一类带治疗项的非局部扩散SIR传染病模型的行波解

一类带治疗项的非局部扩散SIR传染病模型的行波解

邓栋^,¹, 李燕^,²

Traveling Waves in a Nonlocal Dispersal SIR Epidemic Model with Treatment

Deng Dong^,¹, Li Yan^,²

通讯作者: 李燕, E-mail: yanli@xidian.edu.cn

收稿日期: 2018-09-13

基金资助:

陕西省科技厅资助. 2017JQ1024

Received: 2018-09-13

Fund supported:

陕西省科技厅资助. 2017JQ1024

作者简介 About authors

邓栋,E-mail:dd0328a@163.com , E-mail：dd0328a@163.com

摘要

该文主要考虑一类非局部扩散传染病模型的行波解的存在性与不存在性.首先，利用Schauder不动点定理和取极限的方法，得到了行波解的存在性.其次，利用双边拉普拉斯变换和Fubini定理，证明了行波解的不存在性.上述结果表明，最小波速是预测疾病是否传播且以多大速度传播的重要阈值.

关键词： 行波解 ; 非局部扩散 ; 最小波速 ; 双边拉普拉斯变换

Abstract

This paper is concerned with the existence and nonexistence of traveling wave solutions of a nonlocal dispersal epidemic model with treatment. The existence of traveling wave solutions is established by Schauder's fixed point theorem as well as a limiting argument, while the nonexistence of traveling wave solutions is proved by two-sided Laplace transform and Fubini's theorem. From the results, we conclude that the minimal wave speed is an important threshold to predict how fast the disease invades.

Keywords： Traveling waves ; Nonlocal dispersal ; Minimal wave speed ; Two-sided Laplace transform

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本文引用格式

邓栋, 李燕. 一类带治疗项的非局部扩散SIR传染病模型的行波解. 数学物理学报[J], 2020, 40(1): 72-102 doi:

Deng Dong, Li Yan. Traveling Waves in a Nonlocal Dispersal SIR Epidemic Model with Treatment. Acta Mathematica Scientia[J], 2020, 40(1): 72-102 doi:

1 引言和主要结果

众所周知,流感等传染病在现代生活中十分常见,参看文献[1-2, 4, 9, 12, 14, 16].研究各类传染病模型的时空传播动力学具有重要意义.在过去几十年中,扩散传染病模型的行波解得到广泛的研究,参看文献[5-6, 10-11, 13, 19-27, 30].特别地,文献[13, 22, 26-27]等研究了非局部扩散流行病模型的行波解.关于传染病模型行波解的更多工作请参见专著[15],综述论文[17-18]及相关参考文献.

2014年, Zhang和Wang^[30]将扩散引入到带治疗项的流感模型^[12]中,即以下反应扩散方程

$\begin{equation} \left\{\begin{array}{ll} { } \frac{\partial S}{\partial t} = d_{s}\frac{\partial^{2}S}{\partial x^{2}}-\beta(I_{u}+\delta I_{h})S, \\ { } \frac{\partial I_{u}}{\partial t} = d_{u}\frac{\partial^{2}I_{u}}{\partial x^{2}}+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}, \\ { } \frac{\partial I_{h}}{\partial t} = d_{h}\frac{\partial^{2}I_{h}}{\partial x^{2}}+\mu\beta(I_{u}+\delta I_{h})S-k_{h}I_{h}, \\ { } \frac{\partial R}{\partial t} = d_{r}\frac{\partial^{2}R}{\partial x^{2}}+k_{u}I_{u}+k_{h}I_{h}, \end{array}\right. \end{equation}$

(1.1)

并研究了其行波解的存在性和不存在性.其中 $S(t, x)$ , $I_{u}(x, t)$ , $I_{h}(x, t)$ 和 $R(t, x)$ 分别表示易感者,感染而未治愈者,治愈者和恢复者. $0<\mu<1$ , $0<\delta<1$ ,且 $k_{u}<k_{h}$ , $d_{i}(i = s, u, h, r)$ 分别表示易感者,感染未治愈者,治愈者,恢复者的扩散率.文献[30]引入一个辅助系统,构造了一个有界锥,从而利用Schauder不动点定理来证明系统(1.1)行波解的存在性,参见文献[5-7].其次,双边拉普拉斯变换证明了系统(1.1)的行波解的不存在性.

本文主要考虑(1.1)的如下非局部扩散模型

$\begin{equation} \left\{\begin{array}{ll} { } \frac{\partial S}{\partial t} = d_{s}(J*S-S)-\beta(I_{u}+\delta I_{h})S, \\ { } \frac{\partial I_{u}}{\partial t} = d_{u}(J*I_{u}-I_{u})+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}, \\ { } \frac{\partial I_{h}}{\partial t} = d_{h}(J*I_{h}-I_{h})+\mu\beta(I_{u}+\delta I_{h})S-k_{h}I_{h}, \\ { } \frac{\partial R}{\partial t} = d_{r}(J*R-R)+k_{u}I_{u}+k_{h}I_{h}, \end{array}\right. \end{equation}$

(1.2)

其中 $J(x)$ 表示个体在距离 $x$ 处的概率分布且 $J*u-u$ 表示 $u$ 扩散而产生的净增长率. $J*S(t, x)$ , $J*I_{u}(t, x)$ , $J*I_{h}(t, x)$ 和 $J*R(t, x)$ 是关于空间变量 $x$ 的标准卷积算子.本文始终假设 $J$ 满足以下条件

$(J)\ J\in C^{1}({{\Bbb R}} ), \ J(x) = J(-x)\geq0, \int_{{{\Bbb R}} } J{\rm d}x = 1$ 且 $J$ 是紧支的.

我们主要考虑系统(1.2)行波解的存在性和不存在性.系统(1.2)的行波解是一个形如 $(S(\xi), I_{u}(\xi), I_{h}(\xi), R(\xi))$ ,其中 $\xi = x+ct$ ,且满足以下系统

$\begin{equation} \left\{\begin{array}{ll} cS^{'} = d_{s}(J*S-S)-\beta(I_{u}+\delta I_{h})S, \\ cI_{u}^{'} = d_{u}(J*I_{u}-I_{u})+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}, \\ cI_{h}^{'} = d_{h}(J*I_{h}-I_{h})+\mu\beta(I_{u}+\delta I_{h})S-k_{h}I_{h}, \\ cR^{'} = d_{r}(J*R-R)+k_{u}I_{u}+k_{h}I_{h} \end{array}\right. \end{equation}$

(1.3)

及边界条件

$\begin{equation} \left\{\begin{array}{ll} (S(-\infty), I_{u}(-\infty), I_{h}(-\infty), R(-\infty)) = (S^{0}, 0, 0, 0), \\ (S(+\infty), I_{u}(+\infty), I_{h}(+\infty), R(+\infty)) = (S^{1}, 0, 0, S^{0}-S^{1}) \end{array}\right. \end{equation}$

(1.4)

的特殊非负解,其中 $S^0>S^1\ge0$ ,且 $(S^0, 0, 0, 0)$ 和 $(S^1, 0, 0, S^0-S^1)$ 是无病平衡点, $S^0-S^1$ 是感染个体的最终数量.

定义

$R_{0} = R_{u}+R_{h}, \; R_u = \frac{(1-\mu)\beta S^0}{k_u}, \; R_h = \frac{\mu\delta\beta S^0}{k_h}$

及

$\Delta_{u}(\lambda, c) = d_{u}\int^{+\infty}_{-\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d_{u}+R_{u}k_{u}-k_{u}-c\lambda,$

$\Delta_{h}(\lambda, c) = d_{h}\int^{+\infty}_{-\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d_{h}+R_{h}k_{h}-k_{h}-c\lambda,$

其中 $R_{0}$ 是基本再生数.

因

$\Delta_{u}(0, c) = R_{u}k_{u}-k_{u},$

$\Delta_{h}(0, c) = R_{h}k_{h}-k_{h},$

$\Delta_{u}(\lambda, \infty) = \Delta_{h}(\lambda, \infty) = -\infty\; (\forall\lambda>0),$

$\frac{\partial\Delta_{u}(\lambda, c)}{\partial c} = \frac{\partial\Delta_{h}(\lambda, c)}{\partial c} = -\lambda<0\; (\forall\lambda>0),$

$\frac{\partial^{2}\Delta_{u}(\lambda, c)}{\partial\lambda^{2}} = d_{u}\int_{-\infty}^{+\infty}J(y)y^{2}{\rm e}^{-\lambda y}{\rm d}y>0,$

$\frac{\partial^{2}\Delta_{h}(\lambda, c)}{\partial\lambda^{2}} = d_{h}\int_{-\infty}^{+\infty}J(y)y^{2}{\rm e}^{-\lambda y}{\rm d}y>0,$

$\frac{\partial\Delta_{u}(\lambda, c)}{\partial\lambda}\Bigg|_{\lambda = 0} = \left(-d_{u}\int_{-\infty}^{+\infty}J(y)y{\rm e}^{-\lambda y}{\rm d}y-c\right)\Bigg|_{\lambda = 0} = -c,$

$\frac{\partial\Delta_{h}(\lambda, c)}{\partial\lambda}\Bigg|_{\lambda = 0} = \left(-d_{h}\int_{-\infty}^{+\infty}J(y)y{\rm e}^{-\lambda y}{\rm d}y-c\right)\Bigg|_{\lambda = 0} = -c,$

从而 $\Delta_{i}(\lambda, c)$ , $i = u, h$ 具有以下性质

引理1.1 假设 $R_{i}>1$ , $i = u, h$ .则存在 $c_{i}^{*} >0$ 和 $\lambda_{i}^{*}>0$ 满足

$\frac{\partial\Delta_{i}(\lambda, c)}{\partial \lambda}\Big|_{(\lambda_{i}^{*}, c_{i}^{*})} = 0, \; \Delta_{i}(\lambda_{i}^{*}, c_{i}^{*}) = 0, \ i = u, h.$

进一步地

(Ⅰ)若 $c<c_{i}^{*}$ ,则对一切 $\lambda\geq0$ ,有 $\Delta_{i}(\lambda, c)>0$ ;

(Ⅱ)若 $c>c_{i}^{*}$ ,则 $\Delta_{i}(\lambda, c) = 0$ 有两个不同的正解 $\lambda_{i_{1}}^{*}<\lambda_{i_{2}}^{*}$ 满足

$\left\{\begin{array}{ll} \Delta_{i}(\lambda, c)>0, \; \lambda\in(0, \lambda_{i_{1}}^{*})\cup(\lambda_{i_{2}}^{*}, \infty), \\ \Delta_{i}(\lambda, c)<0, \; \lambda\in(\lambda_{i_{1}}^{*}, \lambda_{i_{2}}^{*}), \ i = u, \ h.\\ \end{array}\right.$

将 $\phi(\xi) = {\rm e}^{\lambda\xi}(\nu_{u}, \nu_{h})^{T}$ 带到系统(1.3)的 $(I_u, I_h)$ 方程中,我们得到在 $(S^0, 0, 0, 0)$ 处的线性化特征方程为

$\begin{equation} H(\lambda): = \Delta(\lambda, c)-\gamma = 0, \end{equation}$

(1.5)

其中 $\Delta(\lambda, c) = \Delta_{u}(\lambda, c)\Delta_{h}(\lambda, c)$ 且 $\gamma = R_{u}R_{h}k_{u}k_{h}$ .易知,若 $R_{0}>1$ ,方程(1.5)至少有一个正根.设最小的正根为 $\lambda_{1}$ .易得 $\lim\limits_{\lambda\to+\infty}\Delta(\lambda, c) = +\infty$ 且有以下引理.

引理1.2 假设 $R_{0}>1$ .

(Ⅰ)存在 $c_{1}^{*}>0$ ,使得对 $c>c_{1}^{*}$ 和充分小的 $\omega>0$ ,成立 $\Delta_{u}(\lambda_{1}+\omega, c)<\Delta_{u}(\lambda_{1}, c)<0$ 且 $\Delta_{h}(\lambda_{1}+\omega, c)<\Delta_{h}(\lambda_{1}, c)<0$ .进一步地,存在正常数 $c_{2}^{*}\leq c_{1}^{*}$ ,当 $0<c<c_{2}^{*}$ 时, (1.5)有且只有一个正根.

(Ⅱ)若 $d_{u} = d_{h}$ ,则存在 $c^{*}>0$ ,对 $c>c^{*}$ 和充分小的 $\omega>0$ ,成立 $\Delta_{u}(\lambda_{1}+\omega, c)<\Delta_{u}(\lambda_{1}, c)<0$ 且 $\Delta_{h}(\lambda_{1}+\omega, c)<\Delta_{h}(\lambda_{1}, c)<0$ .进一步地,当 $0<c<c^{*}$ 时,方程(.5)有且只有一个正根.

证 (Ⅰ)因 $R_{u}+R_{h}>1$ ,我们首先讨论以下情形.

情形1 $R_{u}>1$ , $R_{h}>1$ .由引理1.1,若 $c>\overline{c} = \max\{c_{u}^{*}, c_{h}^{*}\}$ ,则 $\Delta_{u}(\lambda, c) = 0$ 和 $\Delta_{h}(\lambda, c) = 0$ 有均两个正根.

情形2 $R_{u}>1$ , $R_{h}<1$ .若 $c>\overline{c} = c_{u}^{*}$ ,则 $\Delta_{u}(\lambda, c) = 0$ 有两个正根,而 $\Delta_{h}(\lambda, c) = 0$ 有一个正根和一个负根.

情形3 $R_{u}<1$ , $R_{h}>1$ .若 $c>\overline{c} = c_{h}^{*}$ ,则 $\Delta_{h}(\lambda, c) = 0$ 有两个正根,而 $\Delta_{u}(\lambda, c) = 0$ 有一个正根和一个负根

情形4 $R_{u}<1$ , $R_{h}<1$ .若 $c>\overline{c} = 0$ ,则 $\Delta_{u}(\lambda, c) = 0$ 和 $\Delta_{h}(\lambda, c) = 0$ 有一个正根和一个负根.

由以上情形,令 $c>\overline{c}$ ,则 $\Delta_{u}(\lambda, c) = 0$ 和 $\Delta_{h}(\lambda, c) = 0$ 都有两个实根.设 $\lambda_{u}^{\pm}, \ \lambda_{h}^{\pm}$ 分别是 $\Delta_{u}(\lambda, c) = 0, \ \Delta_{h}(\lambda, c) = 0$ 的两个实根.令

$\lambda_{M}^{\pm} = \max\{\lambda_{u}^{\pm}, \lambda_{h}^{\pm}\}, \ \lambda_{m}^{\pm} = \min\{\lambda_{u}^{\pm}, \lambda_{h}^{\pm}\},$

则 $\lambda_{M}^{\pm}, \lambda_{m}^{\pm}$ 都是 $\Delta(\lambda, c) = 0$ 的实根.由图 1, $\Delta_{u}(\lambda, c)$ 和 $\Delta_{h}(\lambda, c)$ 对 $c>0$ 都有一个最小值点,分别表示为 $\gamma_{1} = \lambda_{u}^{*}(c)$ 和 $\gamma_{2} = \lambda_{h}^{*}(c)$ .易知

图 1

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图 1 $H(\lambda), \Delta(\lambda, c), \Delta_{u}(\lambda, c), \Delta_{h}(\lambda, c)$ 根的分布

$\begin{eqnarray*} \left\{\begin{array}{ll} \Delta(\lambda, c)>0, \; \lambda\in(-\infty, \lambda_{m}^{-})\cup(\lambda_{M}^{-}, \lambda_{m}^{+})\cup(\lambda_{M}^{+}, +\infty ), \\ \Delta(\lambda, c)<0, \; \lambda\in(\lambda_{m}^{-}, \lambda_{M}^{-}) \cup(\lambda_{m}^{+}, \lambda_{M}^{+}). \end{array}\right. \end{eqnarray*}$

对 $i = u, h,$ $\Delta_{i}(\lambda, c)$ 在 $\lambda\in (-\infty, \lambda_{i}^{*}(c))$ 上递减,在 $\lambda\in (\lambda_{i}^{*}(c), +\infty)$ 上递增.假设 $\lambda_{u}^{*}(c)<\lambda_{h}^{*}(c)$ ,则 $\Delta(\lambda, c)$ 对 $\lambda\in(\lambda_{m}^{-}, \lambda_{M}^{+})$ 在 $(\lambda_{u}^{*}(c), \lambda_{h}^{*}(c))$ 上取到最大值.

因 $\frac{\partial\Delta_{i}(\lambda, c)}{\partial c} = -\lambda<0$ 对 $\lambda>0$ 成立,从而 $\Delta_{u}(\lambda, c)$ 和 $\Delta_{h}(\lambda, c)$ 关于 $c>0$ 递减.若 $c\to+\infty$ ,则 $\Delta_{i}(\lambda, c)\to-\infty$ ,即 $\Delta(\lambda, c)\rightarrow+\infty.$ 从而存在 $c_{t}^{*}>0$ 使得 $H(\lambda) = \Delta(\lambda, c)-\gamma = 0$ 在 $(\lambda_{m}^{-}, \lambda_{M}^{+})$ 上对 $c = c_{t}^{*}$ 只有一个正根,且 $H(\lambda) = \Delta(\lambda, c)-\gamma = 0$ 在 $(\lambda_{m}^{-}, \lambda_{M}^{+})$ 上对 $c>c_{t}^{*}>0$ 有两个正根,其中最小的正根为 $\lambda_{1}$ .令 $c_{s}^{*}>0$ ,满足

$H(\lambda^{*}_{u}(c_{s}^{*})) = \Delta(\lambda^{*}_{u}(c_{s}^{*}), c_{s}^{*})-\gamma = 0.$

定义

$F(\lambda, c): = -d_{u}\int_{-\infty}^{+\infty}J(y) y{\rm e}^{-\lambda y}{\rm d}y-c.$

因

$F(\lambda_{u}^{*}, c) = 0, \; \; \frac{\partial F(\lambda_{u}^{*}, c)}{\partial\lambda_{u}^{*}} = d_{u}\int_{-\infty}^{+\infty}J(y)y^{2}{\rm e}^{-\lambda_{u}^{*} y}{\rm d}y>0,$

利用隐函数定理, $\lambda_{u}^{*}(c)$ 关于 $c>0$ 递增.从而,若取

$c_{1}^{*} = \max\{\overline{c}, c_{t}^{*}, c_{s}^{*}\},$

则对 $c>c_{1}^{*}$ ,有

$H(\lambda^{*}_{u}(c_{s}^{*})) = \Delta(\lambda^{*}_{u}(c_{s}^{*}), c)-\gamma>0.$

即

$\lambda_{1}<\lambda^{*}_{u}(c_{s}^{*})<\lambda_{u}^{*}(c)<\lambda_{h}^{*}(c),$

这表明 $\Delta_{u}(\lambda_{1}+\omega, c)<\Delta_{u}(\lambda_{1}, c)<0$ 且对充分小的 $\omega>0$ , $\Delta_{h}(\lambda_{1}+\omega, c)<\Delta_{h}(\lambda_{1}, c)<0$ .下面证明(Ⅰ)的第二部分.注意

$\Delta_{i}(\lambda, 0) = \Delta_{i}(-\lambda, 0), \ i = u, h$

且

$\frac{\partial\Delta_{i}(\lambda, 0)}{\partial\lambda}\Big|_{\lambda = 0} = \left[- d_{i}\int^{+\infty}_{-\infty}{\rm e}^{-\lambda y}yJ(y){\rm d}y\right]_{\lambda = 0} = 0, \; i = u, h.$

若 $\Delta_{u}(0, c)$ 和 $\Delta_{h}(0, c)$ 都大于零,则 $\Delta(\lambda, 0)$ 在 $\lambda\in(0, +\infty)$ 上递减,即 $H(\lambda) = \Delta(\lambda, 0)-\gamma$ 只有一个正根.同理,若 $\Delta_{u}(0, c)$ 和 $\Delta_{h}(0, c)$ 都是非正的,可得 $H(\lambda)$ 只有一个正根.因而,存在一个正常数 $c_{2}^{*}$ 使得对 $0<c<c_{2}^{*}$ , $H(\lambda)$ 只有一个根.

