4(+\infty)) =(S_{h}^{\infty}, 0, S_{v}^{\infty}, 0), \end{equation} $

且

(2.5) $ \begin{equation} U_1(z), U_3(z)\ \mbox{关于} z\in{{\Bbb R}} 单\mbox{调递减}, \ S_{h}^{0}>S_{h}^{\infty}\geq0, S_{v}^{0}>S_{v}^{\infty}\geq0. \end{equation} $

定理2.1  设$ {\cal R}_{0}>1 $, 则存在$ c_{\ast}>0 $, 使得对任意$ c\geq c_{\ast} $, 系统(1.2) 存在波速为$ c $的有界正行波$ U(z) $, 满足(2.3)–(2.5)式, 其中$ 0<U_2(z)\leq S_{h}^{0}-S_{h}^{\infty} $, $ 0<U_4(z)\leq S_{v}^{0}-S_{v}^{\infty} $且

(2.6) $ \begin{equation} \lim\limits_{z\rightarrow -\infty}\kappa_2^{-1}e^{-\zeta_1z}U_2(z)=1, \lim\limits_{z\rightarrow -\infty}\kappa_4^{-1}e^{-\zeta_1z}U_4(z)=1. \end{equation} $

另外

(2.7) $ \begin{equation} \int_{-\infty}^{+\infty}U_2(z){\rm d}z=\int_{-\infty}^{+\infty}U_1(z)U_4(z-c\tau_1){\rm d}z =\frac{c}{\gamma_2}(S_{h}^{0}-S_{h}^{\infty}), \end{equation} $

(2.8) $ \begin{equation} \int_{-\infty}^{+\infty}U_4(z){\rm d}z=\int_{-\infty}^{+\infty}U_3(z)U_2(z-c\tau_2){\rm d}z =\frac{c}{\gamma_4}(S_{v}^{0}-S_{v}^{\infty}), \end{equation} $

且$ {\cal R}_0^{\infty}:=(k_1k_2S_{h}^\infty S_{v}^\infty/\tilde{\gamma})^{1/2}<1 $, 其中$ c_\ast $, $ \zeta_1 $, $ \kappa_2 $, $ \kappa_4 $均正常数, 由下面引理3.1确定.

定理2.2  若$ {\cal R}_{0}>1 $, $ 0<c<c_{\ast} $, 则系统(1.2) 不存在满足(2.3) 和(2.4)式的正行波解.

定理2.3  若$ {\cal R}_{0}\leq1 $, 则对任意$ c>0 $, 系统(1.2) 不存在满足(2.3) 和(2.4) 式的正行波解.

注2.1  根据定理2.1–2.3可知$ c_\ast $为系统(1.2) 的最小波速.

将(1.2) 式在$ E_0 $处线性化得

(3.1) $ \begin{equation} \left\{\begin{array}{ll} &c\psi'_2(z)=d_2\psi''_2(z)+\tilde{k}_1\psi_4(z-c\tau_1)-\gamma_2\psi_2(z), \\ &c\psi'_4(z)=d_4\psi''_4(z)+\tilde{k}_2\psi_2(z-c\tau_2)-\gamma_4\psi_4(z). \end{array} \right. \end{equation} $

令$ (\psi_2(z), \psi_4(z))^T= (\kappa_2, \kappa_4)^Te^{\lambda z} $, 代入(3.1) 式得

(3.2) $ \begin{equation} \left\{\begin{array}{ll} &c\kappa_2\zeta=d_2\kappa_2\zeta^2+\tilde{k}_1\kappa_4e^{-c\zeta\tau_1}-\gamma_2\kappa_2, \\ &c\kappa_4\zeta=d_4\kappa_4\zeta^2+\tilde{k}_2\kappa_2e^{-c\zeta\tau_2}-\gamma_4\kappa_4. \end{array} \right. \end{equation} $

系统(3.2) 可改写为$ \Sigma(c, \zeta)\Theta=0 $, 其中

$ \begin{eqnarray*} \Sigma(c, \zeta):=\left(\begin{array}{ccc} Q_2(c, \zeta)\; & \tilde{k}_1e^{-c\zeta\tau_1} \\ \tilde{k}_2e^{-c\zeta\tau_2} \; & Q_4(c, \zeta) \end{array}\right), \Theta:=\left(\begin{array}{ccc} \kappa_2\\ \kappa_4 \end{array}\right), Q_j(c, \zeta):=d_j\zeta^2-c\zeta-\gamma_j, j=2, 4. \end{eqnarray*} $

因此, 系统(2.2) 对应的特征方程为

(3.3) $ \begin{equation} H(c, \zeta):=Q_2(c, \zeta)Q_4(c, \zeta) -\tilde{k}e^{-c\zeta(\tau_1+\tau_2)}=0. \end{equation} $

记$ Q(c, \zeta):=Q_{2}(c, \zeta)+Q_{4}(c, \zeta)= (d_2+d_4)\zeta^2-2c\zeta-(\gamma_2+\gamma_4) $, 且

$ \zeta_1^{\pm}=\frac{c\pm\sqrt{c^2+4d_2\gamma_2}}{2d_2}, \zeta_2^{\pm}=\frac{c\pm\sqrt{c^2+4d_4\gamma_4}}{2d_4}, \zeta_3^{\pm}=\frac{c\pm\sqrt{c^2+(d_2+d_4)(\gamma_2+\gamma_4)}}{d_2+d_4}, $

则$ \zeta_1^{\pm} $, $ \zeta_2^{\pm} $和$ \zeta_3^{\pm} $分别是$ Q_2 $, $ Q_4 $和$ Q $的根. 令$ \zeta_m^{+}:=\min\{\zeta_i^{+}, i=1, 2, 3\} $. 显然$ \zeta_m^{+}=\min\{\zeta_1^{+}, \zeta_2^{+}\} $. 利用文献[16, 24–25]中的方法, 易证如下结论(如图 1所示).

引理3.1  设$ {\cal R}_0>1 $, 则存在正常数$ c_{\ast} $和$ \zeta^{\ast} $使得

(3.4) $ \begin{equation} {\left. {\frac{{\partial {H(c, \zeta)}}}{{\partial {\zeta }}}} \right|_{{(c_{\ast}, \zeta^{\ast})}}} = 0, \quad H(c_{\ast}, \zeta^{\ast})=0. \end{equation} $

进一步, 有

$ ({\rm{i}}) $若$ c\in(0, c_{\ast}) $, 则$ H(c, \zeta)<0 $, $ \zeta\in[0, \zeta_m^+) $;

$ ({\rm{ii}}) $若$ c\in(c_{\ast}, \infty) $, 则(3.3) 存在正根$ \zeta_1:=\zeta_{m1}(c) $和$ \zeta_2:=\zeta_{m2}(c) $, 满足$ 0<\zeta_1<\zeta^{\ast}<\zeta_2<\zeta_m^+ $, 使得$ \zeta'_{m1}(c)<0 $, $ \zeta'_{m2}(c)>0 $, $ Q_j(c, \zeta_i)<0 $$ (i=1, 2, j=2, 4) $且

(3.5) $ \begin{equation} H(c, \zeta)=\left\{\begin{array}{ll} &<0, \zeta\in[0, \zeta_1) \cup(\zeta_2, \zeta_2+\vartheta), \\ &>0, \zeta\in(\zeta_1, \zeta_2), \end{array}\quad \right. \end{equation} $

其中$ ' $表示关于$ c $的导数, $ \vartheta>0 $充分小. 令$ \kappa_2=\tilde{k}_1e^{-c\zeta_1\tau_1}>0 $, $ \kappa_4=-Q_2(c, \zeta_1)>0 $, 则$ \Sigma(c, \zeta_1)\Theta=0 $.

为研究系统(2.2) 正解的存在性, 引入辅助系统

(3.6) $ \begin{equation} \left\{\begin{array}{ll} &c\Psi'_1(z)=d_1\Psi''_1(z)+P_1(\Psi)(z), \\ &c\Psi'_2(z)=d_2\Psi''_2(z)+P_2(\Psi)(z), \\ &c\Psi'_3(z)=d_3\Psi''_3(z)+P_3(\Psi)(z), \\ &c\Psi'_4(z)=d_4\Psi''_4(z)+P_4(\Psi)(z), \end{array} \right. \end{equation} $

其中

(3.7) $ \begin{equation} \left\{\begin{array}{ll} &P_1(\Psi)(z)=-k_1\Psi_1(z)\Psi_4(z-c\tau_1), \\ &P_2(\Psi)(z)=k_1\Psi_1(z)\Psi_4(z-c\tau_1) -\gamma_2\Psi_2(z)-\epsilon\Psi_2^2(z), \\ &P_3(\Psi)(z)=-k_2\Psi_3(z)\Psi_2(z-c\tau_2), \\ &P_4(\Psi)(z)=k_2\Psi_3(z)\Psi_2(z-c\tau_2) -\gamma_4\Psi_4(z)-\epsilon\Psi_4^2(z), \end{array} \right. \end{equation} $

这里$ \Psi:=(\Psi_1, \Psi_2, \Psi_3, \Psi_4) $, $ z\in{{\Bbb R}} $, $ \epsilon\in(0, 1] $充分小.

设$ {\cal R}_0>1 $, $ c>c_\ast $. 为应用不动点定理先构造合适上下解(定义见文献[26])

$ \begin{eqnarray*} \begin{array}{ll} &\overline{\Psi}_1(z):=S_{h}^0, \:\quad\quad\quad\quad\quad\quad\quad\underline{\Psi}_1(z) :=\max\{S_{h}^0(1-\chi_1e^{\epsilon_1z}), 0\}, \\ &\overline{\Psi}_2(z):=\min\{\kappa_2e^{\zeta_1z}, \kappa_2{\cal K}^\ast\}, \;\; \underline{\Psi}_2(z) :=\max\{\kappa_2e^{\zeta_1z}(1-\chi_2e^{\epsilon_2z}), 0\}, \\ &\overline{\Psi}_3(z):=S_{v}^0, \:\:\quad\quad\quad\quad\quad\quad\quad\underline{\Psi}_3(z) :=\max\{S_{v}^0(1-\chi_3e^{\epsilon_3z}), 0\}, \\ &\overline{\Psi}_4(z):=\min\{\kappa_4e^{\zeta_1z}, \kappa_4{\cal K}^\ast\}, \;\;\: \underline{\Psi}_4(z) :=\max\{\kappa_4e^{\zeta_1z}(1-\chi_4e^{\epsilon_2z}), 0\}, \\ \end{array} \end{eqnarray*} $

其中$ z\in{{\Bbb R}} $, $ {\cal K}^\ast $, $ \chi_i $ ($ i=1, 2, 3, 4 $) 和$ \epsilon_j $ ($ j=1, 2, 3 $) 将在后文给出.