(II)假设 $d_{u} = d_{h} = d$ 且令

$\kappa = d\int^{+\infty}_{-\infty}{\rm e}^{-\lambda y}J(y){\rm d}y-c\lambda-d.$

则

$H(\lambda) = G(\kappa) = (\kappa+\Delta_{u}(0, c))(\kappa+\Delta_{h}(0, c))-\gamma.$

又 $G(0) = H(0)<0$ ,从而 $G(\kappa)$ 有一个正根 $\kappa^{+}$ 和一个负根 $\kappa^{-}$ .即 $d\int^{+\infty}_{-\infty}{\rm e}^{-\lambda y}J(y){\rm d}y-c\lambda-d-\kappa^{+} = 0$ 有一个正根和负根.令 $c^{*}>0$ ,且 $d\int^{+\infty}_{-\infty}{\rm e}^{-\lambda y}J(y){\rm d}y-c^{*}\lambda-d-\kappa^{-} = 0$ 只有一个正根.

同理,可证若 $c>c^{*}$ 且 $\omega>0$ 充分小,则

$\Delta_{u}(\lambda_{1}+\omega, c)<\Delta_{u}(\lambda_{1}, c)<0,$

$\Delta_{h}(\lambda_{1}+\omega, c)<\Delta_{h}(\lambda_{1}, c)<0,$

且 $d\int^{+\infty}_{-\infty}{\rm e}^{-\lambda y}J(y){\rm d}y-c\lambda-d-\kappa^{-} = 0$ 有两个正根.若 $0<c<c^{*}$ ,则上述方程无实根,从而方程(1.5)只有一个实根.证毕.

现在陈述本文的主要结果.

定理1.1 (ⅰ) (存在性) 设 $R_{0}>1$ 且以下条件之一成立

(H1) $d_{u} = d_{h}$ , $c\geq c^{*}$ ;或

(H2) $d_{u}\neq d_{h}, c>c_{1}^{*}$ .

则系统(1.2)存在波速为c的行波解 $(S(\xi), I_{u}(\xi), I_{h}(\xi), R(\xi))$ 满足(1.3)–(1.4)式.

(ⅱ) (不存在性) 设 $R_{0}>1$ 且以下条件之一成立

(NH1) $d_{u} = d_{h}$ , $0<c<c^{*}$ ;或

(NH2) $d_{u}\neq d_{h}$ , $0<c<c_{2}^{*}$ .

则系统(1.2)不存在行波解.且若 $R_{0}<1$ ,则系统(1.2)的行波解不存在.

定理1.1把文献[5]的行波解的存在性和不存在性从反应扩散模型推广到了非局部扩散情形.受文献[5, 30]的启发,我们利用Schauder不动点定理结合上下解方法和取极限的方法,首先证明了 $R_0>1$ , $c>c^*$ , $d_u = d_h$ 或 $c>c_1^*$ , $d_u\neq d_h$ 情形下行波解的存在性;然后利用双边拉普拉斯变换,得到 $R_0>1$ , $0<c<c^*$ , $d_u = d_h$ 或 $0<c<c_2^*$ , $d_u\neq d_h$ 情形下行波解的不存在性.最后,利用Fubini定理,证明了 $R_0<1$ 时行波解的不存在性.

此外,我们考虑了当 $d_u = d_h$ 且 $c = c*$ 时行波解存在性.受文献[3, 8, 21, 28]启发,通过对超临界波速的递减序列取极限来证明.然而,由于系统(1.2)中出现了治疗项,估计 $I_{u}$ 和 $I_{h}$ 的有界性及S在 $-\infty$ 处的边界渐近行为比较困难.我们通过一精细的分析证明了 $k_{h}I_{u}+k_{u}\delta I_{h}$ 的有界性,从而解决了上述困难.

本文的剩余部分安排如下:第2节致力于证明当波速 $c$ 大于临界速度时,系统(1.2)的行波解的存在性;第3节讨论 $c = c^ {*}$ 时行波解的存在性;第4节主要证明行波解的不存在.

2 行波解的存在性

本节将在假设(H2)或

(H1') $d_{u} = d_{h}$ , $c>c^{*}$ 情形下,通过构造辅助系统,并利用Schauder不动点定理结合上下解方法和取极限的方法证明系统(1.3)存在连接平衡点 $(S^0, 0, 0, 0)$ 和 $S^0, 0, 0, S^0-S^1)$ 的行波解 $(S(\xi), I_{u}(\xi), I_{h}(\xi), R(\xi))$ .本节剩余部分总是假设 $R_{0}>1$ .

因为系统(1.3)的第四个方程是相对独立的,从而只需考虑前三个方程

(2.1)

且满足边界条件

$\begin{equation} \left\{\begin{array}{ll} (S(-\infty), I_{u}(-\infty), I_{h}(-\infty)) = (S^{0}, 0, 0), \\ (S(+\infty), I_{u}(+\infty), I_{h}(+\infty)) = (S^{1}, 0, 0). \end{array}\right. \end{equation}$

(2.2)

对任意小的正常数 $\varepsilon$ ,引入如下含正参数 $\varepsilon$ 的辅助系统

$\begin{equation} \left\{\begin{array}{ll} cS^{'} = d_{s}(J*S-S)-\beta(I_{u}+\delta I_{h})S, \\ cI_{u}^{'} = d_{u}(J*I_{u}-I_{u})+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}-\varepsilon I_{u}^{2}, \\ cI_{h}^{'} = d_{h}(J*I_{h}-I_{h})+\mu\beta(I_{u}+\delta I_{h})S-k_{h}I_{h}-\varepsilon I_{h}^{2}. \end{array}\right. \end{equation}$

(2.3)

2.1 构造上下解

定义

$\underline{S}(\xi) = \max\{{S^{0}-\sigma {\rm e}^{\alpha\xi}, 0}\},$

$\overline{I}(\xi) = V \min\{{\rm e}^{\lambda_{1}\xi}, K\},$

$\underline{I}(\xi) = V \max\{{\rm e}^{\lambda_{1}\xi}(1-M{\rm e}^{\kappa\xi}), 0\},$

其中 $V = (-\Delta_{h}(\lambda_{1}, c), \mu\beta S^{0})^{T}$ , $K>K_{0} = \max\left\{\frac{-\Delta_{h}(\lambda_{1}, c)[(1-\mu)\beta S^{0}-k_{u}]-(1-\mu)\delta(\beta S^{0})^{2}}{\varepsilon [\Delta_{h}(\lambda_{1}, c)]^{2}} 0\right\}$ .

引理2.1 对所有的 $\xi\neq\frac{\ln K}{\lambda_{1}}$ ,函数 $\overline{I}(\xi) = (\overline I_{u}(\xi), \overline I_{h}(\xi))^{T}$ 满足以下不等式

$\begin{eqnarray*} \left\{\begin{array}{ll} c\overline I_{u}^{'}\geq d_{u}(J*\overline I_{u}-\overline I_{u})+(1-\mu)\beta(\overline I_{u}+\delta\overline I_{h})S^{0}-k_{u}\overline I_{u} -\varepsilon\overline I_{u}^{2}, \\ c\overline I_{h}^{'}\geq d_{h}(J*\overline I_{h}-\overline I_{h})+\mu\beta(\overline I_{u}+\delta\overline I_{h})S^{0}-k_{h}\overline I_{h} -\varepsilon\overline I_{h}^{2}. \end{array}\right. \end{eqnarray*}$

引理2.2 对 $d_{s} {c}^{-1}\left\{\int^{+\infty}_{-\infty}{\rm e}^{-\alpha x}J(x){\rm d}x-1\right\}< \alpha < \lambda_{1}$ ,

$\begin{eqnarray*} \sigma>\max\bigg\{S^{0}, S^{0}{\left(\frac{\lambda_{1}}{K}\right)}^{\alpha}, \frac{\beta[\delta\mu\beta S^{0}-\Delta_{h}(\lambda_{1}, c)]S^{0}}{c\alpha-d_{s} \int^{+\infty}_{-\infty}{\rm e}^{-\alpha x}J(x){\rm d}x+d_{s}}\bigg\} \end{eqnarray*}$

及所有的 $\xi\neq\frac{1}{\alpha}\ln(\frac{S^{0}}{\sigma})$ ,函数 $\underline{S} (\xi)$ 满足

$\begin{eqnarray*} c\underline{S}^{'}\leq d_{s}(J*\underline{S}- \underline{S})-\beta(\overline I_{u}+\delta \overline I _{h})\underline{S}. \end{eqnarray*}$

以上两个引理的证明与文献[30]中的引理2.3、引理2.4相似,此处略去证明.

引理2.3 对充分大的 $M>1$ 和充分小的 $\kappa>0$ 及所有的 $\xi\neq1/ \kappa \ln(1/M )$ ,函数 $\underline{I}(\xi) = (\underline{I}_{u}(\xi), \underline{I}_{h}(\xi))^{T}$ 满足

$\begin{equation} \left\{\begin{array}{ll} c\underline{I}_{u}^{'}\leq d_{u}(J*\underline{I}_{u}-\underline{I}_{u})+(1-\mu)\beta(\underline{I}_{u}+\delta\underline{I}_{h})\underline{S}-k_{u}\underline{I}_{u} -\varepsilon\underline{I}_{u}^{2}, \\ c\underline{I}_{h}^{'}\leq d_{h}(J*\underline{I}_{h}-\underline{I}_{h})+\mu\beta(\underline{I}_{u}+\delta\underline{I}_{h})\underline{S}-k_{h}\underline{I}_{h} -\varepsilon\underline{I}_{h}^{2}. \end{array}\right. \end{equation}$

(2.4)

证易知,当 $\xi\geq\frac{1}{\alpha}\ln\frac{S^{0}}{\sigma}$ 时, $\underline{S}(\xi) = 0$ ,且当 $\xi\geq\frac{1}{\kappa}\ln\frac{1}{M}$ 时 $\underline{I}(\xi) = 0$ .从而 $\frac{1}{\kappa}\ln\frac{1}{M}<\frac{1}{\alpha}\ln\frac{S^0}{\sigma}$ 当且仅当 $M>(\frac{\sigma}{S^{0}})^{\frac{\kappa}{\alpha}}$ ,不失一般性,假设 $M>\max\{(\frac{\sigma}{S^{0}})^{\frac{\kappa}{\alpha}}, 1\}$ .注意到

$\begin{eqnarray*} J*\underline{I}\ge V\max\left\{0, {\rm e}^{\lambda_1\xi}\left[\int_{{{\Bbb R}} }J(y){\rm e}^{-\lambda y}{\rm d}y-M{\rm e}^{k\xi}\int_{{{\Bbb R}} }J(y){\rm e}^{-ky}{\rm d}y\right] \right\}. \end{eqnarray*}$

从而,当 $\xi>\frac{1}{\kappa}\ln\frac{1}{M}$ 时, $\underline{I}(\xi) = 0$ 且引理2.3成立.

当 $\xi<\frac{1}{\kappa}\ln\frac{1}{M}$ 时,则 $\xi<\frac{1}{\alpha}\ln\frac{S^{0}}{\sigma}$ 且

$\underline{S}(\xi) = S^{0}-\sigma {\rm e}^{\alpha\xi}>0,$

$\underline{I}_{u}(\xi) = -\Delta_{h}(\lambda_{1}, c){\rm e}^{\lambda_{1}\xi}(1-M{\rm e}^{\kappa\xi})>0,$

$\underline{I}_{h}(\xi) = \mu\beta S^{0}{\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})>0.$

直接计算可得

$\begin{eqnarray*} 0&\le& {\rm e}^{-\lambda_{1}\xi}[d_{h}(J*\underline{I}_{h}-\underline{I}_{h})-c\underline{I}_{h}^{'}+\mu\beta(\underline{I}_{u}+\delta\underline{I}_{h})\underline{S}-k_{h}\underline{I}_{h} -\varepsilon\underline{I}_{h}^{2}]\\ &\le& {\rm e}^{-\lambda_{1}\xi}[d_{h}(J*\underline{I}_{h}-\underline{I}_{u})-c\underline{I}_{u}^{'}+\mu\beta(\underline{I}_{u}+\delta\underline{I}_{h})S^{0}-k_{h}\underline{I}_{h} -\mu\beta(\underline{I}_{u}+\delta\underline{I}_{h})\sigma {\rm e}^{\alpha\xi}-\varepsilon\underline{I}_{h}^{2}]\\ &\le& {\rm e}^{-\lambda_{1}\xi}\Bigg\{\Big[d_{h}\mu\beta S^{0}{\rm e}^{\lambda_{1}\xi}\int^{+\infty}_{-\infty}{\rm e}^{-\lambda x}J(x){\rm d}x-d_{h}M\mu\beta S^{0}{\rm e}^{(\lambda_{1}+\kappa)\xi}\\ &&- c\mu\beta S^{0}\lambda_{1}{\rm e}^{\lambda_{1}\xi}\int^{+\infty}_{-\infty}{\rm e}^{(\lambda_{1}+\kappa) x}J(x){\rm d}x-d_{h}\mu\beta S^{0}{\rm e}^{\lambda_{1}\xi} \\ &&+ d_{h}M\mu\beta S^{0}{\rm e}^{(\lambda_{1}+\kappa)\xi}-c\mu\beta S^{0}\lambda_{1}{\rm e}^{\lambda_{1}\xi}\\ &&+ (\lambda_{1}+\kappa)\mu M\beta S^{0}{\rm e}^{(\lambda_{1}+\kappa)\xi}+\mu\beta[-\Delta_{h}(\lambda_{1}, c){\rm e}^{\lambda_{1}\xi}(1-M{\rm e}^{\kappa\xi})\\ &&+ \delta\mu\beta S^{0}{\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})\Big](S_{0}-\sigma {\rm e}^{\alpha\xi})\\ &&- k_{h}\mu\beta S^{0}{\rm e}^{\lambda_{1}\xi}(1-M{\rm e}^{\kappa\xi})-\varepsilon\mu^{2}\beta^{2}(S^{0})^{2} {\rm e}^{2\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})^{2}\Bigg\}\\ &\le& \mu\beta S^{0}\Bigg[d_{h}\int^{+\infty}_{-\infty}{\rm e}^{-\lambda_{1} x}J(x){\rm d}x-d_{h}M {\rm e}^{\kappa\xi}\int^{+\infty}_{-\infty}{\rm e}^{-(\lambda_{1} +\kappa)x}J(x){\rm d}x-d_{h}+Md_{h}{\rm e}^{\kappa\xi}\\ &&- c\lambda_{1}+(\lambda_{1}+\kappa)M {\rm e}^{\kappa\xi}- k_{h}(1-M {\rm e}^{\kappa\xi})\Bigg]+\mu\beta[-\Delta_{h}(\lambda_{1}, c){\rm e}^{\lambda_{1}\xi}(1-M{\rm e}^{\kappa\xi})\\ &&+ \delta\mu\beta S^{0}{\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})](S_{0}-\sigma {\rm e}^{\alpha\xi})-\varepsilon\mu^{2}\beta^{2}(S^{0})^{2} {\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})^{2}\}\\ &\le& \mu\beta S^{0}\Bigg[d_{h}\int^{+\infty}_{-\infty}{\rm e}^{-\lambda_{1} x}J(x){\rm d}x-d_{h}M {\rm e}^{\kappa\xi}\int^{+\infty}_{-\infty}{\rm e}^{-(\lambda_{1} +\kappa x}J(x){\rm d}x-d_{h}Md_{h}{\rm e}^{\kappa\xi}\\ &&- c\lambda_{1}+(\lambda_{1}+\kappa)M {\rm e}^{\kappa)\xi}- k_{h}(1-M {\rm e}^{\kappa\xi})-\Delta_{h}(\lambda_{1}, c) {\rm e}^{\lambda_{1}\xi}(1-M{\rm e}^{\kappa\xi})+\mu\beta\delta S^{0}\\ &&- \mu\beta\delta S^{0}M{\rm e}^{\kappa\xi}\Bigg]-\mu\beta[\mu\beta\delta S^{0}- \Delta_{h}(\lambda_{1}, c)](1-M {\rm e}^{\kappa\xi})\sigma {\rm e}^{\alpha\xi}\\ &&-\varepsilon\mu^{2}\beta^{2}(S^{0})^{2} {\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})^{2}\\ &\le& {\rm e}^{\kappa\xi}\mu\beta S^{0}M\left[-d_{h}\int^{+\infty}_{-\infty}{\rm e}^{(\lambda_{1}+\kappa) x}J(x){\rm d}x+c(\lambda_{1}+\kappa)-(\mu\beta\delta S^{0}-k_{h})+\Delta_{h}(\lambda_{1}, c)\right]\\ &&- \mu\beta[\mu\beta\delta S^{0}- \Delta_{h}(\lambda_{1}, c)](1-M {\rm e}^{\kappa\xi})\sigma {\rm e}^{\alpha\xi}-\varepsilon\mu^{2}\beta^{2}(S^{0})^{2} {\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})^{2}\\ &\le& {\rm e}^{\kappa\xi}\mu\beta S^{0}M[\Delta_{h}(\lambda_{1}, c)- \Delta_{h}(\lambda_{1}+\kappa, c)]-\mu\beta[\mu\beta\delta S^{0}- \Delta_{h}(\lambda_{1}, c)](1-M{\rm e}^{\kappa\xi})\sigma {\rm e}^{\alpha\xi}\\ &&- \varepsilon\mu^{2}\beta^{2}(S^{0})^{2} {\rm e}^{\lambda_{1}\xi}(1-M {\rm e}^{\kappa\xi})^{2}. \end{eqnarray*}$

又 $\Delta_{h}(\lambda_{1}, c)- \Delta_{h}(\lambda_{1}+\kappa, c)>0$ 且 $1-M{\rm e}^{\kappa\xi}>0$ ,由引理1.2,可取

$\begin{eqnarray*} M>\frac{\varepsilon\mu^{2}\beta^{2}(S^{0})^{2}+\beta\mu\sigma[ \mu\beta\delta S^{0}-\Delta_{h}(\lambda_{1}, c)]}{\mu\beta S^{0}[\Delta_{h}(\lambda_{1}, c)- \Delta_{h}(\lambda_{1}+\kappa, c)] } \end{eqnarray*}$

充分大和充分小的 $\kappa>0$ ,则(2.4)式的第二个不等式成立.由 $M(\lambda_{1})V = 0$ ,可得 $V = (-\Delta_{h}(\lambda_{1}, c), \mu\beta S^{0})^{T}$ .类似地,可证(2.4)式的第一个不等式成立.证毕.

2.2 辅助问题

定义

$\begin{eqnarray*} \Omega = \left\{(S(\cdot), I_{u}(\cdot), I_{h}(\cdot))\in C([-X, X], {{{\Bbb R}} }^{3})\left|\; \begin{array}{ll} & S(-X) = \underline{S}(-X), I_{u}(-X) = \underline{I}_{u}(-X), \\ & I_{h}(-X) = \underline{I}_{h}(-X), \\ & \underline{S}(\xi)\leq S(\xi)\le S^{0}, \\ & \underline{I}_{u}(\xi)\leq{I}_{u}(\xi) \leq\overline{I}_{u}(\xi), \\ &\underline{I}_{h}(\xi) \leq{I}_{h}(\xi)\leq\overline{I}_{h}(\xi), \; \xi\in[-X, X] \end{array} \right.\right\}, \end{eqnarray*}$

其中 $X>\max\{ \frac{1}{\lambda_{1}}\ln\frac{1}{K}, \frac{1}{\alpha}\ln\frac{\sigma}{S^{0}}, \frac{1}{\kappa}\ln M\}$ .

进一步地,对所有的 $(S(\cdot), I_{u}(\cdot), I_{h}(\cdot))\in C([-X, X], {{{\Bbb R}} }^{3})$ ,令

$\begin{eqnarray*} \widetilde{S}(\xi) = \left\{\begin{array}{ll} S(X), \xi>X, \\ S(\xi), |\xi|\le X , \\ \underline{S}(\xi), \xi<-X, \end{array}\right. \; \widetilde{I}_{u}(\xi) = \left\{\begin{array}{ll} {I}_{u}(X) , \xi>X, \\ {I}_{u}(\xi), |\xi|\le X , \\ \underline{I}_{u}(\xi), \xi<-X, \end{array}\right. \; \widetilde{I}_{h}(\xi) = \left\{\begin{array}{ll} {I}_{h}(X), \xi>X, \\ {I}_{h}(\xi), |\xi|\le X, \\ \underline{I}_{h}(\xi), \xi<-X. \end{array}\right. \end{eqnarray*}$

考虑以下系统

$\begin{equation} \left\{ \begin{array}{l} { } cS^{'}(\xi) = d_{s}\int^{+\infty}_{-\infty}\widetilde{S}(\xi-y)J(y){\rm d}y-(d_{s}+\beta(I_{u}+\delta I_{h}))S, \\ { } cI_{u}^{'}(\xi) = d_{u}\int^{+\infty}_{-\infty}\widetilde{I}_{u}(\xi-y)J(y){\rm d}y+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}-d_{u}I_{u}-\varepsilon I_{u}^{2}, \\ { } cI_{h}^{'}(\xi) = d_{h}\int^{+\infty}_{-\infty}\widetilde{I}_{h}(\xi-y)J(y){\rm d}y+\mu\beta(I_{u}+\delta I_{h})S-k_{h}I_{h}-d_{h}I_{h}-\varepsilon I_{h}^{2} \end{array} \right. \end{equation}$

(2.5)

及初值条件

$S(-X) = \underline{S}(-X), \; I_{u}(-X) = \underline{I}_{u}(-X), \; I_{h}(-X) = \underline{I}_{h}(-X).$

定义算子 $F = (F_{s} , F_{u}, F_{h}):\Omega\mapsto C([-X, X], {{{\Bbb R}} }^{3})$ 如下

$\begin{eqnarray*} \left\{\begin{array}{ll} S_{X}(\xi) = F_{s}[ S(\cdot), I_{u }(\cdot), I_{h }(\cdot) ](\xi), \\ I_{u X}(\xi) = F_{u}[ S(\cdot), I_{u }(\cdot), I_{h }(\cdot) ] (\xi), \\ I_{h X}(\xi) = F_{h}[ S(\cdot), I_{u }(\cdot), I_{h }(\cdot) ] (\xi), \\ \end{array}\right. \end{eqnarray*}$

其中 $\xi\in[-X, X]$ .

引理2.4 算子 $F:\Omega\mapsto\Omega$ 是全连续的.

引理2.4的证明类似于文献[27,定理2.6],故略去.

由于 $\Omega$ 是一个闭凸集,利用Schauder不动点定理,存在 $(S_{X}(\xi), I_{uX}(\xi), I_{hX}(\xi))\in\Omega$ 使得

$(S_{X}(\xi), I_{uX}(\xi), I_{hX}(\xi)) = F[S_{X}(\xi), I_{uX}(\xi), I_{hX}(\xi)], \xi\in[-X, X].$

要得到系统(2.3)在边界条件(2.2)下解的存在性,需要对 $S_{X}(\xi), I_{uX}(\xi)$ 和 $I_{hX}(\xi)$ 做以下估计.

引理2.5 对给定 $\varepsilon>0$ ,存在 $C_{\varepsilon}>0$ ,使得对一切 $X>\max\{\frac{1}{\kappa}\ln M, \frac{1}{\alpha}\ln\frac{\sigma}{S^{0}}, \frac{1}{\lambda_{1}}\ln\frac{1}{K}, T \}$

$||S_{X}||_{C^{2}([-X, X]}<C_{\varepsilon}, \; ||I_{uX}||_{C^{2}([-X, X]}<C_{\varepsilon}, \; ||I_{hX}||_{C^{2}([-X, X]}<C_{\varepsilon},$

其中 $T$ 是 ${\rm supp}\ J$ 的半径.

证由 $(S_{X}(\xi), I_{uX}(\xi), I_{hX}(\xi))$ 满足系统(2.5),可得

$\begin{eqnarray*} |S_{X}^{'}(\xi)|& = & \frac{d_{s}}{c}\int^{+\infty}_{-\infty} \widetilde{S}_{X}(\xi-y)J(y){\rm d}y+\frac{d_{s}}{c}|S_{X}|+\frac{\beta}{c}|I_{uX}S_{X}|+ \frac{\beta\delta}{c} |I_{hX}S_{X}|\\ &\leq& \frac{d_{s}}{c}\Bigg|\int^{-X}_{-\infty} \underline{S}_{X}(y)J(\xi-y){\rm d}y +\int^{+\infty}_{X}{S}_{X}(X)J(\xi-y){\rm d}y\\ &&+ \int^{X}_{-X}{S}_{X}(y)J(\xi-y){\rm d}y\Bigg|+\frac{d_{s}}{c}S_{0}+\frac{\beta}{c}\overline{I}_{u}(X)S^{0}+\frac{\beta\delta S^{0}}{c}\overline{I}_{h}(X)\\ &\le& \frac{4d_{s}}{c}S_{0}+\frac{\beta}{c}\overline{I}_{u}(X)S^{0}+\frac{\beta\delta S^{0}}{c}\overline{I}_{h}(X)\\ & = & \frac{1}{c}(4d_{s}S_{0}+\beta\overline{I}_{u}(X)S^{0}+\beta\delta S^{0}\overline{I}_{h}(X)), \end{eqnarray*}$

且

$\begin{eqnarray*} |I_{uX}^{'}(\xi)|&\le&\frac{d_{u}}{c}\Bigg|\int^{+\infty}_{-\infty}\widetilde{I}_{uX}(\xi-y)J(y){\rm d}y\Bigg|+ \frac{(1-\mu)}{c}\beta|I_{uX}||S_{X}|\\ &&+ \frac{(1-\mu)}{c}\beta\delta |I_{hX}||S_{X}|+\frac{k_{u}}{c}|I_{uX}|+\frac{d_{u}}{c}|I_{uX}|+\varepsilon\frac{ |I_{uX}|^{2}}{c}\\ &\le& \frac{d_{u}}{c}\Bigg|\int^{T+\xi}_{-T-\xi}\widetilde{I}_{uX}(y)J(\xi-y){\rm d}y\Bigg|+\frac{1}{c}[(1-\mu)\beta\delta \overline{I}_{h}(X)S^{0}+(1-\mu)\beta \overline{I}_{u}(X)S^{0}\\ &&+ \overline{I}_{u}(X)d_{u}+\overline{I}_{u}(X)k_{u}+\varepsilon\overline{I}_{u}(X)^{2}]\\ &\le& \frac{d_{u}}{c}\overline{I}_{u}(T+X)+\frac{1}{c}[(1-\mu)\beta\delta \overline{I}_{h}(X)S^{0}+(1-\mu)\beta \overline{I}_{u}(X)S^{0}\\ &&+ \overline{I}_{u}(X)d_{u}+\overline{I}_{u}(X)k_{u}+\varepsilon\overline{I}_{u}(X)^{2}]\\ & = & \frac{1}{c}[\overline{I}_{u}(T+X)+(1-\mu)\beta\delta \overline{I}_{h}(X)S^{0}+(1-\mu)\beta \overline{I}_{u}(X)S^{0}\\ &&+ \overline{I}_{u}(X)d_{u}+\overline{I}_{u}(X)k_{u} +\varepsilon\overline{I}_{u}(X)^{2}], \end{eqnarray*}$

其中 ${\rm supp}\ J = [-T, T],$ $0\leq\overline{I}_{u}(\xi)\leq-K\Delta_{h}(\lambda_{1}, c)$ 且 $0\leq\overline{I}_{h}(\xi)\leq \mu K\beta S^{0}.$ 从而,存在一个与 $x$ 无关的常数 $C_{\varepsilon}$ 使得

$|S_{X}^{'}(\xi)|<C_{\varepsilon}, \ |I_{uX}^{'}(\xi)|<C_{\varepsilon}, \ |I_{hX}^{'}(\xi)|<C_{\varepsilon}.$

进一步地,对所有的 $\xi, \eta\in[-X, X]$ ,有

$|S_{X}(\xi)-S_{X}(\eta)|<C_{\varepsilon}|\xi-\eta|, \; |I_{uX}(\xi)-I_{uX}(\eta)|<C_{\varepsilon}|\xi-\eta|, \; |I_{hX}(\xi)-I_{hX}(\eta)|<C_{\varepsilon}|\xi-\eta|.$

现在证明 $S_{X}^{'}(\xi)$ , $I_{uX}^{'}$ 和 $I_{hX}^{'}$ 是利普希茨连续的.由系统(2.5),可得

$\begin{eqnarray*} c|S_{X}^{'}(\xi_{1})- S_{X}^{'}(\xi_{2})|&\le& d_{s}|\int^{+\infty}_{-\infty}\widetilde{S}_{X}(\xi_{1}-y)J(y){\rm d}y-\int^{+\infty}_{-\infty}\widetilde{S}_{X}(\xi_{2}-y)J(y){\rm d}y|\\ &&+ d_{s}|S_{X}(\xi_{1})-S_{X}(\xi_{2})|+\beta|I_{uX}(\xi_{1})S_{X}(\xi_{1})- I_{uX}(\xi_{2})S_{X}(\xi_{2})|\\ &&+ \beta\delta |I_{hX}(\xi_{1})S_{X}(\xi_{1})- I_{hX}(\xi_{2})S_{X}(\xi_{2})|\\ & = & K_{1}+K_{2}+K_{3}+K_{4}. \end{eqnarray*}$

则

$\begin{eqnarray*} K_{1}& = & d_{s}\left|\int^{+\infty}_{-\infty} \widetilde{S}_{X}(\xi_{1}-y)J(y){\rm d}y -\int^{+\infty}_{-\infty}\widetilde{S}_{X} (\xi_{2}-y)J(y){\rm d}y\right|\\ & = & d_{s}\left|\int^{\xi_{1}+T}_{\xi_{1}-T} \widetilde{S}_{X}(y)J(\xi_{1}-y){\rm d}y- \int^{\xi_{2}+T}_{\xi_{2}-T}\widetilde{S}_{X}(y) J(\xi_{2}-y){\rm d}y\right|\\ & = & d_{s}\Bigg|-\int^{\xi_{1}-T}_{\xi_{2}-T}\widetilde{S}_{X}(y)J(\xi_{2}-y){\rm d}y+\int^{\xi_{1}+T}_{\xi_{2}+T}\widetilde{S}_{X}(y)J(\xi_{1}-y){\rm d}y \\ &&+ \int^{\xi_{2}+T}_{\xi_{1}-T}\widetilde{S}_{X}(y) J(\xi_{1}-y){\rm d}y-\int^{\xi_{2}+T}_{\xi_{1}-T} \widetilde{S}_{X}(y)J(\xi_{2}-y){\rm d}y\Bigg|\\ &\le& d_{s}\Bigg[||J||_{L^{\infty}}S^{0}|\xi_{1}-\xi_{2}|+\int^{\xi_{1}+T}_{\xi_{2}+T}\widetilde{S}_{X}(y)J(\xi_{1}-y){\rm d}y\\ &&+ \left|\int^{\xi_{2}+T}_{\xi_{1}-T}\widetilde{S}_{X}(y)J(\xi_{1}-y){\rm d}y-\int^{\xi_{2}+T}_{\xi_{1}-T} \widetilde{S}_{X}(y)J(\xi_{2}-y){\rm d}y\right|\Bigg]\\ &\le& [2d_{s}||J||_{L^{\infty}}S^{0}+d_{s}\parallel J^{'}\parallel_{L^{1}}S^{0}]|\xi_{1}-\xi_{2}|, \end{eqnarray*}$

且

$\begin{eqnarray*} K_{2}& = & d_{s}|S_{X}(\xi_{1})-S_{X}(\xi_{2})|\leq d_{s}C_{\varepsilon}|\xi_{1}-\xi_{2}|, \\ K_{3}& = & \beta|I_{uX}(\xi_{1})S_{X}(\xi_{1})-I_{uX}(\xi_{2})S_{X}(\xi_{1})+I_{uX}(\xi_{2})S_{X}(\xi_{1})- I_{uX}(\xi_{2})S_{X}(\xi_{2})|\\ &\le& \beta S^{0}|I_{uX}(\xi_{1})-I_{uX}(\xi_{2})|+\beta \overline{I_{u}}(X)|S_{X}(\xi_{1})-S_{X}(\xi_{2}|\\ &\le& \beta S^{0}C_{\varepsilon}|\xi_{1}-\xi_{2}|+\beta \overline{I_{u}}(X)C_{\varepsilon}|\xi_{1}-\xi_{2}|\\ & = & \beta(S^{0}+\overline{I_{u}}(X))C_{\varepsilon}|\xi_{1}-\xi_{2}|. \end{eqnarray*}$

同理可得

$K_{4}\leq\beta\delta(S^{0}+\overline{I_{h}}(X))C_{\varepsilon}|\xi_{1}-\xi_{2}|,$

且

$\begin{eqnarray*} |S_{X}^{'}(\xi_{1})-S_{X}^{'}(\xi_{2})| &\leq&\frac{1}{c}[2d_{s}||J||_{L^{\infty}}S^{0}+d_{s}\parallel J^{'}\parallel_{L^{1}}S^{0}+ d_{s}C_{\varepsilon}\\ &&+\beta(S^{0}+\overline{I_{u}}(X))C_{\varepsilon}+\beta\delta(S^{0}+\overline{I_{h}}(X)) C_{\varepsilon}]|\xi_{1}-\xi_{2}|. \end{eqnarray*}$

其他情形可类似证明.证毕.