不难证明如下引理.

引理3.2  函数$ \overline{\Psi}_1(z)=S_{h}^0 $和$ \overline{\Psi}_3(z)=S_{v}^0 $满足

(3.8) $ \begin{equation} c\overline{\Psi}'_1\geq d_1\overline{\Psi}''_1+P_1(\overline{\Psi}_1, \underline{\Psi}_2, \overline{\Psi}_3, \underline{\Psi}_4), c\overline{\Psi}'_3\geq d_3\overline{\Psi}''_3+P_3(\overline{\Psi}_1, \underline{\Psi}_2, \overline{\Psi}_3, \underline{\Psi}_4), z\in{{\Bbb R}} . \end{equation} $

注3.1  引理3.2表明$ \overline{\Psi}_1 $和$ \overline{\Psi}_3 $为系统(3.6) 的上解.

引理3.3  设

(3.9) $ \begin{equation} {\cal K}^\ast>\max\left\{1, \frac{\kappa_4\tilde{k}_1-\gamma_2\kappa_2} {\epsilon\kappa_2^2}, \frac{\kappa_2\tilde{k}_2-\gamma_4\kappa_4} {\epsilon\kappa_4^2}\right\}. \end{equation} $

则$ \overline{\Psi}_j(z) $满足

(3.10) $ \begin{equation} c\overline{\Psi}'_j\geq d_j\overline{\Psi}''_j+P_j(\overline{\Psi}_1, \overline{\Psi}_2, \overline{\Psi}_3, \overline{\Psi}_4), \forall z\neq\hat{z}:=\frac{\ln {\cal K}^\ast}{\zeta_1}, j=2, 4. \end{equation} $

证  当$ z<\hat{z} $时, 则$ \overline{\Psi}_2(z)=\kappa_2e^{\zeta_1z} $和$ \overline{\Psi}_4(z)=\kappa_4e^ {\zeta_1z} $. 因此, 有

$ c\overline{\Psi}'_j- d_2\overline{\Psi}''_j-P_j(\overline{\Psi}_1, \overline{\Psi}_2, \overline{\Psi}_3, \overline{\Psi}_4) =\epsilon\overline{\Psi}_j^2\geq0, j=2, 4. $

当$ z>\hat{z} $时, $ \overline{\Psi}_2(z)=\kappa_2{\cal K}^\ast $和$ \overline{\Psi}_4(z)=\kappa_4{\cal K}^\ast $, 则根据(3.9)式得

$ \begin{eqnarray*} \begin{array}{ll} c\overline{\Psi}'_2-d_2\overline{\Psi}''_2-P_2(\overline{\Psi}_1, \overline{\Psi}_2, \overline{\Psi}_3, \overline{\Psi}_4) =-{\cal K}^\ast(\kappa_4\tilde{k}_1-\gamma_2\kappa_2-\epsilon\kappa_2^2{\cal K}^\ast)\geq0. \end{array} \end{eqnarray*} $

同理$ c\overline{\Psi}'_4-d_4\overline{\Psi}''_2-P_4(\overline{\Psi}_1, \overline{\Psi}_2, \overline{\Psi}_3, \overline{\Psi}_4) \geq0, z>\hat{z} $. 综上知(3.10) 式成立. 引理证毕.

注3.2  引理3.3表明$ \overline{\Psi}_2 $和$ \overline{\Psi}_4 $为系统(3.6) 的上解.

引理3.4  设

(3.11) $ \begin{equation} 0<\epsilon_1<\frac{1}{2}\min\left\{\zeta_1, \frac{c} {d_1}\right\}, \chi_1>\max\left\{1, \frac{\kappa_4k_1 e^{-c\tau_1\zeta_1}}{(c-d_1\epsilon_1)\epsilon_1}\right\}, \end{equation} $

(3.12) $ \begin{equation} 0<\epsilon_3<\frac{1}{2}\min\left\{\zeta_1, \frac{c} {d_3}\right\}, \chi_3>\max\left\{1, \frac{\kappa_2k_2 e^{-c\tau_2\zeta_1}}{(c-d_3\epsilon_3)\epsilon_3}\right\}, \end{equation} $

则$ \underline{\Psi}_1(z) $和$ \underline{\Psi}_3(z) $满足

(3.13) $ \begin{equation} c\underline{\Psi}'_1\leq d_1\underline{\Psi}''_1+P_1(\underline{\Psi}_1, \overline{\Psi}_2, \underline{\Psi}_3, \overline{\Psi}_4), \forall z\neq\underline{z}_{\Psi_1}:=-\frac{\ln \chi_1}{\epsilon_1} \end{equation} $

和

(3.14) $ \begin{equation} c\underline{\Psi}'_3\leq d_3\underline{\Psi}''_3+P_3(\underline{\Psi}_1, \overline{\Psi}_2, \underline{\Psi}_3, \overline{\Psi}_4), \forall z\neq\underline{z}_{\Psi_3}:=-\frac{\ln \chi_3}{\epsilon_3}. \end{equation} $

证  当$ z<\underline{z}_{\Psi_1} $时, 有$ \underline{\Psi}_1(z) =S_{h}^0(1-\chi_1e^{\epsilon_1z}) $. 由(3.9)式知$ {\cal K}^\ast>1 $, 则$ z<-\ln\chi_1/\epsilon_1<0<\ln{\cal K}^\ast/\zeta_1 =\hat{z} $, 进而$ \overline{\Psi}_4(z)= \kappa_4e^{\zeta_1z} $. 由(3.11) 式得

$ c\underline{\Psi}'_1- d_1\underline{\Psi}''_1-P_1(\underline{\Psi}_1, \overline{\Psi}_2, \underline{\Psi}_3, \overline{\Psi}_4)\leq S_{h}^0\left[(d_1\epsilon_1-c)\epsilon_1\chi_1+\kappa_4k_1 e^{-c\tau_1\zeta_1}\right]e^{\epsilon_1z}\leq0. $

当$ z>\underline{z}_{\Psi_1} $时, $ \underline{\Psi}_1(z)=0 $, 从而$ c\underline{\Psi}'_1- d_1\underline{\Psi}''_1-P_1(\underline{\Psi}_1, \overline{\Psi}_2, \underline{\Psi}_3, \overline{\Psi}_4)=0 $. 所以, (3.13) 式成立. 同理可证(3.14) 式. 引理得证.

注3.3  引理3.4表明$ \underline{\Psi}_1 $和$ \underline{\Psi}_3 $为系统(3.6) 的下解.

引理3.5  设$ \epsilon_2<\min\{\zeta_1, \epsilon_1, \epsilon_3\} $, 则存在充分大的正常数$ \chi_2 $和$ \chi_4 $使$ \underline{\Psi}_2(z) $和$ \underline{\Psi}_4(z) $满足

(3.15) $ \begin{equation} c\underline{\Psi}'_2\leq d_2\underline{\Psi}''_2+P_2(\underline{\Psi}_1, \underline{\Psi}_2, \underline{\Psi}_3, \underline{\Psi}_4), \forall z\neq\underline{z}_{\Psi_2}:=-\frac{\ln \chi_2}{\epsilon_2} \end{equation} $

和

(3.16) $ \begin{equation} c\underline{\Psi}'_4\leq d_4\underline{\Psi}''_4+P_4(\underline{\Psi}_1, \underline{\Psi}_2, \underline{\Psi}_3, \underline{\Psi}_4), \forall z\neq\underline{z}_{\Psi_4}:=-\frac{\ln \chi_4}{\epsilon_2}. \end{equation} $

证  引理3.5的证明类似于文献[16, 引理2.5], 不再赘述.

注3.4  因$ P_j(\underline{\Psi}_1, \underline{\Psi}_2, \underline{\Psi}_3, \underline{\Psi}_4)\leq P_j(\overline{\Psi}_1, \underline{\Psi}_2, \overline{\Psi}_3, \underline{\Psi}_4) $ ($ j=2, 4 $), 则由引理3.5知

$ \begin{eqnarray*} c\underline{\Psi}'_2\leq d_2\underline{\Psi}''_2+P_2(\overline{\Psi}_1, \underline{\Psi}_2, \overline{\Psi}_3, \underline{\Psi}_4), z\neq\underline{z}_{\Psi_2}; c\underline{\Psi}'_4\leq d_4\underline{\Psi}''_4+P_4(\overline{\Psi}_1, \underline{\Psi}_2, \overline{\Psi}_3, \underline{\Psi}_4), z\neq\underline{z}_{\Psi_4}. \end{eqnarray*} $

因此, $ \underline{\Psi}_2 $和$ \underline{\Psi}_4 $为系统(3.6) 的下解.

令$ f_i(\Psi)(z):=r_i\Psi_i(z)+P_i(\Psi)(z), i=1, 2, 3, 4 $, 其中$ P_i $由(3.7) 式给出, $ r_i $满足

(3.17) $ \begin{equation} r_1>k_1\kappa_4{\cal K}^\ast, r_2>\gamma_2+2\epsilon\kappa_2{\cal K}^\ast, r_3>k_2\kappa_2{\cal K}^\ast, r_4>\gamma_4+2\epsilon\kappa_4{\cal K}^\ast. \end{equation} $

因此(3.6)式转化为

(3.18) $ \begin{equation} \left\{\begin{array}{ll} &d_1\Psi''_1(z)-c\Psi'_1(z)-r_1\Psi_1(z)+f_1(\Psi)(z)=0, \\ &d_2\Psi''_2(z)-c\Psi'_2(z)-r_2\Psi_2(z)+f_2(\Psi)(z)=0, \\ &d_3\Psi''_3(z)-c\Psi'_3(z)-r_3\Psi_3(z)+f_3(\Psi)(z)=0, \\ &d_4\Psi''_4(z)-c\Psi'_4(z)-r_4\Psi_4(z)+f_4(\Psi)(z)=0.\\ \end{array} \right. \end{equation} $

定义

$ \Gamma=\left\{(\Psi_1, \Psi_2, \Psi_3, \Psi_4)\in C({{\Bbb R}}, {{\Bbb R}} _+^4): \underline{\Psi}_i(z)\leq \Psi_i(z)\leq\overline{\Psi}_i(z), i=1, 2, 3, 4, z\in {{\Bbb R}} \right\} $

和$ \zeta_{i1}=\frac{c-\sqrt{c^{2}+4d_{i}r_{i}}}{2d_{i}}, \zeta_{i2}=\frac{c+\sqrt{c^{2}+4d_{i}r_{i}}}{2d_{i}}, \Lambda_{i}=d_{i}(\zeta_{i2}-\zeta_{i1}), i=1, \ldots, 4 $, $ {{\Bbb R}} _+=[0, +\infty) $且$ r_i $满足(3.17) 式使得$ -\zeta_{i1}>2\zeta_{1} $. 显然$ \zeta_{i1} $和$ \zeta_{i2} $为$ d_i\zeta^{2}-c\zeta-r_i=0 $的根.