2.3 行波解存在性的证明

引理2.6 对给定 $\varepsilon>0$ ,系统(2.3)存在行波解 $(S(\xi), I_{u}(\xi), I_{h}(\xi))$ , $0<S(\xi)<S^{0}$ , $I_{u}(\xi)+I_{h}(\xi)>0$ 满足条件(2.2)且

$0\leq I_{u}(\xi)\leq \frac{(1-\mu)( d_{u}+2k_{u})(S^{0}-S^{1})}{k_{u}},$

$0\leq I_{h}(\xi)\leq \frac{\mu( d_{h}+2k_{h})(S^{0}-S^{1})}{k_{h}}.$

进一步地,有

$\int^{+\infty}_{-\infty}[k_{u}I_{u}(y)+\varepsilon I_{u}^{2}(y)]{\rm d}y = (1-\mu)c(S^{0}-S^{1}),$

$\beta\int^{+\infty}_{-\infty}[I_{u}(y)+\delta I_{h}(y)]\widetilde{S}(y){\rm d}y = c(S^{0}-S^{1}),$

$\int^{+\infty}_{-\infty}[k_{h}I_{h}(y)+\varepsilon I_{h}^{2}(y)]{\rm d}y = \mu c(S^{0}-S^{1}).$

证令 $\{X_{m}\}_{m = 1}^{\infty}$ 是满足

$X_{m}>\max\left\{\frac{1}{\kappa}\ln M, \frac{1}{\alpha}\ln\frac{\sigma}{S^{0}}, \frac{1}{\lambda_{1}}\ln\frac{1}{K}, T \right\}$

的递增序列,且当 $m\to+\infty$ 时, $X_{m}\to+\infty$ .则由引理2.5

$||S_{X_{m}}||_{C^{2}[-X_{m}, X_{m}]}<C_{\varepsilon}, \; ||I_{uX_{m}}||_{C^{2}[-X_{m}, X_{m}]}<C_{\varepsilon}, \; ||I_{hX_{m}}||_{C^{2}[-X_{m}, X_{m}]}<C_{\varepsilon},$

且 $(S_{X_{m}}(\xi)$ , $I_{uX_{m}}(\xi), I_{hX_{m}}(\xi))$ 满足系统(2.5).从而,存在一个序列,表示为 $\{S_{X_{m_{k_{k}}}}\}$ , $\{I_{uX_{m_{k_{k}}}}\}$ 和 $\{I_{hX_{m_{k_{k}}}}\}$ ,且当 $k\rightarrow+\infty$ 时

$\begin{eqnarray*} (S_{X_{m_{k_{k}}}}, I_{uX_{m_{k_{k}}}}, I_{hX_{m_{k_{k}}}})\to(S^{\ast}(\cdot), I_{u}^{\ast}(\cdot), I_{h}^{\ast}(\cdot)). \end{eqnarray*}$

在 $C_{loc}^{1}({{\Bbb R}} ;{{\Bbb R}} ^3)$ 上成立.根据控制收敛定理,

$\begin{eqnarray*} \left\{\begin{array}{ll} { } \lim\limits_{k\to\infty} \int^{+\infty}_{-\infty}J(y)S_{X_{m_{k_{k}}}}(\xi-y){\rm d}y { } = \int^{+\infty}_{-\infty}J(y)S^{\ast}(\xi-y){\rm d}y, \\ { } \lim\limits_{k\to\infty} { } \int^{+\infty}_{-\infty}J(y)I_{uX_{m_{k_{k}}}}(\xi-y){\rm d}y = \int^{+\infty}_{-\infty}J(y)I_{u}^{\ast}(\xi-y){\rm d}y, \\ { } \lim\limits_{k\to\infty} \int^{+\infty}_{-\infty}J(y)I_{hX_{m_{k_{k}}}}(\xi-y){\rm d}y = \int^{+\infty}_{-\infty}J(y)I_{h}^{\ast}(\xi-y){\rm d}y. \end{array}\right. \end{eqnarray*}$

则 $(S^{\ast}(\cdot), I_{u}^{\ast}(\cdot), I_{h}^{\ast}(\cdot))$ 满足系统(2.3)且

$\begin{equation} \underline{S}(\cdot)\leq S^{\ast}(\cdot)<S^{0}, \; \underline{I}_{u}(\cdot)\leq I_{u}^{\ast}(\cdot) \leq\overline{I}_{u}(\cdot), \; \underline{I}_{h}(\cdot)\leq I_{h}^{\ast}(\cdot) \leq\overline{I}_{h}(\cdot)\; \mbox{在}\; {{\Bbb R}} . \end{equation}$

(2.6)

进一步地,由(2.6)式可得

$\begin{eqnarray*} S^{\ast}(-\infty) = S^0, \; I_{u}^{\ast}(-\infty) = 0\; \mbox{且}\; I_{h}^{\ast}(-\infty) = 0. \end{eqnarray*}$

下面,为了叙述方便,令 $S^{\ast} = S, I_{u}^{\ast} = I_{u}$ 且 $I_{h}^{\ast} = I_{h}$ .首先,证明

$\begin{eqnarray*} S(\xi)>0, I_{u}(\xi)+I_{h}(\xi)>0. \end{eqnarray*}$

事实上,假设存在某个 $\xi^{\ast}>0$ 使得 $S(\xi^{\ast}) = 0$ .由系统(2.5)的第一个方程,可得

$d_{s}\int^{T}_{-T}J(y)S(\xi^{\ast}-y){\rm d}y = 0.$

又 $\int^{T}_{-T}J(y){\rm d}y = 1$ 且 $J(y)\geq0$ ,存在 $T_{1}$ 和 $T_{2}$ ,满足 $-T<T_{1}<T_{2}<T$ ,使得对 $x\in[T_{1}, T_{2}]$

$J(x)>0, d_{s}\int^{T_{2}}_{T_{1}}J(y)S(\xi^{\ast}-y){\rm d}y = 0.$

从而,由以上方程可得对 $y\in [T_{1}, T_{2}]$ , $S(\xi^{\ast}-y) = 0$ .重复以上过程,可得对任意 $\xi\in{{\Bbb R}}$ , $S(\xi)\equiv0$ .这与 $S(-\infty) = S^{0}$ 矛盾.同理可得 $I_{u}+I_{h}>0$ .

下面证明

$\begin{eqnarray*} S(+\infty) = S^0, \; I_{u}(+\infty) = 0\; \mbox{且}\; I_{h}(+\infty) = 0. \end{eqnarray*}$

首先证明 $\int^{+\infty}_{-\infty}(I_{u}+\delta I_{h})S{\rm d}\xi<+\infty$ 且 $\liminf\limits_{\xi\rightarrow+\infty}S(\xi)<S^{0}$ .注意到对所有的 $z\in {{\Bbb R}}$ ,有

$\begin{eqnarray*} \int^{z}_{-\infty}(J*S(\xi)-S(\xi)){\rm d}\xi& = & \int^{z}_{-\infty}\int^{+\infty}_{-\infty}J(x)[S(\xi-x)-S(\xi)]{\rm d}x{\rm d}\xi\\ & = & \int^{z}_{-\infty}\int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}S^{'}(\xi-\theta x){\rm d}\theta {\rm d}x{\rm d}\xi\\ & = & \int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}\int^{z}_{-\infty}{S}^{'}(\xi-\theta x){\rm d}\theta {\rm d}x{\rm d}\xi\\ & = &\int^{+\infty}_{-\infty}J(x)x\int^{1}_{0}[S_{0}-S(z-\theta x)]{\rm d}\theta {\rm d}x\\ & = &\int^{+\infty}_{-\infty}J(x)x \int^{1}_{0}S(z+\theta x){\rm d}\theta {\rm d}x\\ &\leq& S^{0}\int^{+\infty}_{-\infty}J(x)|x|{\rm d}x = :\eta_{0}. \end{eqnarray*}$

由系统(2.5)的第一个方程可得

$\begin{eqnarray*} \beta\int^{z}_{-\infty}(I_{u}(\xi)+\delta I_{h}(\xi))S(\xi){\rm d}\xi& = &d_{s}\int^{z}_{-\infty}(J*S(\xi)-S)(\xi){\rm d}\xi+cS^{0}-cS(z)\\ &\leq&d_{s}\eta_{0}+cS^{0}, \end{eqnarray*}$

从而

$\int^{+\infty}_{-\infty}(I_{u}+\delta I_{h})S{\rm d}\xi<+\infty.$

此外, $\liminf\limits_{\xi\rightarrow+\infty}S(\xi)< S^{0}$ .否则, $S(+\infty) = S^{0}$ .从 $-\eta$ 到 $\eta$ 上积分系统(2.5)的第一个方程,有

$\begin{eqnarray*} c[S(\eta)-S(-\eta)]& = &d_{s}\int^{+\infty}_{-\infty}yJ(y)\int^{1}_{0}[S(-\eta-ty)-S(\eta-ty)]{\rm d}t{\rm d}y\\ &&-\beta\int^{\eta}_{-\eta}(I_{u}(\xi)+\delta I_{h}(\xi))S(\xi){\rm d}\xi. \end{eqnarray*}$

令 $\eta\rightarrow+\infty$ ,可得 $\int^{+\infty}_{-\infty}(I_{u}(\xi)+\delta I_{h}(\xi))S(\xi){\rm d}\xi = 0$ ,矛盾.

从 $-\infty$ 到 $\xi$ 上积分系统(2.5)的第二个方程,得

$\begin{eqnarray*} cI_{u}(\xi)& = &d_{u}\int^{\xi}_{-\infty}[J\ast I_{u}(\tau)-I_{u}(\tau)]{\rm d}\tau+(1-\mu)\beta\int^{\xi}_{-\infty}(I_{u}(\tau)+\delta I_{h}(\tau))S(\tau){\rm d}\tau\\ &&-\int^{\xi}_{-\infty}[k_{u}I_{u}(\tau)+\varepsilon I_{u}^{2}(\tau)]{\rm d}\tau. \end{eqnarray*}$

又 $0<\overline{I}_{u}(\xi)\leq-K\Delta_{h}(\lambda_{1}, c)$ ,则 $\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi<+\infty$ .显然, $I_{u}^{'}(\xi)$ 有界,则 $I_{u}(\pm\infty) = 0$ .同理可得 $I_{h}(\pm\infty) = 0$ . $\liminf\limits_{\xi\rightarrow+\infty} S(\xi) = \limsup\limits_{\xi\rightarrow+\infty} S(\xi)<S^{0}$ 的证明类似于文献[26,定理2.8],这里略去证明.

从 $-\infty$ 到 $+\infty$ 上积分(2.5)的第一个方程,得

$c(S^{1}-S^{0}) = d_{s}\int^{+\infty}_{-\infty}(J*S(\xi)-S(\xi)){\rm d}\xi-\beta\int^{+\infty}_{-\infty}(I_{u}(\xi)+\delta I_{h}(\xi))S(\xi){\rm d}\xi,$

且

$\begin{eqnarray*} \int^{+\infty}_{-\infty}(J*S(\xi)-S(\xi)){\rm d}\xi& = & \int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}J(x)[S(\xi-x)-S(\xi)]{\rm d}x{\rm d}\xi\\ & = & \int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}S^{'}(\xi-\theta x){\rm d}\theta {\rm d}x{\rm d}\xi\\ & = & \int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}\int^{+\infty}_{-\infty}{S}^{'}(\xi-\theta x){\rm d}\theta {\rm d}x{\rm d}\xi\\ & = & (S^{1}-S^{0})\int^{+\infty}_{-\infty}J(x)(-x){\rm d}x\\ & = & 0. \end{eqnarray*}$

从而

$\begin{eqnarray*} \beta\int^{+\infty}_{-\infty}[I_{u}(\xi)+\delta I_{h}](\xi)S(\xi){\rm d}\xi = [d_{s}\int^{+\infty}_{-\infty}J(x)(-x){\rm d}x-c](S^{1}-S^{0}) = c(S^{0}-S^{1}), \end{eqnarray*}$

且

$\begin{eqnarray*} \int^{+\infty}_{-\infty}[k_{u}I_{u}(y)+\varepsilon I_{u}^{2}(y)]{\rm d}y = c(1-\mu)(S^{0}-S^{1}). \end{eqnarray*}$

同理

$\begin{eqnarray*} \int^{+\infty}_{-\infty}[k_{h}I_{h}(y)+\varepsilon I_{h}^{2}(y)]{\rm d}y = c\mu(S_{0}-S_{1}). \end{eqnarray*}$

最后证明

$0< I_{u}(\xi)\leq \frac{(1-\mu)( d_{u}+2k_{u})(S^{0}-S^{1})}{k_{u}},$

$0< I_{h}(\xi)\leq \frac{\mu( d_{h}+2k_{h})(S^{0}-S^{1})}{k_{h}}.$

从 $-\infty$ 到 $\xi$ 上积分(2.5)的第二个方程,可得

$\begin{eqnarray*} cI_{u}(\xi)& = &d_{u}\int_{-\infty}^{\xi}(J\ast I_{u}(x)-I_{u}(x)){\rm d}x\\ &&+\int_{-\infty}^{\xi}[(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}-\varepsilon I_{u}^{2}]\\ & = &d_{u}\int_{-\infty}^{+\infty}J(y)\int_{-\infty}^{\xi-y}I_{u}(x){\rm d}x {\rm d}y-d_{u}\int_{-\infty}^{\xi}I_{u}(x){\rm d}x\\ &&+\int_{-\infty}^{\xi}[(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}-\varepsilon I_{u}^{2}]\\ &\leq&d_{u}\int_{-\infty}^{+\infty}I_{u}(x){\rm d}x-d_{u}\int_{-\infty}^{\xi}I_{u}(x){\rm d}x\\ &&+\int_{-\infty}^{\xi}[(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}-\varepsilon I_{u}^{2}]\\ &\leq&d_{u}\int_{-\infty}^{+\infty}I_{u}(x){\rm d}x+\int_{-\infty}^{\xi}[(1-\mu)\beta(I_{u}+\delta I_{h})S+k_{u}I_{u}+\varepsilon I_{u}^{2}]\\ &\leq&(\frac{d_{u}}{k_{u}}+1)\int_{-\infty}^{+\infty}(k_{u}I_{u}(x)+\varepsilon I_{u}^{2}(x)){\rm d}x+\int_{-\infty}^{+\infty}(1-\mu)\beta(I_{u}(x)+\delta I_{h}(x))S(x){\rm d}x\\ &\leq&\frac{c(1-\mu)(d_{u}+k_{u})(S^{0}-S^{1})}{k_{u}}+c(1-\mu)(S^{0}-S^{1}). \end{eqnarray*}$

从而, $0< I_{u}(\xi)\leq \frac{(1-\mu)( d_{u}+2k_{u})(S^{0}-S^{1})}{k_{u}}.$ 同理可得 $0< I_{h}(\xi)\leq \frac{\mu(d_{h}+2k_{h})(S^{0}-S^{1})}{k_{h}}$ .证毕.

定理2.1 假设(H1')或(H2)成立.则系统(2.1)存在满足边界条件(2.2)的波速为 $c$ 的行波解 $(S(\xi), I_{u}(\xi), I_{h}(\xi), R(\xi))$ 且

$\int^{+\infty}_{-\infty}k_{u}I_{u}(y){\rm d}y = c(1-\mu)(S^{0}-S^{1}),$

$\beta\int^{+\infty}_{-\infty}[I_{u}(y)+\delta I_{h}(y)]S(y){\rm d}y = c(S^{0}-S^{1}),$

$\int^{+\infty}_{-\infty}k_{h}I_{h}(y){\rm d}y = c\mu(S^{0}-S^{1}).$

证令系列 $\{\varepsilon_{n}\}$ 满足 $\{0<\varepsilon_{i+1}<\varepsilon_{i}<1 \}$ 且 $\lim\limits_{n\longrightarrow \infty }\varepsilon_{n} = 0$ .根据引理2.6,对任意 $\varepsilon_{n}$ ,系统(2.3)存在满足条件(2.2)的解 $\Phi_{n}(\xi) = (\Phi_{sn}(\xi), \Phi_{un}(\xi), \Phi_{hn}(\xi))$ 使得

$\begin{eqnarray*} \left\{\begin{array}{ll} { } \int^{+\infty}_{-\infty}[k_{u}\Phi_{un}(y)+\varepsilon \Phi_{un}(y)^{2}]{\rm d}y = (1-\mu)c(S_{0}-S_{1}), \\ { } \beta\int^{+\infty}_{-\infty}[\Phi_{un}(y)+\delta \Phi_{hn}(y)]\Phi_{sn}(y){\rm d}y = c(S_{0}-S_{1}), \\ { } \int^{+\infty}_{-\infty}[k_{h}\Phi_{hn}(y)+\varepsilon \Phi_{hn}(y)^{2}]{\rm d}y = \mu c(S_{0}-S_{1}). \end{array}\right. \end{eqnarray*}$

由引理2.5和2.6的证明,可得

$|\Phi_{sn}^{'}(\xi)|< \frac{2 d_{s}S^{0}}{c}+(S^{0}-S^{1}) = (\frac{2 d_{s}}{c}+1)S^{0}-S^{1},$

$|\Phi_{un}^{'}(\xi)|< 2d_{u}[\frac{(1-\mu)( d_{u}+2k_{u})(S^{0}-S^{1})}{ck_{u}}] +2(1-\mu)(S^{0}-S^{1}),$

$|\Phi_{hn}^{'}(\xi)|< 2d_{h}[\frac{\mu (d_{h}+2k_{h})(S^{0}-S^{1})}{ck_{h}}] +2\mu(S^{0}-S^{1}).$

同理,存在与 $\varepsilon$ 无关的常数 $M_{s}$ , $M_{u}$ , $M_{h}$ 使得

$|\Phi_{sn}^{''}(\xi)|<M_{s}, |\Phi_{un}^{''}(\xi)|<M_{u}, |\Phi_{hn}^{''}(\xi)|<M_{h}.$

从而, $\Phi_{n}$ , $\Phi_{n}^{'}$ 和 $\Phi_{n}^{''}$ 在 ${{\Bbb R}}$ 上等度连续且一致有界.由Arzela-Ascoli's定理,存在一个序列 $\{n_{k}\}$ 使得当 $k\rightarrow\infty$ 时,有

$\begin{eqnarray*} && \Phi_{sn_{k}}(\xi)\rightarrow\Psi_{s}(\xi), \; \Phi_{un_{k}}(\xi)\rightarrow\Psi_{u}(\xi), \; \Phi_{hn_{k}}(\xi)\rightarrow\Psi_{h}(\xi) \end{eqnarray*}$

在 $C_{loc}^{1}$ 上一致连续,其中 $\Psi(\xi) = (\Psi_{s}(\xi), \Psi_{u}(\xi), \Psi_{h}(\xi)).$

下证 $\Psi_{s}(-\infty) = S^{0}$ .因 $\Phi_{sn_{k}}(\xi)\rightarrow\Psi_{s}(\xi)$ 在 $[-\eta, 0]$ 的任何有界集中一致有界,则存在与 $\eta$ 无关的 $n_{k}$ 使得对任意 $\varepsilon>0$ 和 $p>k$ ,有

$\begin{eqnarray*} | \Phi_{sn_{p}}(\xi)-\Psi_{s}(\xi)|<\frac{\varepsilon}{2}, \; \xi\in [-\eta, 0]. \end{eqnarray*}$

又 $\Phi_{sn_{p}}(-\infty) = S^{0}$ ,存在充分大的 $\eta$ 使得对 $\xi\leq-\eta$ ,有

$|\Psi_{sn_{p}}(\xi)-S^{0}|<\frac{\varepsilon}{2},$

$|\Psi_{s}(\xi)-S^{0}|<| \Phi_{sn_{p}}(\xi)-\Psi_{s}(\xi)|+|\Psi_{sn_{p}}(\xi)-S^{0}|<\varepsilon.$

从而, $\Psi_{s}(-\infty) = S^{0}$ .同理可得

$\begin{eqnarray*} \left\{\begin{array}{ll} (\Psi_{s}(-\infty), \Psi_{u}(-\infty), \Psi_{h}(-\infty)) = (S^{0}, 0, 0), \\ (\Psi_{s}(+\infty), \Psi_{u}(+\infty), \Psi_{h}(+\infty)) = (S^{1}, 0, 0), \end{array}\right. \end{eqnarray*}$

又 $\Phi_{n_{k}}(\xi)$ 是问题(2.3)和(2.2)的解序列且 $\varepsilon_{n}\rightarrow 0$ ,则

$\begin{eqnarray*} \left\{\begin{array}{ll} c\Psi_{s}^{'}(\xi) = d_{s}(J*\Psi_{s}-\Psi_{s})-\beta(\Psi_{u}+\delta \Psi_{h})\Psi_{s}, \\ c\Psi_{u}^{'}(\xi) = d_{u}(J*\Psi_{u}-\Psi_{u})+(1-\mu)\beta(\Psi_{u}+\delta \Psi_{h})\Psi_{s}-k_{u}\Psi_{u}, \\ c\Psi_{h}^{'}(\xi) = d_{h}(J*\Psi_{h}-\Psi_{h})+\mu\beta(\Psi_{u}+\delta \Psi_{h})\Psi_{s}-k_{h}\Psi_{h} \end{array}\right. \end{eqnarray*}$

满足相应的边界条件.即 $\Psi(\xi)$ 是问题(2.1)与(2.2)的一个解,且对所有的 $\xi\in {{\Bbb R}}$ 满足

$\begin{eqnarray*} \left\{\begin{array}{ll} { } \int^{+\infty}_{-\infty}k_{u}\Psi_{u}(\xi){\rm d}\xi = (1-\mu)c(S^{0}-S^{1}), \\ { } \beta\int^{+\infty}_{-\infty}[\Psi_{u}(\xi)+\delta \Psi_{h}(\xi)]\Psi_{s}(\xi){\rm d}\xi = c(S^{0}-S^{1}), \\ { } \int^{+\infty}_{-\infty}k_{h}\Psi_{h}(\xi){\rm d}\xi = \mu c(S^{0}-S^{1}). \end{array}\right. \end{eqnarray*}$

证毕.