定义算子$ {\cal F}: \Gamma\rightarrow C({{\Bbb R}}, {{\Bbb R}} _+^4) $为

$ {\cal F}(\Psi)(z)=\left({\cal F}_{1}(\Psi)(z), {\cal F}_{2}(\Psi)(z), {\cal F}_{3}(\Psi)(z), {\cal F}_{4}(\Psi)(z)\right)^{T}, $

其中

$ {\cal F}_{i}(\Psi)(z)=\frac{1}{\Lambda_{i}}\left[\int_{-\infty}^{z} e^{\zeta_{i1}(z-\xi)}f_i(\Psi)(\xi){\rm d}\xi+ \int_{z}^{+\infty} e^{\zeta_{i2}(z-\xi)}f_i(\Psi)(\xi){\rm d}\xi\right]. $

易知, 算子$ {\cal F} $有定义且$ d_{i}{\cal F}''_{i}(\Psi)(z)-c{\cal F}'_{i}(\Psi)(z) -r_i{\cal F}_{i}(\Psi)(z)+f_i(\Psi)(z)=0 $, $ \Psi\in\Gamma $. 换言之, $ {\cal F} $的不动点即为系统(3.18) 的解. 易证下面结果.

引理3.6  集合$ \Gamma $是$ C({{\Bbb R}}, {{\Bbb R}} _+^4) $中的有界闭凸集.

引理3.7  算子$ {\cal F} $映$ \Gamma $到$ \Gamma $, 即$ {\cal F}(\Gamma)\subset\Gamma $.

定义Banach空间

$ B_{\rho}({{\Bbb R}}, {{\Bbb R}} _+^4):= \left\{\phi(\cdot)=(\phi_1(\cdot), \phi_2(\cdot), \phi_3(\cdot), \phi_4(\cdot)) \in C({{\Bbb R}}, {{\Bbb R}} _+^4):|\phi|_{\rho} <+\infty\right\} $

赋予范数

$ |\phi|_{\rho}:=\max\left\{\sup\limits_{z\in {{\Bbb R}} }|\phi_1(z)|e^{-\rho|z|}, \sup\limits_{z\in {{\Bbb R}} }|\phi_2(z)|e^{-\rho|z|}, \sup\limits_{z\in {{\Bbb R}} }|\phi_3(z)|e^{-\rho|z|}, \sup\limits_{z\in {{\Bbb R}} }|\phi_4(z)|e^{-\rho|z|}\right\}, $

其中$ \rho>0 $为常数, 满足$ 2\zeta_1<\rho<\min\{-\zeta_{11}, -\zeta_{21}, -\zeta_{31}, -\zeta_{41}\} $.

引理3.8  映射$ {\cal F}=({\cal F}_1, {\cal F}_2, {\cal F}_3, {\cal F}_4):\Gamma\rightarrow C({{\Bbb R}}, {{\Bbb R}} _+^4) $在$ B_{\rho}({{\Bbb R}}, {{\Bbb R}} _+^4) $中关于$ |\cdot|_{\rho} $连续.

证  取$ \phi=(\phi_1, \phi_2, \phi_3, \phi_4) \in\Gamma $和$ \varphi=(\varphi_1, \varphi_2, \varphi_3, \varphi_4) \in\Gamma $满足$ \phi\neq\varphi $. 先证$ {\cal F}_1(\phi)(z) $关于$ |\cdot|_{\rho} $的连续性. 计算知：存在$ C_1>0 $使得

$ \left|f_1(\phi)(\cdot)-f_1(\varphi)(\cdot)\right|_{\rho} \leq C_1[\left|\phi(\cdot)-\varphi(\cdot)\right|_{\rho} +\left|\phi_4(\cdot-c\tau_1)-\varphi_4(\cdot-c\tau_1)\right|_{\rho} ]. $

因此, 有

$ \begin{eqnarray*} &&\left|{\cal F}_1(\phi)(z)-{\cal F}_1(\varphi)(z)\right|e^{-\rho|z|}\\ &\leq&\frac{C_1e^{-\rho|z|}}{\Lambda_1}\left[\left|\phi(\cdot)-\varphi(\cdot)\right|_{\rho} +\left|\phi_4(\cdot-c\tau_1)-\varphi_4(\cdot-c\tau_1)\right|_{\rho}\right]\\ &&\times\left[\int_{-\infty}^{z}e^{\zeta_{11} (z-\xi)+\rho|\xi|}{\rm d}\xi+\int_{z}^{+\infty}e^{\zeta_{12} (z-\xi)+\rho|\xi|}{\rm d}\xi\right]. \end{eqnarray*} $

对于$ z\geq0 $. 因$ \rho<-\zeta_{11}<\zeta_{12} $, 则

$ \begin{eqnarray*} & &\left|{\cal F}_1(\phi)(z)-{\cal F}_1(\varphi)(z)\right|e^{-\rho|z|}\\ &\leq&\frac{C_1}{\Lambda_1}\left[\frac{1}{-\zeta_{11}-\rho} +\frac{\zeta_{12}-\zeta_{11}}{(\rho-\zeta_{11})(\zeta_{12}-\rho)}\right] \left[\left|\phi(\cdot)-\varphi(\cdot)\right|_{\rho} +\left|\phi_4(\cdot-c\tau_1)-\varphi_4(\cdot-c\tau_1)\right|_{\rho}\right]. \end{eqnarray*} $

对于$ z<0 $. 同理可得

$ \begin{eqnarray*} & &\left|{\cal F}_1(\phi)(z)-{\cal F}_1(\varphi)(z)\right|e^{-\rho|z|}\\ &\leq&\frac{C_1}{\Lambda_1}\left[\frac{1}{-\zeta_{11}-\rho} +\frac{2\zeta_{12}}{(\zeta_{12}^2-\rho^2)}\right] \left[\left|\phi(\cdot)-\varphi(\cdot)\right|_{\rho} +\left|\phi_4(\cdot-c\tau_1)-\varphi_4(\cdot-c\tau_1)\right|_{\rho}\right]. \end{eqnarray*} $

所以, 算子$ {\cal F}_1:\Gamma\rightarrow C({{\Bbb R}}, {{\Bbb R}} ) $在$ B_{\rho}({{\Bbb R}}, {{\Bbb R}} ) $中关于$ |\cdot|_{\rho} $连续. 类似地, 可证$ {\cal F}_j $的连续性, $ j=2, 3, 4 $. 引理证毕.

引理3.9  映射$ {\cal F}=({\cal F}_1, {\cal F}_2, {\cal F}_3, {\cal F}_4):\Gamma\rightarrow \Gamma $在$ B_{\rho}({{\Bbb R}} , {{\Bbb R}} _+^4) $中关于$ |\cdot|_{\rho} $是紧的.

证  对$ \forall \Psi(\cdot)=(\Psi_1(\cdot), \Psi_2(\cdot), \Psi_3(\cdot), \Psi_4(\cdot))\in\Gamma $, 有

$ \frac{{\rm d}{\cal F}_{1}(\Psi)(z)}{{\rm d}z}= \frac{1}{\Lambda_{1}}\left[\zeta_{11}\int_{-\infty}^{z} e^{\zeta_{11}(z-\xi)}f_1(\Psi)(\xi){\rm d}\xi+ \zeta_{12}\int_{z}^{+\infty} e^{\zeta_{12}(z-\xi)}f_1(\Psi)(\xi){\rm d}\xi\right]. $

因$ \Psi(\cdot)\in\Gamma $, 则存在$ C_2>0 $使得$ |f_1(\Psi)(z)|\leq C_2 $, $ z\in{{\Bbb R}} $. 因此, 有

$ \left|\frac{{\rm d}{\cal F}_{1}(\Psi)(z)}{{\rm d}z}\right| \leq\frac{C_2}{\Lambda_{1}}\left[\left|\zeta_{11}\right|\int_{-\infty}^{z} e^{\zeta_{11}(z-\xi)}{\rm d}z+\lambda_{12}\int_{z}^{+\infty} e^{\zeta_{12}(z-\xi)}{\rm d}z\right]=\frac{2C_2}{\Lambda_{1}}, {\quad} \forall z\in{{\Bbb R}} . $

从而$ \left|{\rm d}{\cal F}_{1}(\Psi)(z)/{\rm d}z\right| $有界. 类似可得$ \left|{\rm d}{\cal F}_{j}(\Psi)(z)/{\rm d}z\right| $的有界性, $ j=2, 3, 4 $, 于是$ {\cal F}(\Gamma) $在任意紧区间上关于$ |\cdot|_{\rho} $等度连续且一致有界.