3 $c=c^{*}$ 情形下行波解的存在性

假设 $R_{0}>1$ 且 $d_{u} = d_{h} = d$ .本节主要证明系统(1.3)有一个波速为 $c = c^{*}$ 非平凡的正解 $(S(\xi), I_{u}(\xi), I_{h}(\xi), R(\xi))$ ,由于(1.3)的第四个方程是相对独立的,只需考虑系统(2.1)的临界波速 $c = c^{*}$ .下面总是假设 $(S(\xi), I_{u}(\xi), I_{h}(\xi))$ 是系统(2.1)的波速为 $c>0$ 的正解.

引理3.1 存在某个正常数 $C$ 使得

$\begin{eqnarray*} \int_{-\infty}^{+\infty}J(y)\frac{k_{h}I_{u}(\xi-y)+k_{u}\delta I_{h}(\xi-y)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}{\rm d}y<C, \quad \Bigg| \frac{k_{h}I_{u}^{'}(\xi)+k_{u}\delta I_{h}^{'}(\xi)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}\Bigg|<C, \quad\xi\in {{\Bbb R}} . \end{eqnarray*}$

证首先,将系统(2.1)第二个和第三个方程分别乘以 $k_{h}$ 和 $k_{u}\delta$ ,然后相加,可得

$\begin{eqnarray} c(k_{h}I_{u}^{'}(\xi)+k_{u}\delta I_{h}^{'}(\xi)) & = &d[J*(k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi))-(k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi))]\\&& +[(1-\mu)\beta k_{h}S(\xi)+\mu\beta\delta k_{u}S(\xi)-k_{u}k_{h}](I_{u}(\xi)+\delta I_{h}(\xi)). \end{eqnarray}$

(3.1)

令 $w(\xi) = \frac{k_{h}I_{u}^{'}(\xi)+k_{u}\delta I_{h}^{'}(\xi)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}$ ,则

$\begin{eqnarray*} w(\xi)& = &\frac{d}{c}\int_{-\infty}^{+\infty}J(y)\frac{k_{h}I_{u}(\xi-y)+k_{u}\delta I_{h}(\xi-y)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}{\rm d}y-\frac{d}{c}\\ &&+[(1-\mu)\beta k_{h}S(\xi)+\mu\beta\delta k_{u}S(\xi)-k_{u}k_{h}]\frac{I_{u}(\xi)+\delta I_{h}(\xi)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}\\ & = &\frac{d}{c}\int_{-\infty}^{+\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}w(s){\rm d}s}{\rm d}y-\frac{d}{c}+[(1-\mu)\beta k_{h}S(\xi)\\ &&+\mu\beta\delta k_{u}S(\xi)-k_{u}k_{h}]\frac{I_{u}(\xi)+\delta I_{h}(\xi)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}\\ &\geq&\frac{d}{c}\int_{-\infty}^{+\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}w(s){\rm d}s}{\rm d}y-\frac{d}{c}-\frac{k_{u}k_{h}(I_{u}(\xi)+\delta I_{h}(\xi))}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}\\ &\geq&\frac{d}{c}\int_{-\infty}^{+\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}w(s){\rm d}s}{\rm d}y-\frac{d}{c}-k_{h}, \end{eqnarray*}$

最后一个不等式由

$\begin{eqnarray*} \frac{1}{k_{h}}<\frac{I_{u}(\xi)+\delta I_{h}(\xi)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}<\frac{1}{k_{u}}, \forall \xi\in {{\Bbb R}} \end{eqnarray*}$

可得.令 $\varpi = \frac{d}{c}$ , $\rho = \frac{d}{c}+k_{h}$ 且 $K(\xi) = \exp\{\rho\xi+\int_{0}^{\xi}w(s){\rm d}s\}$ .通过简单计算,得到

$\begin{equation} K^{'}(\xi) = (\rho+w(\xi))K(\xi)\geq\varpi\int_{-\infty}^{+\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}w(s){\rm d}s}{\rm d}yK(\xi), \end{equation}$

(3.2)

这表明 $K(\xi)$ 是非增的,且 $K(-\infty) = 0$ .令 $T_{0}>0$ ,使得 $0<2T_{0}<T$ ,其中 $T$ 表示 ${\rm supp} J$ 的半径.在(3.2)式两边从 $-\infty$ 到 $\xi$ 上积分,可得

$\begin{eqnarray*} K(\xi)& = &\varpi\int_{-\infty}^{\xi}\int_{-\infty}^{+\infty}J(y){\rm e}^{\int_{x}^{x-y}w(s){\rm d}s}{\rm d}yK(x){\rm d}x\\ & = &\varpi\int_{-\infty}^{+\infty}J(y){\rm e}^{\rho y}\int_{-\infty}^{\xi}K(x){\rm d}x{\rm d}y\\ &\geq&\varpi T_{0}\int_{-\infty}^{+\infty}J(y){\rm e}^{\rho y}K(\xi-T_{0}-y){\rm d}y \end{eqnarray*}$

且

$\begin{equation} \int_{-\infty}^{0}J(y){\rm e}^{\rho y}\frac{K(\xi-T_{0}-y)}{K(\xi)}{\rm d}y\leq (\varpi T_{0})^{-1}. \end{equation}$

(3.3)

在(3.2)式两边从 $\xi-T_{0}$ 到 $\xi$ 上积分,可得

$\begin{eqnarray*} K(\xi)-K(\xi-T_{0})&\geq&\varpi\int_{\xi-T_{0}}^{\xi}\int_{-\infty}^{+\infty}J(y){\rm e}^{\int_{x}^{x-y}w(s){\rm d}s}{\rm d}yK(x){\rm d}x\\ & = &\varpi\int_{-\infty}^{+\infty}J(y){\rm e}^{\rho y}\int_{\xi-T_{0}}^{\xi}K(x){\rm d}x{\rm d}y\\ &\geq&\varpi T_{0}\int_{-\infty}^{-2T_{0}}J(y){\rm e}^{\rho y}{\rm d}yK(\xi+T_{0}), \end{eqnarray*}$

定义

$\begin{eqnarray*} \theta_{0}: = \frac{1}{\varpi T_{0}\int_{-\infty}^{-2T_{0}}J(y){\rm e}^{\rho y}{\rm d}y}, \end{eqnarray*}$

可得

$\begin{equation} K(\xi+T_{0})\leq \theta_{0}K(\xi), \xi\in {{\Bbb R}} . \end{equation}$

(3.4)

又

$\begin{eqnarray*} \int_{-\infty}^{+\infty}J(y)\frac{k_{h}I_{u}(\xi-y)+k_{u}\delta I_{h}(\xi-y)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}{\rm d}y = \int_{-\infty}^{+\infty}J(y){\rm e}^{\rho y}\frac{K(\xi-y)}{K(\xi)}{\rm d}y, \end{eqnarray*}$

从而,由(3.3)和(3.4)式,易得

$\begin{eqnarray*} &&\int_{-\infty}^{+\infty}J(y)\frac{k_{h}I_{u}(\xi-y)+k_{u}\delta I_{h}(\xi-y)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}{\rm d}y \\& = &\int_{-\infty}^{0}J(y){\rm e}^{\rho y}\frac{K(\xi-y)}{K(\xi)}{\rm d}y+\int_{0}^{+\infty}J(y){\rm e}^{\rho y}\frac{K(\xi-y)}{K(\xi)}{\rm d}y\\ &\leq&\int_{-\infty}^{0}J(y){\rm e}^{\rho y}\frac{K(\xi-y)}{K(\xi)}{\rm d}y+\int_{0}^{+\infty}J(y){\rm e}^{\rho y}{\rm d}y\\ &\leq&\theta_{0}\int_{-\infty}^{0}J(y){\rm e}^{\rho y}\frac{K(\xi-T_{0}-y)}{K(\xi)}{\rm d}y+\int_{0}^{+\infty}J(y){\rm e}^{\rho y}{\rm d}y\\ &\leq&\frac{\theta_{0}}{\varpi T_{0}}+\int_{0}^{+\infty}J(y){\rm e}^{\rho y}{\rm d}y. \end{eqnarray*}$

又因为

$\begin{eqnarray*} |w(\xi)|&\leq&\frac{d}{c}\int_{-\infty}^{+\infty}J(y)\frac{k_{h}I_{u}(\xi-y)+k_{u}\delta I_{h}(\xi-y)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}{\rm d}y+\frac{d}{c}\\ &&+[(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta k_{u}S^{0}-k_{u}k_{h}]\frac{I_{u}(\xi)+\delta I_{h}(\xi)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}\\ &\leq&\frac{d}{c}\int_{-\infty}^{+\infty}J(y)\frac{k_{h}I_{u}(\xi-y)+k_{u}\delta I_{h}(\xi-y)}{k_{h}I_{u}(\xi)+k_{u}\delta I_{h}(\xi)}{\rm d}y+\frac{d}{c}\\ &&+\frac{[(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta k_{u}S^{0}-k_{u}k_{h}]}{k_{u}}, \end{eqnarray*}$

可证得最终结论.证毕.

引理3.2 令 $\{(S_{k}, I_{uk}, I_{hk}, c_{k})\}$ 是系统(2.1)的波速为 $\{c_{k}\}$ 的行波解,其中 $c_{k}\in (c^{*}, c^{*}+1)$ .若存在序列 $\{\eta_{n}\}$ 使得当 $k\rightarrow +\infty$ 时, $I_{uk}(\eta_{k})\rightarrow\infty$ 或 $I_{hk}(\eta_{k})\rightarrow\infty$ ,则当 $k\rightarrow+\infty$ 时, $S_{k}(\eta_{k})\rightarrow0$ .

证假设存在序列 $\{\eta_{k}\}$ ,为简便仍表示为 $\{\eta_{k}\}$ ,使得对一切 $k\in {\Bbb N}$ 和某个正常数 $\epsilon$ , $S_{k}(\eta_{k})\geq\epsilon$ ,且当 $k\rightarrow+\infty$ 时, $I_{uk}(\eta_{k})\rightarrow +\infty$ .注意 $0<S_{k}<S^{0}$ , $I_{uk}+I_{hk}>0$ ,由(2.1)式的第一个方程可得 $S_{k}^{'}(\xi)\leq\frac{2d_{s}S^{0}}{c^{*}}$ .则存在正常数 $\nu$ 使得对所有的 $\xi\in[\eta_{k}-\nu, \eta_{k}]$ 和 $k\in {\Bbb N}$ , $S_{k}(\xi)\geq \frac{\epsilon}{2}$ .再由引理3.1,存在一个和 $k$ 无关的正常数 $C_{1}$ 使得 $\Big| \frac{k_{h}I_{uk}^{'}(\xi)+k_{u}\delta I_{hk}^{'}(\xi)}{k_{h}I_{uk}(\xi)+k_{u}\delta I_{hk}(\xi)}\Big|<C_{1}$ .从而,对任意 $k$ 和 $\xi\in[\eta_{k}-\nu, \eta_{k}]$ ,有

$\begin{eqnarray*} \frac{k_{h}I_{uk}^{'}(\eta_{k})+k_{u}\delta I_{hk}^{'}(\eta_{k})}{k_{h}I_{uk}(\xi)+k_{u}\delta I_{hk}(\xi)} = \exp\Bigg\{\int_{\xi}^{\eta_{k}} \frac{k_{h}I_{uk}^{'}(s)+k_{u}\delta I_{hk}^{'}(s)}{k_{h}I_{uk}(s)+k_{u}\delta I_{hk}(s)}{\rm d}s\Bigg\}\leq {\rm e}^{C_{1}\nu}. \end{eqnarray*}$

且由 $k\rightarrow+\infty$ 时, $I_{uk}(\eta_{k})\rightarrow+\infty$ ,不难得到当 $k\rightarrow +\infty$ 时,有

$\begin{eqnarray*} \min\limits_{\xi\in[\eta_{k}-\nu, \eta_{k}]}(k_{h}I_{uk}(\xi)+k_{u}\delta I_{hk}(\xi))\geq {\rm e}^{-\nu C_{1}}(k_{h}I_{uk}(\eta_{k})+k_{u}\delta I_{hk}(\eta_{k})) \rightarrow+\infty. \end{eqnarray*}$

另一方面,由系统(1.2)的第一个方程,有

$\begin{eqnarray*} \max\limits_{[\eta_{k}-\nu, \xi]} S_{k}^{'}(\xi)&\leq&\frac{2d_{s}S^{0}}{c_{k}}-\frac{\beta\epsilon}{2}(I_{uk}(\xi)+\delta I_{hk}(\xi))\\ &\leq&\frac{2d_{s}S^{0}}{c_{k}}-\frac{\beta\epsilon}{2k_{h}}(k_{h}I_{uk}(\xi)+k_{u}\delta I_{hk}(\xi))\\ &\leq&\frac{2d_{s}S^{0}}{c_{k}}-\frac{\beta\epsilon}{2k_{h}}\min\limits_{\xi\in[\eta_{k}-\nu, \eta_{k}]}(k_{h}I_{uk}(\xi)+k_{u}\delta I_{hk}(\xi))\rightarrow-\infty, \end{eqnarray*}$

这与 $0<S_{k}(\xi)<S^{0}$ 矛盾.证毕.

下述引理是文献[29,命题3.7]的特殊情形.

引理3.3 假设 $h$ 是常数.令 $r(\xi)$ 是满足

$\begin{eqnarray*} cr(\xi) = d\int_{-\infty}^{\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}r(s){\rm d}s}{\rm d}y+h, \xi\in {{\Bbb R}} \end{eqnarray*}$

的可测非常数函数,则 $r(\pm\infty): = \lim\limits_{\xi\rightarrow\pm\infty}r(\xi)$ 存在且是以下特征方程

$\begin{eqnarray*} cr(\pm\infty) = d\int_{-\infty}^{\infty}J(y){\rm e}^{-r(\pm\infty)y}{\rm d}y+h, r(-\infty)<r(+\infty) \end{eqnarray*}$

的正根.

由引理3.3,可得到一个重要引理.