固定整数$ n\in{\Bbb N}^\ast $, 定义算子$ {\cal F}^n $, 有

$ \begin{eqnarray*} {\cal F}^n(\Psi)(z)=\left\{\begin{array}{ll} {\cal F}(\Psi)(-n), \;&z\in(-\infty, -n], \\ {\cal F}(\Psi)(z), &z\in[-n, n], \\ {\cal F}(\Psi)(n), &z\in[n, +\infty). \end{array} \right. \end{eqnarray*} $

由Ascoli-Arzelà引理知, $ {\cal F}^n(\Psi)(z) $关于$ \Psi(\cdot)\in\Gamma $等度连续且一致有界. 因此, $ {\cal F}^n:\Gamma\rightarrow \Gamma $在$ C({{\Bbb R}}, {{\Bbb R}} _+^4) $上关于上确界范数是紧的. 于是, $ {\cal F}^n $在$ B_{\rho}({{\Bbb R}}, {{\Bbb R}} _+^4) $中关于$ |\cdot|_{\rho} $紧. 又因

$ \left|{\cal F}_{1}(\Psi)(z)\right| \leq\frac{C_2}{\Lambda_{1}}\left[\int_{-\infty}^{z} e^{\zeta_{11}(z-\xi)}{\rm d}\xi+\int_{z}^{+\infty} e^{\zeta_{12}(z-\xi)}{\rm d}\xi\right]=\frac{C_2}{d_{1}|\zeta_{11}| \zeta_{12}}, {\quad} z\in{{\Bbb R}}, $

则

$ \begin{eqnarray*} \left|{\cal F}_1^n(\Psi)(\cdot)-{\cal F}_1(\Psi)(\cdot)\right|_{\rho} &=&\sup\limits_{z\in(-\infty, -n]\cup[n, +\infty)} \left|{\cal F}_1^n(\Psi)(z)- {\cal F}_1(\Psi)(z)\right|e^{-\rho|z|}\\ &\leq&\frac{2C_2}{d_{1}|\zeta_{11}|\zeta_{12}} e^{-\rho n}\rightarrow0, {\quad} n\rightarrow +\infty. \end{eqnarray*} $

同理, $ |{\cal F}_j^n(\Psi) $$ (\cdot)-{\cal F}_j (\Psi)(\cdot)|_{\rho}\rightarrow0, n\rightarrow +\infty, j=2, 3, 4 $. 进而, 当$ n\rightarrow +\infty $时, $ |{\cal F}^n(\Psi)(\cdot) $$ -{\cal F} (\Psi)(\cdot)|_{\rho}\rightarrow0 $. 运用紧算子逼近定理(见文献[27, 命题2.12])知, $ {\cal F}^n $在$ \Gamma $中关于$ |\cdot|_\rho $收敛到$ {\cal F} $, $ n\rightarrow +\infty $. 这表明算子$ {\cal F} $在$ B_{\rho}({{\Bbb R}}, {{\Bbb R}} ^4) $中关于$ |\cdot|_{\rho} $是紧的. 引理得证.

命题3.1  设$ {\cal R}_0>1 $, 则对任意$ c>c_\ast $, 系统(3.6) 存在一有界正解$ \Psi(z) $满足(2.3)式和$ \Psi(+\infty) $$ =(\Psi_1^{\infty}, 0, \Psi_3^{\infty}, 0) $, 其中$ \Psi_1'(z)<0 $, $ \Psi_3'(z)<0 $, $ \Psi_1(z)<S_{h}^0 $, $ \Psi_3(z)<S_{v}^0 $, $ \Psi_2(z)\leq S_h^0-\Psi_{1}^{\infty} $, $ \Psi_4(z)\leq S_v^0-\Psi_{3}^{\infty} $, $ z\in{{\Bbb R}} $, 这里$ \Psi_1^{\infty} $, $ \Psi_3^{\infty}\geq0 $, 满足$ \Psi_1^{\infty}<S_{h}^{0} $且$ \Psi_3^{\infty}<S_{v}^{0} $. 进一步, 有$ \lim\limits_{z\rightarrow -\infty}\kappa_2^{-1} e^{-\zeta_1z}\Psi_2(z)=1, \lim\limits_{z\rightarrow -\infty}\kappa_4^{-1}e^{-\zeta_1z}\Psi_4(z)=1 $, 和

(3.19) $ \begin{equation} \int_{-\infty}^{+\infty}\Psi_2(z){\rm d}z<\frac{c}{\gamma_2}(S_{h}^{0}-\Psi_{1}^{\infty}), \int_{-\infty}^{+\infty}\Psi_4(z){\rm d}z<\frac{c}{\gamma_4}(S_{v}^{0}-\Psi_{3}^{\infty}). \end{equation} $

证  结合引理3.6–3.9, 利用Schauder不动点定理知, 系统(3.6) 存在非负有界解$ \Psi(\cdot)\in\Gamma $, 显然它是$ {\cal F} $的不动点. 因$ \Psi\in\Gamma $, 所以有$ \Psi(-\infty)=(S_{h}^0, 0, S_{v}^0, 0) $. 又因

$ \kappa_2e^{\zeta_1z}(1-\chi_2e^{\epsilon_2z})\leq\Psi_2(z)\leq\kappa_2e^{\zeta_1z}, \kappa_4e^{\zeta_1z}(1-\chi_4e^{\epsilon_2z})\leq\Psi_4(z)\leq\kappa_4e^{\zeta_1z}, $

则$ \kappa_2^{-1}e^{-\zeta_1z}\Psi_2(z)\rightarrow1 $, $ \kappa_4^{-1}e^{-\zeta_1z}\Psi_4(z)\rightarrow1 $, $ z\rightarrow -\infty $. 根据反证法和最大值原理^[28]可知: $ 0<\Psi_1(z)<S_{h}^0 $, $ 0<\Psi_3(z)<S_{v}^0 $, $ \Psi_j(z)>0 $, $ z\in{{\Bbb R}} $, $ j=2, 4 $.

(ⅰ) 为证$ \Psi_1(z) $和$ \Psi_3(z) $关于$ z $单调递减. 因$ \Psi(z) $是$ {\cal F} $的不动点, 由洛必达法则^[29]得$ \Psi'_i(-\infty)=0 $, $ i=1, 2, 3, 4 $. 在系统(3.6)中, 令$ z\rightarrow -\infty $, 并结合$ \Psi(-\infty)=(S_{h}^0, 0, S_{v}^0, 0) $知$ \Psi''_i(-\infty)=0 $. 对(3.6) 的第一个方程两端同乘$ e^{-cz/d_1} $得

$ [\Psi'_1(z)e^{-cz/d_1}]'=\frac{k_1}{d_1}e^{-cz/d_1} \Psi_1(z)\Psi_4(z-c\tau_1), z\in{{\Bbb R}} . $

因$ \Psi_1(z) $有界, 则对上式两端在$ z $到$ +\infty $上积分得

$ \Psi'_1(z)=-\frac{k_1}{d_1}\int_z^{+\infty}e^{-c(\xi-z)/d_1} \Psi_1(\xi)\Psi_4(\xi-c\tau_1){\rm d}\xi<0, z\in{{\Bbb R}} . $

因此, $ \Psi_1(z) $关于$ z $单调递减. 又因$ 0<\Psi_1(z)\leq S_h^0 $, 则由单调有界原理知, 存在$ \Psi_{1}^{\infty}\geq0 $满足$ \Psi_1^{\infty}<S_h^0 $, 使得$ \Psi_1(z)\rightarrow \Psi_1^{\infty} $, $ z\rightarrow +\infty $. 同理, 可证$ \Psi_3(z) $关于$ z $单调递减且存在常数$ \Psi_3^{\infty}\geq0 $满足$ \Psi_3^{\infty}<S_v^0 $, 使得$ \Psi_3(z)\rightarrow \Psi_3^{\infty} $, $ z\rightarrow +\infty $.

(ⅱ) 为证$ \Psi_2(+\infty)=0 $且$ \Psi_4(+\infty)=0 $. 对(3.6)式的第一个方程在$ -\infty $到$ z $上积分得

(3.20) $ \begin{equation} d_1\Psi'_1(z)=c[\Psi_1(z)-S_{h}^{0}]+k_1\int_{-\infty}^{z}\Psi_1(\xi) \Psi_4(\xi-c\tau_1){\rm d}\xi. \end{equation} $

因$ \Psi'_1 $和$ \Psi_1 $有界, 则$ \int_{-\infty}^{+\infty}\Psi_1(\xi) \Psi_4(\xi-c\tau_1){\rm d}\xi<+\infty $. 接着, 对系统(3.6) 的第二个方程在$ -\infty $到$ z $上积分得

(3.21) $ \begin{equation} c\Psi_2(z)=d_2\Psi'_2(z)+k_1\int_{-\infty}^{z}\Psi_1(\xi)\Psi_4(\xi-c\tau_1){\rm d}\xi -\gamma_2\int_{-\infty}^{z}\Psi_2(\xi){\rm d}\xi-\epsilon\int_{-\infty}^{z}\Psi_2^2(\xi){\rm d}\xi. \end{equation} $

根据$ \Psi_2 $和$ \Psi'_2 $的有界性以及$ \int_{-\infty}^{+\infty}\Psi_1(\xi)\Psi_4(\xi-c\tau_1){\rm d}\xi<+\infty $, 可知$ \int_{-\infty}^{+\infty}\Psi_2(\xi){\rm d}\xi<+\infty $, 从而$ \lim\limits_{z\rightarrow +\infty}\Psi_2(z)=0 $. 同理, 可证$ \lim\limits_{z\rightarrow +\infty}\Psi_4(z)=0 $. 因此, $ \Psi(+\infty)=(\Psi_1^{\infty}, 0, \Psi_3^{\infty}, 0) $.

(ⅲ) 为证(3.19)式. 因$ \Psi_2={\cal F}_2(\Psi) $, 则$ \Psi'_2(+\infty)=0 $. 结合(3.20)–(3.21)式知

$ \begin{eqnarray*} \gamma_2\int_{-\infty}^{+\infty}\Psi_2(\xi){\rm d}\xi+\epsilon\int_ {-\infty}^{+\infty}\Psi_2^2(\xi){\rm d}\xi =k_1\int_{-\infty}^{+\infty}\Psi_1(\xi)\Psi_4(\xi-c\tau_1){\rm d}\xi =c(S_h^0-\Psi_{1}^{\infty}). \end{eqnarray*} $

由$ \epsilon>0 $知, $ \int_{-\infty}^{+\infty}\Psi_2(\xi){\rm d}\xi< c(S_h^0-\Psi_{1}^{\infty})/\gamma_2 $. 同理, $ \int_{-\infty}^{+\infty}\Psi_4(\xi){\rm d}\xi< c(S_v^0-\Psi_{3}^{\infty})/\gamma_4 $. 因此, (3.19) 式成立.

(ⅳ) 最后证$ \Psi_2(\cdot) $和$ \Psi_4(\cdot) $关于$ \epsilon $的一致有界性. 定义泛函

$ {\cal P}(z):=\frac{1}{c} \int_{-\infty}^{z}[\gamma_2\Psi_2(\xi)+\epsilon\Psi_2^2(\xi)]{\rm d}\xi+ \frac{1}{c}\int_{z}^{+\infty}e^{c(z-\xi)/d_2}[\gamma_2\Psi_2(\xi)+\epsilon\Psi_2^2(\xi)]{\rm d}\xi, z\in{{\Bbb R}} . $

显然$ c{\cal P}'(z)-d_2{\cal P}''(z)=\gamma_2\Psi_2(z)+\epsilon\Psi_2^2(z) $, $ {\cal P}(-\infty)=0 $, $ {\cal P}(+\infty)\leq S_{h}^{0} -\Psi_{1}^{\infty} $且$ {\cal P}'(\pm\infty) $$ =0 $. 记$ {\cal H}(z):=\Psi_2(z)+{\cal P}(z) $, 则$ {\cal H}'(+\infty)=0 $, $ c{\cal H}'(z)-d_2{\cal H}''(z)=k_1\Psi_1(z)\Psi_4(z-c\tau_1) $. 因此, 类似上面过程得

$ {\cal H}'(z)=\frac{k_1}{d_2}\int_{z}^{+\infty} e^{c(z-\xi)/d_2}\Psi_1(\xi) \Psi_4(\xi-c\tau_1){\rm d}\xi>0, \forall z\in{{\Bbb R}}, $

即$ {\cal H}(z) $关于$ z $单调递增. 又因$ {\cal H}(+\infty)\leq S_{h}^{0}-\Psi_{1}^{\infty} $, $ {\cal H}(z)\geq0 $, 则$ \Psi_2(z)\leq S_{h}^{0}-\Psi_{1}^{\infty} $, $ z\in{{\Bbb R}} $. 同理, $ \Psi_4(z)\leq S_{v}^{0}-\Psi_{3}^{\infty} $, $ z\in{{\Bbb R}} $. 命题得证.

本节完成定理2.1的证明.

定理2.1的证明  设$ \Psi:=(\Psi_1, \Psi_2, \Psi_3, \Psi_4)\in\Gamma $为由命题3.1确定的系统(3.6) 的有界正解, 则

(4.1) $ \begin{equation} 0<\Psi_1<S_{h}^0, 0<\Psi_3<S_{v}^0, 0<\Psi_2\leq S_h^0-S_h^{\infty}, 0<\Psi_4\leq S_v^0-S_v^{\infty}. \end{equation} $

取序列$ \{\epsilon_n\} $, 满足$ 0<\epsilon_{j+1}<\epsilon_j<1 $, $ \epsilon_n\rightarrow0 $, $ n\rightarrow +\infty $. 于是, 根据命题3.1, 对每个$ \epsilon=\epsilon_n $, 系统(3.6) 均存在有界正解$ \Psi_n:=(\Psi_{1, n}, \Psi_{2, n}, \Psi_{3, n}, \Psi_{4, n})\in\Gamma $, 满足(4.1) 且

$ \lim\limits_{z\rightarrow -\infty}\kappa_2^{-1}e^{-\zeta_1z}\Psi_{2, n}(z)=1, {\qquad} \lim\limits_{z\rightarrow -\infty}\kappa_4^{-1}e^{-\zeta_1z}\Psi_{4, n}(z)=1. $

因$ \Psi_n(\cdot) $是$ {\cal F} $的不动点, 则$ \Psi'_n(\cdot) $关于$ n $一致有界. 从而由(3.6) 知, $ \Psi''_n(\cdot) $和$ \Psi'''_n(\cdot) $关于$ n $一致有界. 因此, 根据Ascoli-Arzelà引理, 存在$ U:=(U_1, U_2, U_3, U_4) $和子序列$ \{\epsilon_{n_{p}}\} $满足$ \epsilon_{n_{p}}\rightarrow0 $ ($ p\rightarrow +\infty $) 使得$ \Psi_{n_{p}}(z)\rightarrow U(z), \Psi'_{n_{p}}(z)\rightarrow U'(z), \Psi''_{n_{p}}(z)\rightarrow U''(z) $在$ {{\Bbb R}} $的任意有界闭区间上一致成立且逐点收敛. 因此, 有

$ \left\{\begin{array}{ll} cU'_1(z)=d_1U''_1(z)-k_1 U_1(z)U_4(z-c\tau_1), \\ cU'_2(z)=d_2U''_2(z)+k_1U_1(z)U_4(z-c\tau_1)-\gamma_2 U_2(z), \\ cU'_3(z)=d_3U''_3(z)-k_2 U_3(z) U_2(z-c\tau_2), \\ cU'_4(z)=d_4U''_4(z)+k_2 U_3(z) U_2(z-c\tau_2)-\gamma_4 U_4(z). \end{array} \right. $

即$ U(z) $是系统(2.2) 的解并满足(2.3)式. 类似于命题3.1的证明知：$ U_1(z) $和$ U_3(z) $关于$ z $单调递减, $ U_1(+\infty)=S_{h}^{\infty}<S_{h}^{0} $且$ U_3(+\infty)=S_{v}^{\infty}<S_{v}^{0} $. 进一步可证：$ \lim\limits_{z\rightarrow -\infty}\kappa_2 ^{-1}e^{-\zeta_1z}U_2 $$ (z)=1 $, $ \lim\limits_{z\rightarrow -\infty}\kappa_4^{-1} e^{-\zeta_1z}U_4(z)=1 $, 从而(2.6) 式得证. 再根据命题3.1的证明得

$ \gamma_2\int_{-\infty}^{+\infty}U_2(z){\rm d}z=k_1\int_{-\infty}^{+\infty}U_1(z) U_4(z-c\tau_1){\rm d}z=c(S_h^0-S_h^{\infty}), U_2(+\infty)=0, $

$ \gamma_4\int_{-\infty}^{+\infty}U_4(z){\rm d}z=k_2\int_{-\infty}^{+\infty}U_3(z) U_2(z-c\tau_2){\rm d}z=c(S_v^0-S_v^{\infty}), U_4(+\infty)=0. $

因此, $ U(\cdot) $满足(2.4), (2.5), (2.7) 和(2.8)式, $ 0<U_2(z)\leq S_{h}^{0}-S_{h}^{\infty} $, $ 0<U_4(z)\leq S_{v}^{0}-S_{v}^{\infty} $.

下证$ {\cal R}_0^{\infty}<1 $. 假设$ {\cal R}_0^{\infty}\geq1 $. 因$ V^{-1}F $非负不可约, 则存在正向量$ {\bf l}=(l_1, l_2)^{T} $使得

$ \begin{eqnarray*} \left(\begin{array}{ccc} 0 {\quad} &{ } \frac{k_1S_{h}^\infty}{\gamma_2} \\ { } \frac{k_2S_{v}^\infty}{\gamma_4} {\quad} & 0 \end{array}\right) \left(\begin{array}{ccc} l_1 \\ l_2 \end{array}\right)={\cal R}_0^{\infty} \left(\begin{array}{ccc} l_1 \\ l_2 \end{array}\right)\geq \left(\begin{array}{ccc} l_1 \\ l_2 \end{array}\right). \end{eqnarray*} $

从而$ k_1S_{h}^\infty l_2\gamma_2^{-1}-l_1\geq0 $, $ k_2S_{v}^\infty l_1\gamma_4^{-1}-l_2\geq0 $, 即

(4.2) $ \begin{equation} \frac{k_1S_{h}^\infty}{\gamma_2}\geq \frac{l_1}{l_2}, {\quad} \frac{k_2S_{v}^\infty}{\gamma_4}\geq \frac{l_2}{l_1}. \end{equation} $

由$ U_j(\pm\infty)=0 $和$ U'_j(\pm\infty)=0 $ ($ j=2, 4 $) 知, 对(2.2) 式的第二和第四个方程在$ {{\Bbb R}} $上积分得

(4.3) $ \begin{equation} \left\{\begin{array}{ll} { } k_1\int_{-\infty}^{+\infty}U_1(z) U_4(z-c\tau_1){\rm d}z-\gamma_2\int_{-\infty}^{+\infty}U_2(z){\rm d}z=0, \\ { } k_2\int_{-\infty}^{+\infty}U_3(z) U_2(z-c\tau_2){\rm d}z-\gamma_4\int_{-\infty}^{+\infty}U_4(z){\rm d}z=0. \end{array} \right. \end{equation} $

因$ U_1(\cdot) $在$ {{\Bbb R}} $上单调递减, 则

(4.4) $ \begin{equation} \int_{-\infty}^{+\infty} U_1(z)U_4(z-c\tau_1){\rm d}z >S_{h}^\infty\int_{-\infty}^{+\infty} U_4(z){\rm d}z. \end{equation} $

同理$ \int_{-\infty}^{+\infty}U_3(z) U_2(z-c\tau_2){\rm d}z>S_{v}^\infty\int_{-\infty}^{+\infty}U_2(z){\rm d}z $. 于是, 由(4.3)式得

(4.5) $ \begin{equation} \int_{-\infty}^{+\infty}U_2(z){\rm d}z-\frac{k_1S_{h}^\infty}{\gamma_2} \int_{-\infty}^{+\infty}U_4(z){\rm d}z>0, \int_{-\infty}^{+\infty}U_4(z){\rm d}z-\frac{k_2S_{v}^\infty}{\gamma_4} \int_{-\infty}^{+\infty}U_2(z){\rm d}z>0. \end{equation} $

将(4.2) 式代入(4.5) 式得

$ l_2\int_{-\infty}^{+\infty}U_2(z){\rm d}z-l_1\int_{-\infty}^{+\infty}U_4(z){\rm d}z>0, l_1\int_{-\infty}^{+\infty}U_4(z){\rm d}z-l_2\int_{-\infty}^{+\infty}U_2(z){\rm d}z>0. $

矛盾. 因此, $ {\cal R}_0^{\infty}<1 $. 定理2.1证毕.

本小节证明$ {\cal R}_0>1 $, $ 0<c<c_\ast $时行波的不存在性, 如下引理是必要的.