引理3.4 设 $c>0$ 且 $H(\cdot)$ 是在 ${{\Bbb R}}$ 上满足 $h_{1}<\liminf\limits_{\xi\rightarrow\pm\infty} H(\xi)\leq\limsup\limits_{\xi\rightarrow\pm\infty} H(\xi)<h_{2}$ 的连续函数,其中 $h_{1}$ , $h_{2}$ 都是常数.令 $r(\xi)$ 是满足

$\begin{eqnarray*} cr(\xi) = d\int_{-\infty}^{\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}r(s){\rm d}s}{\rm d}y+H(\xi), \xi\in {{\Bbb R}} \end{eqnarray*}$

的可测函数.则

$r_{1}(-\infty)<\liminf\limits_{\xi\rightarrow-\infty}r(\xi)\leq\limsup\limits_{\xi\rightarrow-\infty}r(\xi)<r_{2}(-\infty),$

$r_{2}(+\infty)<\liminf\limits_{\xi\rightarrow+\infty}r(\xi)\leq\limsup\limits_{\xi\rightarrow+\infty}r(\xi)<r_{1}(+\infty),$

其中 $r_{i}(\pm\infty)$ 满足以下特征方程

$cr_{i}(\pm\infty) = d\int_{-\infty}^{\infty}J(y){\rm e}^{-r_{i}(\pm\infty)y}{\rm d}y+h_{i}, \; r_{i}(-\infty)<r_{i}(+\infty) \ , \ i = 1, 2.$

证由引理3.3,存在 $r_{i}(\pm\infty)$ 使得

$\begin{eqnarray*} cr_{i}(\pm\infty) = d\int_{-\infty}^{\infty}J(y){\rm e}^{-r_{i}(\pm\infty)y}{\rm d}y+h_{i}, i = 1, 2. \end{eqnarray*}$

因 $h_1<h_2$ ,进一步可得 $r_{1}(-\infty)<r_{2}(-\infty)<r_{2}(+\infty)<r_{1}(+\infty)$ .对一切 $h_{1}<h<h_{2}$ ,令 $F(x): = d\int_{-\infty}^{\infty}J(y){\rm e}^{-xy}{\rm d}y- cx+h$ , $x\in {{\Bbb R}}$ .存在 $r(\pm\infty)$ 满足 $F(r(\pm\infty)) = 0$ .注意

$\begin{eqnarray*} F(0) = 1+h, \frac{\partial F(x)}{\partial x}\bigg|_{x = 0} = -c<0, \frac{\partial^{2} F(x)}{\partial x^{2}} = d\int_{-\infty}^{\infty}J(y)y^{2}{\rm e}^{-xy}{\rm d}y>0, \forall x\in {{\Bbb R}} , \end{eqnarray*}$

可得

$\begin{equation} r_{1}(-\infty)<r(-\infty)<r_{2}(-\infty), \quad r_{2}(+\infty)<r(+\infty)<r_{1}(+\infty). \end{equation}$

(3.5)

又 $h_{1}<\liminf\limits_{\xi\rightarrow+\infty} H(\xi)\leq\limsup\limits_{\xi\rightarrow+\infty} H(\xi)<h_{2}$ ,从而对所有满足 $\liminf\limits_{\xi\rightarrow+\infty}H(\xi)\leq\widetilde{h}\leq\limsup\limits_{\xi\rightarrow+\infty}H(\xi)$ 的常数 $\widetilde{h}$ ,存在序列 $\{\zeta_{n}\}$ ,满足当 $n\rightarrow+\infty$ 时 $\zeta_{n}\rightarrow+\infty$ ,使得对所有的 $n\in {\Bbb N}$ , $H(\zeta_{n}) = \widetilde{h}$ .进一步地

$\begin{eqnarray*} cr(\zeta_{n}) = d\int_{-\infty}^{\infty}J(y){\rm e}^{\int_{\zeta_{n}}^{\zeta_{n}-y}r(s){\rm d}s}{\rm d}y+\widetilde{h}, \forall n\in {\Bbb N}. \end{eqnarray*}$

从而由引理3.3,当 $n\rightarrow+\infty$ , $r(\zeta_{n})\rightarrow \widetilde{r}(+\infty)$ ,其中 $\widetilde{r}(+\infty)$ 满足

$\begin{eqnarray*} c\widetilde{r}(+\infty) = d\int_{-\infty}^{\infty}J(y){\rm e}^{-\widetilde{r}(+\infty)y}{\rm d}y+\widetilde{h}. \end{eqnarray*}$

注意 $h_{1}<\widetilde{h}<h_{2}$ ,由(3.5)式,有

$\begin{equation} r_{2}(+\infty)<\widetilde{r}(+\infty)<r_{1}(+\infty). \end{equation}$

(3.6)

下面证明

$\begin{equation} r_{2}(+\infty)<\liminf\limits_{\xi\rightarrow+\infty}r(\xi)\leq\limsup\limits_{\xi\rightarrow+\infty}r(\xi)<r_{1}(+\infty). \end{equation}$

(3.7)

对任意序列 $\{\eta_n\}$ 满足当 $n\rightarrow\infty$ , $\eta_n\rightarrow\infty$ ,有

$\begin{equation} cr(\eta_{n}) = d\int_{-\infty}^{\infty}J(y){\rm e}^{\int_{\eta_{n}}^{\eta_{n}-y}r(s){\rm d}s}{\rm d}y+H(\eta_{n}), \xi\in {{\Bbb R}} . \end{equation}$

(3.8)

对(3.8)式,讨论可以分为以下两种情况.

情形1 $H(\eta_n)$ 收敛.

若存在一个常数 $h_3$ 使得当 $n\rightarrow\infty$ , $H(\eta_n)\rightarrow h_3$ .则

$\begin{equation} h_1<h_3<h_2. \end{equation}$

(3.9)

由引理3.3, (3.6)和(3.9)式,可得

$\lim\limits_{n\rightarrow+\infty}r(\eta_{n}) = \overline{r}_3,$

其中 $\overline{r}_3 = {\overline{r}(+\infty)}$ 且 $\overline{r}(\xi)$ 满足

$c\overline{r}(\xi) = d\int_{-\infty}^{\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}\overline{r}(s){\rm d}s}{\rm d}y+h_{3}, \xi\in {{\Bbb R}} .$

结合(3.5)和(3.8)式,不难得到

$r_{2}(+\infty)<\overline{r}_3<r_{1}(+\infty).$

情形2 $H(\eta_n)$ 不收敛.对序列 $\{\eta_n\}$ ,则有

$h_{1}<\liminf\limits_{n\rightarrow \infty} H(\eta_n)\leq\limsup\limits_{n\rightarrow \infty} H(\eta_n)<h_{2}.$

选取 $H(\eta_n)$ 的任意收敛子列 $H(\eta_{n_k})$ 且由情形1,可得

$r_{2}(+\infty)<\lim\limits_{\eta_{n_k}\rightarrow+\infty}{r(\eta_{n_k})}<r_{1}(+\infty).$

即

$r_{2}(+\infty)<\liminf\limits_{\eta_n\rightarrow+\infty}{r(\eta_n)}\leq\limsup\limits_{\eta_n\rightarrow+\infty}{r(\eta_n)}<r_{1}(+\infty).$

根据情形1和情形2, (3.7)式成立.

同理可得

$r_{1}(-\infty)<\liminf\limits_{\xi\rightarrow-\infty}{r(\xi)}\leq\limsup\limits_{\xi\rightarrow-\infty}{r(\xi)}<r_{2}(-\infty).$

证毕.

现在用渐近法来证明结论.不失一般性,可以选取一列严格递减序列 $\{c_{k}\}$ ,使得对所有的 $k$ , $c_{k}\in (c^{*}, c^{*}+1]$ 且当 $k\rightarrow+\infty$ 时, $c_{k}\rightarrow c^{*}$ .对所有的 $k$ ,令 $(S_{k}(\xi), I_{uk}(\xi), I_{hk}(\xi), c_{k})$ 是(1.1)式的一个正解.再令 ${I_{k}(\xi)} = k_{h}I_{uk}(\xi)+k_{u}\delta I_{hk}(\xi)$ ,则可得以下引理.

引理3.5 序列 ${I_{k}(\xi)}$ 在 ${{\Bbb R}}$ 上一致有界.

证假设当 $k\rightarrow+\infty$ 时,存在一个序列 $\{\xi_{k}\}$ 使得 $I_{k}(\xi_{k})\rightarrow+\infty$ .则由引理3.2可得当 $k\rightarrow+\infty$ 时, $S_{k}(\xi_{k})\rightarrow 0$ .又 $I_{k}(\pm\infty) = 0$ 且 $I_{k}(\xi)>0$ ,不失一般性,可设对一切 $k\in {\Bbb N}$ $I_{k}(\xi_{k}) = \max\limits_{\xi\in {{\Bbb R}} }I_{k}(\xi)$ .则由(3.1)式可得

$\begin{eqnarray*} 0 = c_{k}I^{'}_{k}(\xi_{k}))& = &d\int_{-\infty}^{+\infty}J(y)(I_{k}(\xi_{k}-y)-I_{k}(\xi_{k})){\rm d}y+[(1-\mu)\beta k_{h}S_{k}(\xi_{k})\\ &&+\mu\beta\delta k_{u}S_{k}(\xi_{k})-k_{u}k_{h}][I_{uk}(\xi_{k})+\delta I_{hk}(\xi_{k})]\\ &\leq&[(1-\mu)\beta k_{h}S_{k}(\xi_{k})+\mu\beta\delta k_{u}S_{k}(\xi_{k})-k_{u}k_{h}]I_{k}(\xi_{k}), \end{eqnarray*}$

当 $k$ 充分大时,上式不成立.证毕.

定理3.1 若 $R_{0}>1$ 且 $c = c^{*}$ ,则对所有的 $\xi \in {{\Bbb R}}$ ,系统(2.1)存在行波解 $(S_{*}(\xi), I_{u*}(\xi),$ $I_{h*}(\xi))$ ,满足 $S_{*}(-\infty) = S^{0}$ , $S_{*}(+\infty) = S^{1}<S^{0}$ , $I_{u*}(\pm\infty) = I_{h*}(\pm\infty) = 0$ 且 $0<S_{*}(\xi)<S^{0}$ , $I_{u*}(\xi)+I_{h*}(\xi)>0$ .

证注意到 $S_{k}(-\infty) = S^{0}$ , $I_{uk}(\pm\infty) = I_{hk}(\pm\infty) = 0$ 且 $I_{uk}(\xi)$ , $I_{hk}(\xi)>0$ .在 ${{\Bbb R}}$ 上积分(3.1)式,得

$\begin{eqnarray*} 0& = &c_{k}\int_{-\infty}^{+\infty}I_{k}^{'}(\xi){\rm d}\xi \\& = &d\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}[J(y)I_{k}(\xi-y)-I_{k}(\xi)]{\rm d}y{\rm d}\xi+\int_{-\infty}^{+\infty}[(1-\mu)\beta k_{h}S_{k}(\xi)\\ &&+\mu\beta\delta k_{u}S_{k}(\xi)-k_{u}k_{h}][I_{uk}(\xi)+\delta I_{hk}(\xi)]{\rm d}\xi\\ & = &\int_{-\infty}^{+\infty}[(1-\mu)\beta k_{h}S_{k}(\xi)+\mu\beta\delta k_{u}S_{k}(\xi)-k_{u}k_{h}][I_{uk}(\xi)+\delta I_{hk}(\xi)]{\rm d}\xi, \end{eqnarray*}$

这表明 $\inf\limits_{\xi\in {{\Bbb R}} }S_{k}(\xi)<\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ 对所有的 $k$ 成立.从而,存在序列 $\{\eta_{k}\}$ 使得 $S_{k}(\xi_{k}) = \frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ .

将 $\eta_{k}$ 平移到原点,不失一般性,假设 $S_{k}(0) = \frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ .由 $S_{k}(\xi)$ , $I_{uk}(\xi)$ 和 $I_{uk}(\xi)$ 在 ${{\Bbb R}}$ 上是一致有界的,从而 $\parallel S_{k}\parallel_{C^{2}}$ , $\parallel I_{uk}\parallel_{C^{2}}$ 且 $\parallel I_{hk}\parallel_{C^{2}}$ 都是一致有界的.利用Arzela-Ascoli定理,存在某个序列当 $k\rightarrow+\infty$ , $S_{k}\rightarrow S_{*}$ , $I_{uk}\rightarrow I_{u*}$ 且 $I_{hk}\rightarrow I_{h*}$ 在 $C_{loc}^{1}({{\Bbb R}} )$ 上成立,其中 $(S_{*}(\xi), I_{u*}(\xi), I_{h*}(\xi))$ 满足

$\begin{equation} \left\{\begin{array}{rl} { } cS_{*}^{'}(\xi) = &{ } d_{s}\int_{-\infty}^{+\infty}J(y)(S_{*}(\xi-y)-S_{*}(\xi)){\rm d}y-\beta(I_{u*}(\xi)+\delta I_{h*}(\xi))S_{*}(\xi), \\ { } cI_{u*}^{'}(\xi) = &{ } d\int_{-\infty}^{+\infty}J(y)(I_{u*}(\xi-y)-I_{u*}(\xi)){\rm d}y\\ &+(1-\mu)\beta(I_{u*}(\xi)+\delta I_{h*}(\xi))S_{*}(\xi)-k_{u}I_{u*}(\xi), \\ { } cI_{h*}^{'}(\xi) = &{ } d\int_{-\infty}^{+\infty}J(y)(I_{h*}(\xi-y)-I_{h*}(\xi)){\rm d}y\\ &{ } +\mu\beta(I_{u*}(\xi)+\delta I_{h*}(\xi))S_{*}(\xi)-k_{h}I_{h*}(\xi). \end{array}\right. \end{equation}$

(3.10)

且 $0<S_{*}(\xi)<S^{0}$ , $0<I_{*}<M_{*}$ , $S_{*}(0) = \frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ ,其中 $M^{*}$ 是 ${{\Bbb R}}$ 上的正常数, $I_{*} = k_{h}I_{u*}+k_{u}\delta I_{h*}$ .

同理,由引理2.6中的讨论,可得

$\begin{eqnarray*} \int_{-\infty}^{+\infty}(I_{u*}(\xi)+\delta I_{h*}(\xi))S_{*}(\xi){\rm d}\xi<+\infty, \int_{-\infty}^{+\infty}I_{u*}(\xi){\rm d}\xi<+\infty, \int_{-\infty}^{+\infty}I_{h*}(\xi){\rm d}\xi<+\infty \end{eqnarray*}$

使得 $I_{u*}(\pm\infty) = I_{h*}(\pm\infty) = 0$ .

接下来,分两步证明.

步骤1 $S_{*}(\xi)>0$ , $I_{*}(\xi)>0$ .假设存在 $\eta_{0}$ 使得 $S_{*}(\eta_{0}) = 0$ ,则 $S_{*}'(\eta_{0}) = 0$ .由(3.10)式的第一个方程, $d_{s}\int_{-\infty}^{+\infty}J(y)S_{*}(\eta_{0}-y){\rm d}y = 0$ .则对任意 $y\in {{\Bbb R}}$ , $S_{*}(y) = 0$ .而 $S_{*}(0) = \frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ ,矛盾.

现在证明 $I_{*}(\xi)>0$ .假设 $I_{*}(\xi_{1}) = 0$ ,则对任意 $\xi\in {{\Bbb R}}$ ,由(3.10)式可得 $I_{*}(\xi) = 0$ .从而由(3.10)式的第一个方程可得

$\begin{equation} c^{*}S_{*}^{'}(\xi) = d_{s}\int_{-\infty}^{+\infty}J(y)(S_{*}(\xi-y)-S_{*}(y)){\rm d}y. \end{equation}$

(3.11)

利用(3.11)式和文献[29,命题3.6],存在正常数 $C_{1}$ 和 $C_{2}$ 使得

$\begin{eqnarray*} S(\xi) = C_{1}{\rm e}^{\lambda_{1*}}+C_{2}{\rm e}^{\lambda_{2*}\xi}, \end{eqnarray*}$

其中 $\lambda_{1*} < \lambda_{2*}$ 是以下方程的根

$\begin{eqnarray*} c^{*}\lambda = d_{s}\int_{-\infty}^{\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d_{s}. \end{eqnarray*}$

注意 $0<S_{*}(\xi)<S^{0}$ ,则对某个常数 $C_{1}\in[0, S^{0}]$ , $S_{*}(\xi)\equiv C_{1}$ .再由 $S_{*}(0) = \frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ ,可得对任意 $\xi\in {{\Bbb R}}$ , $S_{*}(\xi) = \frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ .

由 $0<S_{k}(-\infty) = S^{0}>\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ ,可以选取 $\xi_{k}<0$ 使得

$\begin{equation} S_{k}(\xi)>\frac{\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}+S^{0}}{2}, S_{k}(\xi_{k}) = \frac{\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}+S^{0}}{2} \end{equation}$

(3.12)

对所有的 $\xi< \xi_{k}$ 和 $k\in{\Bbb N}$ 成立.

此外, $\{\xi_{k}\}$ 是下有界的.反之,假设 $k\rightarrow+\infty$ 时, $\xi_{k}\rightarrow-\infty$ .定义

$\begin{eqnarray*} \varphi_{k}(\xi): = S_{k}(\xi+\xi_{k}), \psi_{k}(\xi): = \frac{I_{k}(\xi+\xi_{k})}{I_{k}(\xi_{k})}, \psi_{ik}(\xi): = \frac{I_{ik}(\xi+\xi_{k})}{I_{k}(\xi_{k})}, \xi\in {{\Bbb R}} , i = u, h. \end{eqnarray*}$

则 $(\varphi_{k}(\xi), \psi_{k}(\xi))$ 满足

$\begin{eqnarray*} \left\{\begin{array}{rl} c_{k}\varphi_{k}'(\xi) = &{ } d_{1}\int_{-\infty}^{+\infty}J(y)(\varphi_{k}(\xi-y)-\varphi_{k}(\xi)){\rm d}y-\beta \varphi_{k}(\xi)(I_{uk}(\xi_{k}+\xi)+\delta I_{hk}(\xi_{k}+\xi)), \\ c_{k}\psi_{k}^{'}(\xi) = &{ } d\int_{-\infty}^{+\infty}[J(y)\psi_{k}(\xi-y)-\psi_{k}(\xi)]{\rm d}y\\ &+[(1-\mu)\beta k_{h}\varphi_{k}(\xi)+\mu\beta\delta k_{u}\varphi_{k}(\xi) -k_{u}k_{h}](\psi_{uk}(\xi)+\delta \psi_{hk}(\xi)). \end{array}\right. \end{eqnarray*}$

注意 $\frac{I_{k}'(\xi)}{I_{k}(\xi)}$ 在 ${{\Bbb R}}$ 上是一致有界的且由引理3.1,有

$\begin{eqnarray*} \psi_{k}(\xi) = \frac{I_{k}(\xi_{k}+\xi)}{I_{k}(\xi_{k})} = \exp\Bigg\{\int_{\xi_{k}}^{\xi_{k}+\xi}\frac{I_{k}'(x)}{I_{k}(x)}{\rm d}x\Bigg\}, \frac{ \psi_{k}(\xi)}{k_{h}} <\psi_{uk}(\xi)+\delta \psi_{hk}(\xi)<\frac{\psi_{k}(\xi)}{k_{u}}. \end{eqnarray*}$

则 $\psi_{k}$ , $\psi_{uk}$ 和 $\psi_{hk}$ 在 ${{\Bbb R}}$ 上局部一致有界的.

另一方面,由 $0<\varphi_{k}(\xi)<S^{0}$ ,假设对 $i = u, k$ ,当 $k\rightarrow+\infty$ 时, $\varphi_{k}\rightarrow\varphi_{\infty}$ , $\psi_{k}\rightarrow\psi_{\infty}$ 且 $\psi_{ik}\rightarrow\psi_{i\infty}$ 在 $C_{loc}^{1}$ 上成立.则由 $\lim\limits_{k\rightarrow+\infty}I_{k}(\xi_{k}) = 0$ ,可得 $(\varphi_{\infty}, \psi_{\infty})$ 满足

$\begin{equation} \left\{\begin{array}{rl} c^{*}\varphi_{\infty}'(\xi) = &{ } d_{1}\int_{-\infty}^{+\infty}J(y)(\varphi_{\infty}(\xi-y)-\varphi_{\infty}(\xi)){\rm d}y, \\ c^{*}\psi_{\infty}^{'}(\xi) = &{ } d\int_{-\infty}^{+\infty}[J(y)\psi_{\infty}(\xi-y)-\psi_{\infty}(\xi)]{\rm d}y\\ &+[(1-\mu)\beta k_{h}\varphi_{\infty}(\xi) +\mu\beta\delta k_{u}\varphi_{\infty}(\xi)-k_{u}k_{h}](\psi_{u\infty}(\xi)+\delta \psi_{h\infty}(\xi)). \end{array}\right. \end{equation}$

(3.13)

同理,由文献[29,命题3.6],由(3.13)式的第一个方程可得

$\varphi_{\infty}(\xi)\equiv C_{*} = \frac{\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}+S^{0}}{2}.$

另一方面,令 $v = \frac{\psi_{\infty}'}{\psi_{\infty}}$ ,由(3.13)式的第二个方程可得

$\begin{eqnarray*} c^{*}v(\xi)& = &d\int_{-\infty}^{\infty}J(y){\rm e}^{\int_{\xi}^{\xi-y}v(s){\rm d}s}{\rm d}y-d\\ &&+[(1-\mu)\beta k_{h}C_{*}+\mu\beta\delta k_{u}C_{*} -k_{u}k_{h}]\frac{\psi_{u\infty}(\xi)+\delta \psi_{h\infty}(\xi)}{k_{h}\psi_{u\infty}(\xi)+k_{u}\delta\psi_{h\infty}(\xi)}, \quad\xi\in {{\Bbb R}} . \end{eqnarray*}$

从而,由引理3.4,对充分大的 $M>0$ 成立

$\begin{equation} \kappa_{1}\leq v(\xi)\leq \kappa_{2}, \xi<-M, \\ \kappa_{3}\leq v(\xi)\leq \kappa_{4}, \xi>M, \end{equation}$

(3.14)

其中 $\kappa_{1}$ , $\kappa_{4}$ 是以下方程

$\begin{eqnarray*} c^{*}\lambda = d\int_{-\infty}^{\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d+\frac{[(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta kS^{0} -k_{u}k_{h}]}{2k_{h}} \end{eqnarray*}$

的两个正根. $\kappa_{2}$ , $\kappa_{3}$ 是以下方程

$\begin{eqnarray*} c^{*}\lambda = d\int_{-\infty}^{\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d+\frac{[(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta kS^{0} -k_{u}k_{h}]}{2k_{u}}. \end{eqnarray*}$

的两个正根.再由文献[29,命题3.6],易得存在正常数 $C_{i}(i = 2, 3, 4, 5)$ ,使得

$\begin{equation} \begin{array}{l} C_{2}{\rm e}^{\kappa_{1}\xi}\leq \psi_{\infty}(\xi)\leq C_{3}{\rm e}^{\kappa_{2}\xi}, \xi<-M, \\ C_{4}{\rm e}^{\kappa_{3}\xi}\leq \psi_{\infty}(\xi)\leq C_{5}{\rm e}^{\kappa_{4}\xi}, \xi>M. \end{array} \end{equation}$

(3.15)

下一步,令 $\widetilde{\nu_{k}}(\xi) = \frac{I_{k}'(\xi)}{I_{k}(\xi)}$ .注意 $S_{k}(\pm\infty)$ 存在,利用引理3.4

$\begin{equation} \begin{array}{l} \kappa_{1k}'\leq \widetilde{\nu_{k}}(\xi)\leq \kappa_{2k}', \xi<-M_{k}, \\ \kappa_{3k}'\leq \widetilde{\nu_{k}}(\xi)\leq \kappa_{4k}', \xi>M_{k} \end{array} \end{equation}$

(3.16)

对充分大的 $M_{k}>0$ 成立.其中 $\kappa_{1k}' < \kappa_{4k}'$ 是以下方程

$\begin{eqnarray*} c_{k}\lambda = d\int_{-\infty}^{\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d+\frac{[(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta kS^{0} -k_{u}k_{h}]}{k_{h}} \end{eqnarray*}$

的两个正根,且 $\kappa_{2k}'<\kappa_{3k}'$ 是以下方程

$\begin{eqnarray*} c_{k}\lambda = d\int_{-\infty}^{\infty}J(y){\rm e}^{-\lambda y}{\rm d}y-d+\frac{[(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta kS^{0} -k_{u}k_{h}]}{k_{u}} \end{eqnarray*}$

的两个正根.令 $k\rightarrow+\infty$ ,则对 $i = 1, 2, 3, 4$ , $\kappa_{ik}'\rightarrow\kappa_{i}'$ .另一方面

$\begin{eqnarray*} \varphi_{k}(\xi) = \frac{I_{k}(\xi_{k}+\xi)}{I(\xi_{k})} = {\rm e}^{\int_{\xi_{k}}^{\xi_{k}+\xi}\frac{I_{k}'(s)}{I_{k}(s)}{\rm d}s} = {\rm e}^{\int_{\xi_{k}}^{\xi_{k}+\xi}\widetilde{\nu_{k}}(s){\rm d}s}, \end{eqnarray*}$

从而,由(3.16)式可得

$\begin{eqnarray*} {\rm e}^{\kappa_{1}'\xi}<\varphi_{\infty}(\xi)<{\rm e}^{\kappa_{2}'\xi}. \end{eqnarray*}$

由于 $\kappa_{2}^{'}<\kappa_{3}$ 且 $\kappa_{1}'<\kappa_{4}$ ,这与(3.15)式的第二个不等式矛盾.从而存在某个常数使得对所有的 $k$ , $\xi_{k}>-\overline{C}$ .则

$\begin{eqnarray*} S_{k}(-\overline{C})>\frac{\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}+S^{0}}{2}, \end{eqnarray*}$

这表明 $S_{*}(-\overline{C})>\frac{\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}+S^{0}}{2}$ .又 $(1-\mu)\beta k_{h}S^{0}+\mu\beta\delta k_{u}S^{0}>k_{u}k_{h}$ ,易得 $S_{*}(-\overline{C})>\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ .注意到 $S_{*}(\xi)\equiv\frac{k_{u}k_{h}}{(1-\mu)\beta k_{h}+\mu\beta\delta k_{u}}$ ,矛盾.即 $I_{*}(\xi)>0$ 在 ${{\Bbb R}}$ 上成立.

步骤2 $S_{*}(\xi)<S^{0}$ , $\forall\xi\in{{\Bbb R}}$ , $S_{*}(-\infty) = S^{0}$ 且 $S_{*}(+\infty) = S^{1}<S^{0}$ .

首先,假设存在 $\widehat{\xi}>0$ 使得 $S_{*}(\widehat{\xi}) = S^{0}$ ,则 $S_{*}'(\widehat{\xi}) = 0$ .另外, $S_{*}(\xi)$ 满足

$\begin{eqnarray*} 0 = c^{*}S_{*}^{'}(\widehat{\xi})& = &d_{s}(\int_{-\infty}^{+\infty}J(y)S_{*}(\widehat{\xi}-y)-S_{*}(\xi)){\rm d}y-\beta(I_{u*}(\widehat{\xi})+\delta I_{h*}(\widehat{\xi}))S_{*}(\widehat{\xi})\\ &\leq&-\beta(I_{u*}(\widehat{\xi})+\delta I_{h*}(\widehat{\xi}))S_{*}(\widehat{\xi})\\ &\leq&\frac{-\beta I_{*}(\widehat{\xi})S_{*}(\widehat{\xi})}{k_{h}}, \end{eqnarray*}$

这与 $I_{*}(\xi)>0$ 矛盾.

其次,要证 $S_{*}(-\infty) = S^{0}$ ,只需证 $\liminf\limits_{\xi\rightarrow-\infty}S_{*}(\xi) = S^{0}$ .假设 $\underline{S} = \liminf\limits_{\xi\rightarrow-\infty}S_{*}(\xi)<S^{0}$ ,则存在某个序列 $\{\eta_{n}\}$ 满足当 $n\rightarrow+\infty$ , $\eta_{n}\rightarrow-\infty$ 使得 $S_{*}(\eta_{n})\rightarrow \underline{S}_{*}$ .令 $S_{*n}(\xi): = S_{*}(\xi+\eta_{n})$ 且 $I_{*n}(\xi): = I_{*}(\xi+\eta_{n})$ .因为 $\parallel S_{*n}(\xi) \parallel_{C^{2}}$ 一致有界,通过抽取子列,可以假设 $S_{*n}(\xi)\rightarrow S_{\infty}$ 在 $C_{loc}^{1}({{\Bbb R}} )$ 上成立.注意 $\lim\limits_{n\rightarrow+\infty}I_{*n}(\xi) = 0$ 在 $R$ 上是局部一致的,则 $S_{\infty}(\xi)$ 满足

$\begin{eqnarray*} c^{*}S_{\infty}'(\xi) = d_{s}\int_{-\infty}^{+\infty}J(y)(S_{\infty}(\xi-y)-S_{\infty}(\xi)){\rm d}y, \quad\xi\in {{\Bbb R}} . \end{eqnarray*}$

由 $S_{\infty}(0) = \underline{S}_{*}$ ,由(3.11)式类似的讨论表明 $S_{\infty}(\xi) = \underline{S}_{*}$ .从而,当 $n\rightarrow+\infty$ 时, $S_{n}(\xi+\eta_{n})\rightarrow \underline{S}_{*}$ 在 ${{\Bbb R}}$ 上是局部一致收敛的.现在,从 $-\infty$ 到 $\eta_{n}$ 上积分 $(S, I_{uk}, I_{hk}, c_{k})$ ,可得

$\begin{eqnarray} c_{k}[S_{k}(\eta_{n})-S^{0}] & = &d_{s}\int_{-\infty}^{\eta_{n}}\int_{-\infty}^{+\infty}J(y)(S_{k}(\eta_{n}-y)-S_{k}(\xi)){\rm d}y{\rm d}\xi \\&&-\int_{-\infty}^{\eta_{n}}\beta(I_{uk}(\xi)+\delta I_{hk}(\xi))S_{k}(\xi){\rm d}\xi. \end{eqnarray}$

(3.17)

又

$\begin{eqnarray*} \int^{\eta_{n}}_{-\infty}(J*S_{k}(\xi)-S_{k}(\xi)){\rm d}\xi & = & \int^{\eta_{n}}_{-\infty}\int^{+\infty}_{-\infty}J(x)[S_{k}(\xi-x)-S_{k}(\xi)]{\rm d}x{\rm d}\xi\\ & = & \int^{\eta_{n}}_{-\infty}\int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}S_{k}^{'}(\xi-\theta x){\rm d}\theta {\rm d}x{\rm d}\xi\\ & = & \int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}\int^{\eta_{n}}_{-\infty}{S}_{k}^{'}(\xi-\theta x){\rm d}\theta {\rm d}x{\rm d}\xi\\ & = &\int^{+\infty}_{-\infty}J(x)(-x)\int^{1}_{0}[S_{k}(\eta_{n}-\theta x)-S^{0}]{\rm d}\theta {\rm d}x\\ & = &\int^{+\infty}_{-\infty}J(x)x\int^{1}_{0}S_{k}(\eta_{n}+\theta x){\rm d}\theta {\rm d}x, \end{eqnarray*}$

令 $k\rightarrow+\infty$ ,则

$\begin{eqnarray*} c_{*}[S_{*}(\eta_{n})-S^{0}]& = &d_{s}\int_{-\infty}^{\eta_{n}}\int_{-\infty}^{+\infty}J(y)(S_{*}(\eta_{n}-y)-S_{*}(\xi)){\rm d}y{\rm d}\xi\\ &&-\int_{-\infty}^{\eta_{n}}\beta(I_{*}(\xi)+\delta I_{*}(\xi))S_{*}(\xi){\rm d}\xi. \end{eqnarray*}$

又 $S_{*}, I_{*}$ 在 ${{\Bbb R}}$ 上有界且 $\int_{-\infty}^{+\infty}\beta(I_{*}(\xi)+\delta I_{*}(\xi))S_{*}(\xi){\rm d}\xi<+\infty$ ,令 $n\rightarrow+\infty$ ,则

$0\leq c^{*}[\underline{S}_{*}-S^{0}]<0,$

矛盾.另一方面,由引理2.5的类似讨论可得 $S_{*}(+\infty) = S^{1}<S^{0}$ .证毕.

4 行波解的不存在性

在本节中,主要考虑行波解的不存在性.准确地说,通过双边拉普拉斯变换得到了当 $R_{0}>1$ 且波速小于临界值时行波解的不存在性,而当 $R_{0}<1$ 时,主要利用Fubini定理得到了行波解的不存在性.

4.1 $R_{0}>1$ 情形下的不存在性

现在证明以下引理.

引理4.1 假设(NH1)或(NH2)成立.若 $(S(\xi), I_{u}(\xi), I_{h}(\xi), R(\xi))$ 是系统(2.1)的一个解,则存在一个正常数 $\varpi$ 使得

$\sup\limits_{\xi\in {{\Bbb R}} }\{I_{u}(\xi){\rm e}^{-\varpi\xi}\}<+\infty, \ \sup\limits_{\xi\in {{\Bbb R}} }\{I_{h}(\xi){\rm e}^{-\varpi\xi}\}<+\infty,$

$\sup\limits_{\xi\in {{\Bbb R}} }\{I_{u}^{'}(\xi){\rm e}^{-\varpi\xi}\}<+\infty, \ \sup\limits_{\xi\in {{\Bbb R}} }\{I_{h}^{'}(\xi){\rm e}^{-\varpi\xi}\}<+\infty.$

证由 $S(-\infty) = S^{0}$ ,可得 $\xi_{\varrho}<0$ 使得 $S(\xi)>\varrho S^{0}$ 对 $0<\varrho<1$ 和 $\xi<\xi_{\varrho}$ 成立.从而,

$\begin{eqnarray*} cI_{u}^{'}(\xi) & = & d_{u}(J*I_{u}-I_{u})+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}\\ &\ge& d_{u}(J*I_{u}-I_{u})+\varrho(1-\mu)\beta S^{0}(I_{u}+\delta I_{h})-k_{u}I_{u}. \end{eqnarray*}$

以上不等式两边同乘 $k_{h}$ ,可得

$\begin{eqnarray} ck_{h}I_{u}^{'}(\xi)&\ge& d_{u}k_{h}(J*I_{u}-I_{u})+\varrho k_{h}(1-\mu)\beta S^{0}(I_{u}+\delta I_{h})-k_{u}k_{h}I_{u}\\ & = &d_{u}k_{h}(J*I_{u}-I_{u})+[\varrho k_{h}(1-\mu)\beta S^{0}-k_{u}k_{h}]I_{u}+\varrho k_{h}(1-\mu)\delta\beta S^{0}I_{h}. \end{eqnarray}$

(4.1)

令 $K_{u}(\xi) = \int^{\xi}_{-\infty}I_{u}(t){\rm d}t$ 且 $K_{h}(\xi) = \int^{\xi}_{-\infty}I_{h}(t){\rm d}t$ .利用Fubini定理,有

$\begin{eqnarray*} \int^{\xi}_{-\infty}J*I_{u}(t){\rm d}t& = & \int^{\xi}_{-\infty}\int^{+\infty}_{-\infty}J(y)I_{u}(t-y){\rm d}y{\rm d}t\\ & = & \int^{+\infty}_{-\infty} \int^{\xi}_{-\infty}J(y)I_{u}(t-y){\rm d}t{\rm d}y\\ & = & \int^{+\infty}_{-\infty}J(y)\int^{\xi-y}_{-\infty}I_{u}(t){\rm d}t{\rm d}y\\ & = & \int^{+\infty}_{-\infty}J(y)K_{u}(\xi-y){\rm d}y\\ & = & J*K_{u}(\xi). \end{eqnarray*}$

从 $-\infty$ 到 $\xi$ 上积分(4.1)式,有

$\begin{eqnarray} ck_{h}I_{u}(\xi)&\geq& d_{u}k_{h}(J*K_{u}(\xi)-K_{u}(\xi))+[\varrho k_{h}(1-\mu)\beta S^{0}-k_{u}k_{h}]K_{u}(\xi)\\ &&+ \varrho k_{h}(1-\mu)\delta\beta S^{0}K_{h}(\xi). \end{eqnarray}$

(4.2)

进一步地,通过简单计算可得

$\begin{eqnarray*} \int^{\xi}_{-\infty}[J*K_{u}-K_{u}]{\rm d}t& = & \int^{\xi}_{-\infty}\int^{+\infty}_{-\infty}J(y)[K_{u}(t-y)-K_{u}(y)]{\rm d}t{\rm d}y\\ & = & \int^{\xi}_{-\infty}\int^{+\infty}_{-\infty}J(y)[K_{u}(t-y)-K_{u}(y)]{\rm d}y{\rm d}t\\ & = & \int^{\xi}_{-\infty}\int^{+\infty}_{-\infty}J(y)(-y)\int^{1}_{0}K_{u}^{'}(t-\theta y){\rm d}\theta {\rm d}y{\rm d}t\\ & = & \int^{1}_{0}\int^{+\infty}_{-\infty}J(y)(-y)\int^{\xi}_{-\infty}K_{u}^{'}(t-\theta y){\rm d}t {\rm d}y{\rm d}\theta\\ & = & \int^{+\infty}_{-\infty}J(y)(-y)\int^{1}_{0}K_{u}(\xi-\theta y){\rm d}\theta {\rm d}y. \end{eqnarray*}$

从而对任意 $\xi\in{{\Bbb R}}$ , $J*K_{u}(\xi)-K_{u}(\xi)$ 在 $(-\infty, \xi]$ 上可积.因此,由(4.2)式, $K_{u}(\xi)$ 和 $K_{h}(\xi)$ 在 $(-\infty, \xi]$ 上对任意的 $\xi\in{{\Bbb R}}$ 可积.从 $-\infty$ 到 $\xi$ 上积分(4.2)式,可得

$\begin{eqnarray*} ck_{h}K_{u}(\xi) &\ge& d_{u}k_{h}\int^{+\infty}_{-\infty}J(y)(-y)\int^{1}_{0}K_{u}(\xi-\theta y){\rm d}\theta {\rm d}y\\ &&+ [\varrho k_{h}(1-\mu)\beta S^{0}-k_{u}k_{h}]\int^{\xi}_{-\infty}K_{u}(t){\rm d}t+ \varrho k_{h}(1-\mu)\delta\beta S^{0}\int^{\xi}_{-\infty}K_{h}(t){\rm d}t. \end{eqnarray*}$

易得 $xK_{u}(\xi-\theta x)$ 关于 $\theta\in[0, 1]$ 递减,则 $xK_{u}(\xi-\theta x)\leq xK_{u}(\xi)$ 且