引理5.1  设$ {\cal R}_{0}>1 $和$ 0<c<c^\ast $. 记$ U(z) $是系统(2.2) 满足(2.3) 和(2.4) 的有界正解, 则存在常数$ \omega^{\ast}>0 $使得

$ \sup\limits_{z\in{{\Bbb R}} }\left\{ [S_{h}^{0}-U_1(z)]e^ {-\omega^{\ast}z}\right\}<+\infty, {\quad} \sup\limits_{z\in{{\Bbb R}} }\left\{ [S_{v}^{0}-U_3(z)]e^ {-\omega^{\ast}z}\right\}<+\infty, $

且

$ \sup\limits_{z\in{{\Bbb R}} }\left\{ U_j(z)e^{-\omega^{\ast}z}\right\}<+\infty, {\quad} j=2, 4. $

证  由$ U_1(-\infty)=S_{h}^{0} $, $ U_3(-\infty)=S_{v}^{0} $, $ {\cal R}_{0}>1 $知, 存在一点$ z_0^\ast<0 $使得

$ U_1(z)>\frac{S_{h}^{0}}{2}+ \frac{\tilde{\gamma}}{2k_1\tilde{k}_2}, {\quad} U_3(z)>\frac{S_{v}^{0}}{2}+ \frac{\tilde{\gamma}}{2k_2\tilde{k}_1}, {\quad} z\leq z_0^\ast. $

其中$ \tilde{\gamma}=\gamma_2\gamma_4 $, $ \tilde{k}_1=k_1S_{h}^{0} $且$ \tilde{k}_2=k_2S_{v}^{0} $. 由系统(2.2) 的第二和第四个方程知, 对任意的$ z\leq z_0^\ast $, 有

(5.1) $ \begin{eqnarray} cU'_2(z)&=&d_2U''_2(z)+k_1U_1(z)U_4(z-c\tau_1)-\gamma_2U_2(z){}\\ &\geq& d_2U''_2(z)+\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2} [U_4(z-c\tau_1)-U_4(z)] +\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}U_4(z)-\gamma_2U_2(z), \end{eqnarray} $

(5.2) $ \begin{eqnarray} cU'_4(z)&=&d_4U''_4(z)+k_2U_3(z) U_2(z-c\tau_2)-\gamma_4U_4(z){}\\ &\geq& d_4U''_4(z)+ \frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1} [U_2(z-c\tau_2)-U_2(z)] +\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1}U_2(z)-\gamma_4U_4(z). \end{eqnarray} $

定义

$ N_1(z):=\int_{-\infty}^{z}U_4(\xi){\rm d}\xi, N_2(z):=\int_{-\infty}^{z}U_2(\xi){\rm d}\xi, z\in{{\Bbb R}} . $

对(5.1) 和(5.2)式在$ -\infty $到$ z $ ($ \leq z_0^\ast $) 上积分得

(5.3) $ \begin{equation} \frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}N_1(z)\leq cU_2(z)- d_2U'_2(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}[N_1(z-c\tau_1)-N_1(z)] +\gamma_2N_2(z), \end{equation} $

(5.4) $ \begin{equation} \frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1}N_2(z)\leq cU_4(z)- d_4U'_4(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1}[N_2(z-c\tau_2)-N_2(z)] +\gamma_4N_1(z). \end{equation} $

将(5.4)式代入(5.3) 式得

$ \begin{eqnarray*} \frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}N_1(z) &\leq&cU_2(z)- d_2U'_2(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}[ N_1(z-c\tau_1)-N_1(z)] +\frac{2\tilde{k}_1\tilde{\gamma}}{\tilde{k}_1\tilde{k}_2+\tilde{\gamma}}N_1(z)\\ &&+\frac{2\tilde{k}_1\gamma_2}{\tilde{k}_1\tilde{k}_2+\tilde{\gamma}} \{cU_4(z)- d_4U'_4(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1}[N_2(z-c\tau_2)-N_2(z)]\}. \end{eqnarray*} $

因此

(5.5) $ \begin{eqnarray} \frac{(\tilde{k}_1\tilde{k}_2-\tilde{\gamma})^2} {2\tilde{k}_2(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} N_1(z)&\leq&cU_2(z)- d_2U'_2(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}[N_1(z-c\tau_1)-N_1(z)]{}\\ &&+\frac{2\tilde{k}_1\gamma_2}{\tilde{k}_1\tilde{k}_2+\tilde{\gamma}} \{cU_4(z)- d_4U'_4(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1}[N_2(z-c\tau_2)-N_2(z)]\}.{\qquad} \end{eqnarray} $

同理可得

(5.6) $ \begin{eqnarray} \frac{(\tilde{k}_1\tilde{k}_2-\tilde{\gamma})^2} {2\tilde{k}_2(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} N_2(z)&\leq&cU_4(z)- d_4U'_4(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_1}[N_2(z-c\tau_2)-N_2(z)]{}\\ &&+\frac{2\tilde{k}_2\gamma_4}{\tilde{k}_1\tilde{k}_2+\tilde{\gamma}} \{cU_2(z)- d_2U'_2(z)-\frac{(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} {2\tilde{k}_2}[N_1(z-c\tau_1)-N_1(z)]\}.{\qquad} \end{eqnarray} $

令$ N(z)=N_1(z)+N_2(z) $. 因此, 结合(5.5) 和(5.6) 式知存在$ A>0 $使得

$ \begin{eqnarray*} \frac{(\tilde{k}_1\tilde{k}_2-\tilde{\gamma})^2} {2(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} \int_{0}^{+\infty}N(z-\xi){\rm d}\xi\leq cAN(z), {\quad} \forall z\leq z_{0}^{\ast}. \end{eqnarray*} $

又因$ N(z) $关于$ z $单调递增, 所以对任意$ a>0 $, 有

$ \frac{(\tilde{k}_1\tilde{k}_2-\tilde{\gamma})^2} {2(\tilde{k}_1\tilde{k}_2+\tilde{\gamma})} N(z-a)\leq cAN(z), {\quad} \forall z\leq z_{0}^{\ast}. $

所以, 存在充分大的$ a^\ast>0 $和$ \theta^\ast\in(0, 1) $使得

$ N(z-a^\ast)\leq \theta^\ast N(z), {\quad} \forall z\leq z_{0}^{\ast}. $

令$ h(z)=N(z)e^{-\omega_0z} $, 其中$ 0<\omega_0=-(a^\ast)^{-1} \ln\theta^{\ast}<\zeta_1 $. 因此, 有

$ h(z-a^\ast)=N(z-a^\ast)e^{-\omega_0(z-a^\ast)}\leq\theta^{\ast}N(z) e^{-\omega_0(z-a^\ast)}=h(z), {\quad} \forall z\leq z_{0}^{\ast}. $

由$ \lim\limits_{z\rightarrow +\infty}h(z)=0 $知, 存在$ h^\ast>0 $使得$ h(z)\leq h^\ast $, $ z\in{{\Bbb R}} $. 进而有$ N(z)\leq h^\ast e^{\omega_0z} $, $ z\in{{\Bbb R}} $. 进一步有$ \int_{-\infty}^{z}N(\xi){\rm d}\xi\leq h^\ast e^{\omega_0z}\omega_0^{-1} $, $ z\in{{\Bbb R}} $. 所以, 根据上述讨论得

$ \sup\limits_{z\in{{\Bbb R}} }\{U_j(z)e^{-\omega_0z}\}<+\infty, {\quad} \sup\limits_{z\in{{\Bbb R}} }\{U'_j(z)e^{-\omega_0z}\}<+\infty, {\quad} \sup\limits_{z\in{{\Bbb R}} }\{U''_j(z)e^{-\omega_0z}\}<+\infty, {\quad} j=2, 4. $

对(2.2) 式的第一个方程从$ -\infty $到$ z $积分得

$ c[U_1(z)-S_{h}^{0}]=d_1U'_1(z)- k_1\int_{-\infty}^zU_1(\xi)U_4(\xi-c\tau_1){\rm d}\xi. $

因$ 0<\omega_0<\zeta_1 $, 则存在$ C_3>0 $使得

(5.7) $ \begin{equation} k_1\int_{-\infty}^zU_1(\xi)U_4(\xi-c\tau_1){\rm d}\xi \leq C_3 e^{\omega_{0}z}, {\quad} z\in{{\Bbb R}} . \end{equation} $

设$ B_1(z)=S_{h}^{0}-U_1(z) $, 则$ B_1(z)\geq0 $, $ z\in{{\Bbb R}} $且

$ d_1B'_1(z)-cB_1(z)=- k_1\int_{-\infty}^zU_1(\xi)U_4(\xi-c\tau_1){\rm d}\xi. $

利用常数变易法得

$ B_1(z)=B_1(0)e^{\frac{c}{d_1}z}+\frac{k_1}{d_1} \int_{z}^{0}e^{\frac{c}{d_1}(z-x)}\int_{-\infty}^{x} U_1(\xi)U_4(\xi-c\tau_1){\rm d}\xi {\rm d}x, {\quad} z\in{{\Bbb R}} . $

因此, 存在$ \omega_{0}^{\ast} $满足$ \omega_{0}^{\ast}=\min\{\omega_{0}, c/d_1\} $使得$ B_1(z)\rightarrow O(e^{\omega_{0}^{\ast}z}) $, $ z\rightarrow -\infty $. 因$ 0\leq B_1(z)\leq S_{h}^{0} $, 则$ \sup\limits_{x\in{{\Bbb R}} }\{B_1(z)e^ {-\omega_{0}^{\ast}z}\}<+\infty $, 即$ \sup\limits_{z\in{{\Bbb R}} }\{[S_{h}^{0}-U_1(z)] e^{-\omega_{0}^{\ast}z}\}<+\infty $. 同理$ \sup\limits_{z\in{{\Bbb R}} }\{[S_{v}^{0}-U_3(z)] e^{-\omega_{0}^{\ast\ast}z}\}<+\infty $, 其中$ \omega_{0}^{\ast\ast} =\min\{\omega_{0}, c/d_3\} $. 取$ \omega^{\ast} =\min\{\omega_{0}, \omega_{0}^{\ast}, \omega_{0}^{\ast\ast}\} $, 得

$ \sup\limits_{z\in{{\Bbb R}} }\left\{ [S_{h}^{0}-U_1(z)]e^ {-\omega^{\ast}z}\right\}<+\infty, {\quad} \sup\limits_{z\in{{\Bbb R}} }\left\{ [S_{v}^{0}-U_3(z)]e^ {-\omega^{\ast}z}\right\}<+\infty, $

$ \sup\limits_{z\in{{\Bbb R}} }\left\{ U_2(z)e^{-\omega^{\ast}z}\right\}<+\infty, {\quad} \sup\limits_{z\in{{\Bbb R}} }\left\{ U_4(z)e^{-\omega^{\ast}z}\right\}<+\infty. $

引理得证.

下面给出定理2.2的证明.

定理2.2的证明  反证. 对$ 0<c<c_{\ast} $, 假设系统(1.2) 存在一有界正解

$ (S_{h}(x, t), I_{h}(x, t), S_{v}(x, t), I_{v}(x, t))=(U_1(z), U_2(z), U_3(z), U_4(z)), z=x+ct. $

对任意$ \kappa>0 $和非负函数$ V(z) $, 定义双边Laplace变换

$ {\cal P}[V(\cdot)](\zeta):=\int_{-\infty}^{+\infty} e^{-\zeta z}V(z){\rm d}z. $

记

$ {\cal P}[V(\cdot)](\zeta):={\cal P}^{-}[V(\cdot)](\zeta)+ {\cal P}^{+}[V(\cdot)](\zeta), $

其中

$ {\cal P}^{-}[V(\cdot)](\zeta):=\int_{-\infty}^{0} e^{-\zeta z}V(z){\rm d}z, {\cal P}^{+}[V(\cdot)](\zeta):=\int_{0}^{+\infty} e^{-\zeta z}V(z){\rm d}z. $

根据文献[24]知, $ {\cal P}[V(\cdot)](\zeta) $和$ {\cal P}^{-}[V(\cdot)](\zeta) $关于$ \zeta $有相同的收敛性. 因$ {\cal P}^{-}[V(\cdot)](\zeta) $是$ \zeta>0 $的单调增函数, 则存在$ \zeta^{\ast\ast}>0 $使$ {\cal P}[V(\cdot)] $在$ [0, \zeta^{\ast\ast}) $上有定义, 其中$ \zeta^{\ast\ast}<+\infty $, $ \lim\limits_{\zeta \rightarrow \zeta^{\ast\ast-}} $$ {\cal P}[V(\cdot)](\zeta)=+\infty $或$ \zeta^{\ast\ast}=+\infty $.

令$ {\cal P}_j(\zeta)={\cal P}[U_j(\cdot)](\zeta) $, $ \zeta\in [0, \zeta_j^{\ast\ast}), j=2, 4 $. 由引理5.1知, $ \zeta_j^{\ast\ast}\leq\omega^{\ast}<\zeta_1 $, $ j=2, 4 $. 直接计算得

$ \begin{eqnarray*} \int_{-\infty}^{+\infty}e^{-\zeta z}U_4(z-c\tau_1){\rm d}z={\cal P}_4(\zeta) e^{-c\tau_1\zeta}. \end{eqnarray*} $

同理得$ \int_{-\infty}^{+\infty}e^{-\zeta z} U_2(z-c\tau_2){\rm d}z={\cal P}_2(\zeta)e^{-c\tau_2\zeta} $. 因此, 由系统(2.2) 的第二和第四个方程得

$ \begin{eqnarray*} \left\{\begin{array}{ll} d_2U''_2(z)-cU'_2(z)+k_1S_{h}^{0} U_4(z-c\tau_1)-\gamma_2U_2(z)=k_1[S_{h}^{0}- U_1(z)]U_4(z-c\tau_1), \\ d_4U''_4(z)-cU'_4(z)+k_2S_{v}^{0} U_2(z-c\tau_2)-\gamma_4U_4(z)=k_2[S_{v}^{0}- U_3(z)]U_2(z-c\tau_2).\\ \end{array} \right. \end{eqnarray*} $

对上面两方程作Laplace变换得

(5.8) $ \begin{equation} \left\{\begin{array}{ll} Q_{2}(c, \zeta){\cal P}_2(\zeta)+k_1S_{h}^{0} e^{-c\tau_1\zeta}{\cal P}_4(\zeta)=\ell_4(\zeta), \\ Q_{4}(c, \zeta){\cal P}_4(\zeta)+k_2S_{v}^{0} e^{-c\tau_2\zeta}{\cal P}_2(\zeta)=\ell_2(\zeta), \\ \end{array} \right. \end{equation} $

其中

(5.9) $ \begin{equation} \ell_2(\zeta)=k_2\int_{-\infty}^{+\infty}e^{-\zeta z}[S_{v}^{0}- U_3(z)]U_2(z-c\tau_2){\rm d}z, \end{equation} $

和

(5.10) $ \begin{equation} \ell_4(\zeta)=k_1\int_{-\infty}^{+\infty}e^{-\zeta z}[S_{h}^{0}- U_1(z)]U_4(z-c\tau_1){\rm d}z. \end{equation} $

这里$ \zeta>0 $满足$ \zeta<\min\{\omega_0+\omega_0^{\ast}, \omega_0+\omega_0^{\ast\ast}\} $.

不难知$ \zeta_j^{\ast\ast}<+\infty $, $ j=2, 4 $, 从而$ \zeta_0^{\ast\ast}:=\zeta_2^{\ast\ast} =\zeta_4^{\ast\ast}<\zeta_m^+ $. 由(5.9) 和(5.10) 式知$ \ell_2(\zeta_0^{\ast\ast})<+\infty $且$ \ell_4(\zeta_0^{\ast\ast})<+\infty $. 由(5.8)式得

(5.11) $ \begin{equation} Q_{2}(c, \zeta){\cal P}_2(\zeta)= \ell_4(\zeta)-k_1S_{h}^{0} e^{-c\tau_1\zeta}{\cal P}_4(\zeta), \end{equation} $

且

(5.12) $ \begin{equation} Q_{4}(c, \zeta){\cal P}_4(\zeta)= \ell_2(\zeta)-k_2S_{v}^{0} e^{-c\tau_2\zeta}{\cal P}_2(\zeta). \end{equation} $

将(5.11) 式和(5.12) 式相乘得

$ H(c, \zeta_0^{\ast\ast})= \lim\limits_{\zeta\rightarrow \zeta_0^{\ast\ast-}} \frac{\ell_2(\zeta)\ell_4(\zeta)-\tilde{k}_1e^{-c\tau_1\zeta}\ell_2(\zeta) {\cal P}_4(\zeta)-\tilde{k}_2e^{-c\tau_2\zeta}\ell_4(\zeta) {\cal P}_2(\zeta)}{{\cal P}_2(\zeta){\cal P}_4(\zeta)}=0, $

这显然与引理3.1(ⅰ) 矛盾. 因而, 结论成立. 定理2.2得证.

本小节证明当$ {\cal R}_0\leq1 $时行波解的不存在性.

定理2.3的证明  反证, 假设系统存在一有界正解$ U(z) $, 满足(2.3) 和(2.4)式.

(ⅰ) 首先考虑$ {\cal R}_0<1 $. 因$ U(z) $是(2.2) 式的不动点, 所以有

(5.13) $ \begin{equation} U_{2}(z)=\frac{k_1}{\tilde{\Lambda}_{2}}\left[\int_{-\infty}^{z} e^{\tilde{\zeta}_{21}(z-\xi)}U_1(\xi)U_4(\xi-c\tau_1){\rm d}\xi+ \int_{z}^{+\infty} e^{\tilde{\zeta}_{22}(z-\xi)}U_1(\xi)U_4(\xi-c\tau_1){\rm d}\xi\right], \end{equation} $

且

(5.14) $ \begin{equation} U_{4}(z)=\frac{k_2}{\tilde{\Lambda}_{4}}\left[\int_{-\infty}^{z} e^{\tilde{\zeta}_{41}(z-\xi)}U_3(\xi)U_2(\xi-c\tau_2){\rm d}\xi+ \int_{z}^{+\infty} e^{\tilde{\zeta}_{42}(z-\xi)}U_3(\xi)U_2(\xi-c\tau_2){\rm d}\xi\right], \end{equation} $

其中$ \tilde{\zeta}_{j1}=\frac{c-\sqrt{c^{2}+4d_{j}\gamma_{j}}}{2d_{j}}, \tilde{\zeta}_{j2}=\frac{c+\sqrt{c^{2}+4d_{j}\gamma_{j}}}{2d_{j}}, \tilde{\Lambda}_{j}=d_{j}(\tilde{\zeta}_{j2}-\tilde{\zeta}_{j1}), j=2, 4 $. 直接计算得

(5.15) $ \begin{eqnarray} U_2(z) &=&\frac{k_1}{\tilde{\Lambda}_{2}}\int_{0}^{+\infty} e^{\tilde{\zeta}_{21}\xi}U_1(z-\xi) U_4(z-\xi+c\tau_1){\rm d}\xi{}\\ &&+\frac{k_1}{\tilde{\Lambda}_{2}}\int_{-\infty}^{0} e^{\tilde{\zeta}_{22}\xi}U_1(z-\xi) U_4(z-\xi+c\tau_1){\rm d}\xi. \end{eqnarray} $

同理

(5.16) $ \begin{eqnarray} U_4(z) &=&\frac{k_2}{\tilde{\Lambda}_{4}}\int_{0}^{+\infty} e^{\tilde{\zeta}_{41}\xi}U_3(z-\xi) U_2(z-\xi+c\tau_2){\rm d}\xi{}\\ &&+\frac{k_2}{\tilde{\Lambda}_{4}}\int_{-\infty}^{0} e^{\tilde{\zeta}_{42}\xi}U_3(z-\xi) U_2(z-\xi+c\tau_2){\rm d}\xi. \end{eqnarray} $

对(5.15) 式两端积分得

(5.17) $ \begin{equation} \int_{-\infty}^{+\infty}U_2(z){\rm d}z =\frac{k_1}{\gamma_{2}}\int_{-\infty}^{+\infty} U_1(z)U_4(z-c\tau_1){\rm d}z\leq\frac{k_1 S_{h}^0}{\gamma_{2}}\int_{-\infty}^{+\infty}U_4(z){\rm d}z. \end{equation} $

同理, 对(5.16) 式有

(5.18) $ \begin{equation} \int_{-\infty}^{+\infty}U_4(z){\rm d}z =\frac{k_2}{\gamma_{4}}\int_{-\infty}^{+\infty} U_3(z)U_2(z-c\tau_2){\rm d}z\leq\frac{k_2 S_{v}^0}{\gamma_{4}}\int_{-\infty}^{+\infty}U_2(z){\rm d}z. \end{equation} $

将(5.18) 式代入(5.17) 式得

$ \int_{-\infty}^{+\infty}U_2(z){\rm d}z \leq\frac{k_1S_{h}^0}{\gamma_{2}}\cdot\frac{k_2S_{v}^0} {\gamma_{4}}\int_{-\infty}^{+\infty} U_2(z){\rm d}z=({\cal R}_0)^{2}\int_{-\infty}^{+\infty} U_2(z){\rm d}z<\int_{-\infty}^{+\infty} U_2(z){\rm d}z. $

矛盾, 即假设不成立, 从而当$ {\cal R}_0<1 $, $ c>0 $时, 系统不存在有界正行波解.

(ⅱ) 再考虑$ {\cal R}_0=1 $. 记$ U_{j, \sup}:=\sup\limits_{z\in{{\Bbb R}} }\{U_j(z)\} $, $ j=2, 4 $. 因此, 存在关于特征值$ {\cal R}_0 $的正特征向量$ {\bf W}=(w_1, w_2)^{T} $使得$ {\cal M}{\bf W}={\bf W} $, 其中$ {\cal M}:=V^{-1}F $ (见第2节). 简单计算得

(5.19) $ \begin{equation} \gamma_2=\frac{k_1S_{h}^0w_2}{w_1}, {\quad} \gamma_4=\frac{k_2S_{v}^0w_1}{w_2}. \end{equation} $

取序列$ \{z_{n}\}\subset{{\Bbb R}} $, $ n\in{\Bbb N}^\ast $使得$ \lim\limits_{n\rightarrow +\infty}U_2(z_{n})=U_{2, \sup }=\sup\limits_{z\in{{\Bbb R}} }\{U_2(z)\}:=\bar{P} $.

为了推出矛盾, 下证$ \bar{P}=0 $. 假设$ \bar{P}>0 $. 考虑如下函数序列$ U_{n}(\cdot):=(U_{1, n}(\cdot), U_{2, n}(\cdot), $$ U_{3, n}(\cdot), U_{4, n}(\cdot))= (U_{1}(\cdot+z_{n}), U_{2}(\cdot+z_{n}), U_{3}(\cdot+z_{n}), U_{4}(\cdot+z_{n})) $满足(2.3)式, $ n\in{\Bbb N}^\ast $. 根据$ U_{n}(z) $的有界性和椭圆估计知, 存在子序列$ U_{n_j}(z) $和$ (U_1^{\ast}(z), U_2^{\ast}(z), $$ U_3^{\ast}(z), U_4^{\ast}(z)) $使得在$ C_{{\rm loc}}^2({{\Bbb R}} ) $中, 有

$ \lim\limits_{j\rightarrow +\infty}U_{n_j}(z)=(U_1^{\ast}(z), U_2^{\ast}(z), U_3^{\ast}(z), U_4^{\ast}(z)), $

且$ (U_1^{\ast}(z), U_2^{\ast}(z), U_3^{\ast}(z), U_4^{\ast}(z)) $满足

(5.20) $ \begin{equation} \left\{\begin{array}{ll} d_1U_{1}^{\ast''}(z)-cU_{1}^{\ast'}(z)-k_1U_{1}^{\ast}(z) U_{4}^{\ast}(z-c\tau_1)=0, \\ d_2U_{2}^{\ast''}(z)-cU_{2}^{\ast'}(z)+k_1U_{1}^{\ast}(z) U_{4}^{\ast}(z-c\tau_1)-\gamma_2U_{2}^{\ast}(z)=0, \\ d_3U_{3}^{\ast''}(z)-cU_{3}^{\ast'}(z)-k_2U_{3}^{\ast}(z) U_{2}^{\ast}(z-c\tau_2)=0, \\ d_4U_{4}^{\ast''}(z)-cU_{4}^{\ast'}(z)+k_2U_{3}^{\ast}(z) U_{2}^{\ast}(z-c\tau_2)-\gamma_4U_{4}^{\ast}(z)=0, \end{array} \right. \end{equation} $

其中$ U_{2}^{\ast}(0)=\bar{P} $, $ U_{2}^{\ast}(z)\leq\bar{P} $, $ 0\leq U_{1}^{\ast}(z)\leq S_{h}^0 $, $ 0\leq U_{3}^{\ast}(z)\leq S_{v}^0 $. 类似于上述分析, 对(5.20) 式的第四个方程利用比较原理并结合(5.19) 式得$ k_2S_{v}^0\bar{P}\geq\gamma_{4}U_{4, \sup} =k_2S_{v}^0w_1w_2^{-1}U_{4, \sup } $. 于是$ U_{4, \sup}\leq\bar{P}w_2/w_1 $. 因此, 得

$ 0\leq d_2U_{2}^{\ast''}(0)+k_1U_{1}^{\ast}(0) \frac{\bar{P}w_2}{w_1}-\frac{k_1S_{h}^0w_2}{w_1}\bar{P} =d_2U_{2}^{\ast''}(0)+\left[U_{1}^{\ast}(0)-S_{h}^0\right] \frac{k_1w_2}{w_1}\bar{P}. $

因$ U_{2}^{\ast''}(0)\leq0 $知$ U_{1}^{\ast}(0)\geq S_{h}^0 $. 由最大值原理知$ U_{1}^{\ast}(z)\equiv S_{h}^0 $, $ z\in{{\Bbb R}} $. 将其代入(5.20) 式的第一个方程得$ U_{4}^{\ast}(z)\equiv 0 $, $ z\in{{\Bbb R}} $. 从而由(5.20)式的第四个方程得$ U_{2}^{\ast}(z)\equiv 0 $或$ U_{3}^{\ast}(z)\equiv0 $, $ z\in{{\Bbb R}} $. 若$ U_{3}^{\ast}(z)\equiv0 $, 则与$ U_{3}^{\ast}(-\infty)=S_{v}^0 $矛盾. 因此$ U_{2}^{\ast}(z)\equiv 0 $, 于是$ \bar{P}\equiv0 $, 显然与假设$ \bar{P}>0 $矛盾. 从而定理2.3得证.

本节探讨登革热病毒潜伏期及染病人和蚊子的扩散率对临界波速$ c_\ast $的影响.

根据引理3.1知, 存在$ c_\ast, \zeta^\ast>0 $使得$ H(c_\ast, \zeta^\ast)=0 $, 其中

$ H(c, \lambda)=(d_2\zeta^2-c\zeta-\gamma_2)(d_4\lambda^2-c\lambda-\gamma_4) -\tilde{k}e^{-c(\tau_1+\tau_2)\zeta}=0. $

(A) 先探究$ c_\ast=c_\ast(\tau_1, \tau_2) $关于$ \tau_i $的连续依赖性. 不难得

$ \frac{\partial c_\ast(\tau_1, \tau_2)}{\partial\tau_i} =\frac{\tilde{k}\tau_ic_\ast e^{-c_\ast(\tau_1+\tau_2)\zeta^\ast}} {Q_2(c_\ast, \zeta^\ast)+Q_4(c_\ast, \zeta^\ast)- \tilde{k}(\tau_1+\tau_2)e^{-c_\ast(\tau_1+\tau_2)\zeta^\ast}}. $

因$ Q_j(c_\ast, \zeta^\ast)<0, j=2, 4 $, 则$ \partial c_\ast(\tau_1, \tau_2)/\partial\tau_i<0 $, 即$ c_\ast $是$ \tau_i $的减函数. 根据文献[4, 8, 28], 取$ d_2=0.1 $, $ d_4=0.3 $, $ \mu_h=0.001 $, $ \alpha_h=0.1667 $, $ \mu_v=0.0001 $, $ k_1=0.00682 $, $ k_2=0.0015 $, 则$ c_\ast $和$ \tau_i $的关系见图 2(a). 由图 2(a)可知: 当$ \tau_i $充分小时, 曲线急剧下降, 这表明适当延长病毒内部和外部潜伏期可在一定程度上降低疾病传播速度, 从而降低传播风险. 然而, 当$ \tau_i $持续增大时, $ c_\ast $增加速度减慢, 这似乎又表明仅仅通过化学或物理手段延长病毒潜伏期并不能有效控制登革热.

(B) 再探讨$ c_\ast $关于$ d_j $的连续依赖性, $ j=2, 4 $. 类似于(A) 得$ \partial c_\ast(d_2, d_4)/\partial d_j $$ >0 $. 因此, $ c_\ast $关于$ d_j $单调递增. 取$ k_1=0.00682 $, $ k_2=0.0015 $, $ \gamma_2=0.1677 $, $ \gamma_4=0.0001 $. 图 2(b)表明$ c_\ast $是关于$ d_j $的增函数. 从生物学角度, 染病人群和蚊子的地理位置移动在一定程度上会加快疾病传播. 同时, 由图 2(b)可知, $ d_4 $对$ c_\ast $的影响大于$ d_2 $对$ c_\ast $的影响, 即染病蚊子的扩散对疾病传播有较大影响, 因此需更加注重对蚊虫的控制, 减少蚊虫扩散.

由于登革病毒在人和蚊子体内具有潜伏期, 并且人群和蚊子在空间中的随机移动需要耗费时间. 受此启发, 为了考虑不同扩散性和潜伏期对疾病传播的影响, 本文研究了一类具有离散时滞的反应扩散登革热模型行波的存在性和不存在性, 证明了(ⅰ) 当$ {\cal R}_0>1 $, $ c>c_\ast $时, 系统(1.2) 存在满足条件(2.3)–(2.8) 式的正行波解；(ⅱ) 当$ {\cal R}_0>1 $, $ 0<c<c_\ast $或$ {\cal R}_0\leq1 $, $ c>0 $时, 系统(1.2) 不存在满足上述条件的行波解. 同时, 分析了扩散系数和潜伏期对最小波速$ c_\ast $的影响. 具体地, 延长病毒内部或外部潜伏期可降低疾病传播速度, 但仅通过延长潜伏期不能有效控制登革热, 如图 2(a)所示; 染病人群和蚊子的地理移动会增加疾病传播速度, 如图 2(b)所示. 因而, 应减少患病人群移动并侧重于蚊虫控制, 将有利于疾病控制.