$\begin{eqnarray*} && [\varrho k_{h}(1-\mu)\beta S^{0}-k_{u}k_{h}]\int^{\xi}_{-\infty}K_{u}(t){\rm d}t+\varrho k_{h}(1-\mu)\delta\beta S^{0}\int^{\xi}_{-\infty}K_{h}(t){\rm d}t\\ &<& ck_{h}K_{u}(\xi)+d_{u}k_{h}\int^{+\infty}_{-\infty}J(y)y{\rm d}yK_{u}(\xi). \end{eqnarray*}$

又 $\int^{+\infty}_{-\infty}J(y)y{\rm d}y = 0$ ,则

$\begin{eqnarray} [\varrho k_{h}(1-\mu)\beta S^{0}-k_{u}k_{h}]\int^{\xi}_{-\infty}K_{u}(t){\rm d}t+\varrho k_{h}(1-\mu)\delta\beta S^{0}\int^{\xi}_{-\infty}K_{h}(t){\rm d}t<ck_{h}K_{u}(\xi). \end{eqnarray}$

(4.3)

同理可得

$\begin{eqnarray} [\varrho k_{u}\mu\beta\delta^{2} S^{0}-\delta k_{u}k_{h}]\int^{\xi}_{-\infty}K_{h}(t){\rm d}t+\varrho k_{u}\mu\beta\delta S^{0}\int^{\xi}_{-\infty}K_{u}(t){\rm d}t<c\delta k_{u}K_{h}(\xi). \end{eqnarray}$

(4.4)

将(4.3)式和(4.4)式相加,有

$\begin{equation} \left\{\begin{array}{ll} { } [\varrho k_{h}(1-\mu)\beta S^{0}+\varrho k_{u}\mu\beta\delta S^{0}-k_{u}k_{h}]\int^{\xi}_{-\infty}K_{u}(t){\rm d}t+\delta[\varrho k_{u}\mu\beta\delta S^{0}\\ { } +\varrho k_{h}(1-\mu)\beta S^{0}-k_{u}k_{h}]\int^{\xi}_{-\infty}K_{h}(t){\rm d}t<ck_{h}K_{u}(\xi)+c\delta k_{u} K_{h}(\xi). \end{array}\right. \end{equation}$

(4.5)

又 $R_{0}>1$ ,即 $k_{h}(1-\mu)\beta S^{0}+k_{u}\mu\beta\delta S^{0}-k_{u}k_{h}>0$ ,存在 $0<\varrho_{1}<1$ 使得

$\varrho_{1} k_{h}(1-\mu)\beta S^{0}+\varrho_{1} k_{u}\mu\beta\delta S^{0}-k_{u}k_{h}>0.$

令 $\overline{K} = K_{u}+K_{h}$ .则由(4.5)式存在常数 $a>0$ 和 $b>0$ 使得

$a\int_{-\infty}^{\xi}\overline{K}(x){\rm d}x\leq b \overline{K}(\xi), \; \forall \xi<\xi_{\varrho_{1}}.$

从而存在一个正常数 $\overline{M}>0$ 使得

$\int^{\xi}_{-\infty}\overline{K}{\rm d}t = \int^{+\infty}_{0}\overline{K}(\xi-\zeta){\rm d}\zeta\geq\int^{\overline{M}}_{0}\overline{K}(\xi-\zeta){\rm d}\zeta \geq \overline{M}\overline{K}(\xi-\overline{M}).$

因此,对任意 $\xi\in {{\Bbb R}}$ ,存在某个充分大的 $\overline{M}$ 使得

$\overline{K}(\xi-\overline{M})<1/3 \overline{K}(\xi).$

令 $W(x) = \overline{K}(x){\rm e}^{-\varpi x}$ ,其中 $\varpi = \frac{1}{\overline{M}}\ln3$ ,则

$W(\xi-\overline{M}) = \overline{K}(\xi-\overline{M}){\rm e}^{-\varpi(\xi-\overline{M})}<1/3\overline{K}{\rm e}^{-\varpi(\xi-\overline{M})} = W(\xi).$

从而,由 $W(x)$ 的定义,当 $\xi\rightarrow-\infty$ 时, $W(x)$ 有界.即存在一个正常数 $W_{0}$ 使得

$\overline{K}(x)\leq W_{0}{\rm e}^{\varpi x}$

对一切 $x\in {{\Bbb R}}$ 成立.从而

$K_{u}(x)\leq W_{0}{\rm e}^{\varpi x}, K_{h}(x)\leq W_{0}{\rm e}^{\varpi x}.$

又

$cI_{u}(\xi)<d_{u}(J*K_{u}-K_{u})+(1-\mu)\beta(K_{u}+\delta K_{h})S_{0}-k_{u}K_{u},$

可得

$\begin{eqnarray*} |I_{u}(\xi)|&<& \frac{1}{c}[W_{0}{\rm e}^{\varpi \xi}\int^{+\infty}_{-\infty}J(y)|y|{\rm e}^{-\varpi y}{\rm d}y+(1-\mu)\beta(W_{0}{\rm e}^{\varpi\xi}+\delta W_{0}{\rm e}^{\varpi\xi})S_{0}+k_{u}W_{0}{\rm e}^{\varpi\xi}]\\ & = & \frac{1}{c}[W_{0}\int^{+\infty}_{-\infty}J(y)|y|{\rm e}^{-\varpi y}{\rm d}y+(1-\mu)\beta(W_{0}+\delta W_{0})S_{0}+k_{u}W_{0}]{\rm e}^{\varpi\xi}, \end{eqnarray*}$

即 $\sup\limits_{\xi\in {{\Bbb R}} }\{I_{u}(\xi){\rm e}^{-\varpi\xi}\}<+\infty$ .

同理可证, $\sup\limits_{\xi\in {{\Bbb R}} }\{I_{h}(\xi){\rm e}^{-\varpi\xi}\}<+\infty$ , $\sup\limits_{\xi\in {{\Bbb R}} }\{I_{u}^{'}(\xi){\rm e}^{-\varpi\xi}\}<+\infty$ , $\sup\limits_{\xi\in {{\Bbb R}} }\{I_{h}^{'}(\xi){\rm e}^{-\varpi\xi}\}<+\infty$ .证毕.

定理4.1 假设(NH1)或(NH2)成立,则系统(1.3)不存在波速为 $c$ 的非平凡正解.

证首先,对 $\lambda\in {\Bbb C}$ 和一个非负函数 $W(\xi)$ 定义双边拉普拉斯变换为

$\begin{eqnarray*} L[W(\cdot)](\lambda) = \int^{+\infty}_{-\infty}W(\cdot){\rm e}^{-\lambda\xi}{\rm d}\xi. \end{eqnarray*}$

显然 $L[W(\cdot)](\lambda)$ 在 $[0, \lambda_{1}^{*})$ 上有定义,且 $\lambda_{1}^{*}$ 满足 $\lim\limits_{\lambda\rightarrow\lambda_{1}^{*-}}L[W(\cdot)](\lambda) = +\infty$ 或 $\lambda_{1}^{\ast} = +\infty$ .

令

$\triangle_{u}(\lambda) = L[I_{u}(\cdot)](\lambda), \lambda\in[0 , \lambda_{u}^{*}),$

$\triangle_{h}(\lambda) = L[I_{h}(\cdot)](\lambda), \lambda\in[0 , \lambda_{h}^{*}).$

根据引理4.1, $\lambda_{u}^{*}\geq\varpi$ , $\lambda_{h}^{*}\geq\varpi$ .则系统(1.3)可以表示为

$\begin{equation} \left\{ \begin{array}{rl} &d_{u}(J*I_{u}(\xi)-I_{u}(\xi))- cI_{u}^{'}(\xi)+[(1-\mu)\beta S^{0}-k_{u}]I_{u}(\xi)\\ = &(1-\mu)[ S^{0}-S(\xi)]\beta(I_{u}+\delta I_{h})-(1-\mu)\beta\delta S^{0}I_{h}(\lambda), \\ & d_{h}(J*I_{h}(\xi)-I_{h}(\xi))- cI_{h}^{'}(\xi)+[\mu\beta \delta S^{0}-k_{h}]I_{h}(\xi)\\ = &\mu[ S^{0}-S(\xi)]\beta(I_{u}+\delta I_{h})-\mu\beta S^{0}I_{u}(\lambda). \end{array} \right. \end{equation}$

(4.6)

令 $g(\xi) = \beta[S^{0}-S(\xi)](I_{u}(\xi)+\delta I_{h}(\xi))$ 且对(4.6)式做双边拉普拉斯变换,可得

$\begin{equation} \left\{ \begin{array}{l} \Delta_{u}(\lambda, c)\triangle_{u}(\lambda) = (1-\mu)G(\lambda)-(1-\mu)\beta\delta S^{0}\triangle_{h}(\lambda), \\ \Delta_{h}(\lambda, c)\triangle_{h}(\lambda) = \mu G(\lambda)-\mu\beta S^{0}\triangle_{u}(\lambda), \end{array} \right. \end{equation}$

(4.7)

其中 $G(\lambda) = L[g(\cdot)](\lambda)$ 和 $0<\lambda<\varpi$ .进一步地,第一个等式乘以(4.7)式的第二个等式,有

$\begin{equation} H(\lambda)\triangle_{u}(\lambda)\triangle_{h}(\lambda) = \mu(1-\mu)G(\lambda)[G(\lambda)- \beta S^{0}(\triangle_{u}(\lambda)+ \delta\triangle_{h}(\lambda))]. \end{equation}$

(4.8)

下面证明 $\lambda_{u}<\infty$ , $\lambda_{h}<\infty$ .根据(4.7)式,可得

$\begin{eqnarray*} \triangle(\lambda) = [d_{u}\int^{+\infty}_{-\infty}{\rm e}^{-\lambda y}J(y){\rm d}y-c\lambda-k_{h}]\triangle_{h}(\lambda) +\mu\beta L[S(\cdot)I_{u}+\delta S(\cdot)I_{h}](\lambda) = 0. \end{eqnarray*}$

又 $L[S(\cdot)I_{u}+\delta S(\cdot)I_{h}](\lambda)>0$ 且 $\triangle_{h}(\lambda)>0$ 对一切 $\lambda\in[0$ , $\lambda_{h}^{*})$ 成立.若 $\lambda_{h} = +\infty$ ,则 $\triangle(+\infty) = +\infty$ ,矛盾.从而 $\lambda_{u}<\infty$ 且 $\lambda_{h}<\infty$ .

进一步地,下证 $\lambda_{u}^{*} = \lambda_{h}^{*}$ .否则,若 $\lambda_{u}^{*}<\lambda_{h}^{*}$ ,则

$\begin{eqnarray*} \lim\limits_{\lambda\longrightarrow\lambda_{u}^{*-}}\triangle_{u}(\lambda) = +\infty, \lim\limits_{\lambda\longrightarrow\lambda_{u}^{*-}}\triangle_{h}(\lambda) = \triangle_{h}(\lambda_{u}^{*}). \end{eqnarray*}$

则由 $G(\lambda_{u}^{*})<+\infty$ ,这与(4.7)式的第二个等式矛盾.从而, $\lambda_{u}^{*}>\lambda_{h}^{*}$ .同理可证 $\lambda_{u}^{*}<\lambda_{h}^{*}$ 和 $\lambda_{u}^{*} = \lambda_{h}^{*} = \lambda_{1}^{*}$ .此外,考虑 $\Delta_{u}(\lambda_{1}^{*}, c)$ 和 $\Delta_{h}(\lambda_{1}^{*}, c)$ .假设 $\Delta_{h}(\lambda_{1}^{*}, c)\geq0$ ,则

$\begin{eqnarray*} \Delta_{h}(\lambda_{1}^{*}, c)\triangle_{h}(\lambda_{1}^{*})+\mu\beta S^{0}\triangle_{u}(\lambda_{1}^{*}) = +\infty>\mu G(\lambda_{1}^{*}), \end{eqnarray*}$

这与(4.7)式的第二个等式矛盾.从而, $\Delta_{u}(\lambda_{1}^{*}, c)<0$ 且 $\Delta_{h}(\lambda_{1}^{*}, c)<0$ .由引理1.2, $H(\lambda_{1}^{*})<0$ .令 $\overline{S}(\xi) = S^{0}-S(\xi)$ .由文献[26,定理3.1],存在正常数 $\omega$ 使得当 $\xi\rightarrow\pm\infty$ 时, $\overline{S}(\xi){\rm e}^{-\omega\xi}\rightarrow 0$ ,且 $G(\lambda) = L[g(.)](\lambda) = L[\overline{S}(\xi)(I_{u}(\xi)+\delta I_{h}(\xi)](\lambda)$ 在 $[0, \lambda_{1}^{*}+\omega)$ 上有定义.从而,由引理1.2, $G(\lambda_{1}^{*})<+\infty.$ 进一步地

$\lim\limits_{\lambda\longrightarrow\lambda_{1}^{*-}}\frac{G(\lambda)[G(\lambda)-\beta S^{0}(\triangle_{u}(\lambda)+\delta \triangle_{h}(\lambda)) ])}{\triangle_{u}(\lambda)\triangle_{h}(\lambda)} = 0,$

这与(4.8)式矛盾.证毕.

4.2 $R_{0}<1$ 情形下的不存在性

定理4.2 假设 $R_{0}<1$ .则系统(1.3)不存在 $c>0$ 的非平凡的有界正解.

证假设对所有 $c>0$ ,存在非平凡的有界正解 $(S(x+ct), I_{u}(x+ct), I_{h}(x+ct))$ 满足 $S(-\infty) = S^{0}, S(+\infty) = S^{1}$ 且 $I_{u}(\pm\infty) = I_{h}(\pm\infty) = 0.$ 在 ${{\Bbb R}}$ 上积分下列方程

$\begin{eqnarray*} \left\{\begin{array}{ll} cI_{u}^{'}(\xi) = d_{u}(J*I_{u}-I_{u})+(1-\mu)\beta(I_{u}+\delta I_{h})S-k_{u}I_{u}, \\ cI_{h}^{'}(\xi) = d_{h}(J*I_{h}-I_{h})+\mu\beta(I_{u}+\delta I_{h})S-k_{h}I_{h}, \\ \end{array}\right. \end{eqnarray*}$

可得

$\begin{equation} \left\{\begin{array}{ll} { } k_{u}\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi = d_{u}\int^{+\infty}_{-\infty}(J*I_{u}-I_{u}){\rm d}\xi+(1-\mu)\beta\int^{+\infty}_{-\infty}(I_{u}+\delta I_{h})S{\rm d}\xi, \\ { } k_{h}\int^{+\infty}_{-\infty}I_{h}(\xi){\rm d}\xi = d_{h}\int^{+\infty}_{-\infty}(J*I_{h}-I_{h}){\rm d}\xi+\mu\beta\int^{+\infty}_{-\infty}(I_{u}+\delta I_{h})S{\rm d}\xi. \end{array}\right. \end{equation}$

(4.9)

由Fubini定理,可得

$\int^{+\infty}_{-\infty}J*I_{u}{\rm d}\xi\leq\int^{+\infty}_{-\infty}I_{u}{\rm d}\xi,$

$\int^{+\infty}_{-\infty}J*I_{h}{\rm d}\xi\leq\int^{+\infty}_{-\infty}I_{h}{\rm d}\xi.$

把以上不等式代入(4.9)式,得到

$\begin{eqnarray*} \left\{\begin{array}{ll} { } (k_{u}-(1-\mu)\beta S^{0})\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi \leq(1-\mu)\beta S^{0}\delta\int^{+\infty}_{-\infty}I_{h}(\xi){\rm d}\xi, \\ { } (k_{h}-\mu \beta \delta S^{0})\int^{+\infty}_{-\infty}I_{h}(\xi){\rm d}\xi \leq\mu\beta S^{0}\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi. \end{array}\right. \end{eqnarray*}$

从而

$\begin{eqnarray*} \int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi &\leq&\frac{(1-\mu)\beta S^{0}\delta}{k_{u}-(1-\mu)\beta S^{0}}\int^{+\infty}_{-\infty}I_{h}(\xi){\rm d}\xi\\ &\leq& \frac{ (1-\mu)\beta S^{0}\delta }{[k_{u}-(1-\mu)\beta S^{0}]} \frac{ \mu\beta S^{0} }{[k_{h}-\mu\beta \delta S^{0}]} \int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi.\\ \end{eqnarray*}$

又 $R_{u}+R_{h}<1$ ,有

$\mu k_{u}\beta\delta S^{0}+(1-\mu) k_{h}\beta S^{0}<k_{u}k_{h}.$

即

$\mu(1-\mu)\beta^{2} S_{0}^{2}\delta<[k_{u}-(1-\mu)\beta S^{0}][k_{h}-\mu\beta \delta S^{0}].$

从而

$\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi<\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi.$

矛盾.证毕.

参考文献

原文顺序

文献年度倒序

文中引用次数倒序

被引期刊影响因子

[1]

Arino

, Brauer

, Driessche

, et al.

A model for influenza with vaccination and antiviral treatment

J Theoret Biol, 2008, 253, 118- 130

DOI:10.1016/j.jtbi.2008.02.026 [本文引用: 1]

[2]

Alexander

M E

, Bowman

, Moghzdas

, et al.

A vaccination model for transmission dynamics of influenza

SIAM J Appl Dyn Syst, 2004, 3 (4): 503- 524

DOI:10.1137/030600370 [本文引用: 1]

[3]

Chen

, Guo

J S

Uniqueness and existence of traveling waves for discrete quasilinear monostable dynamics

Math Ann, 2003, 326, 123- 146

DOI:10.1007/s00208-003-0414-0 [本文引用: 1]

[4]

Diekmann

, Heesterbeek

J A

Mathematical Epidemiology of Infective Diseases:Model Building, Analysis and Interpretation. New York: Wiley, 2000

[本文引用: 1]

[5]

Ducrot

, Magal

Travelling wave solution for an infection-age strutured model with diffusion

Proc Roy Soc Edinburgy Sec A, 2009, 139, 459- 482

DOI:10.1017/S0308210507000455 [本文引用: 4]

[6]

Ducrot

, Magal

, Ruan

Travelling wave solutions in Mltigroup Age-Structured Epidemic Models

Arch Rational Mech Anal, 2010, 195, 311- 331

DOI:10.1007/s00205-008-0203-8 [本文引用: 1]

[7]

Ducrot

, Magal

Travelling wave solution for an infection-age strutured epidemic model with external supplies

Nonliearity, 2011, 24, 2891- 2911

DOI:10.1088/0951-7715/24/10/012 [本文引用: 1]

[8]

S C

, Guo

J S

Uniqueness and existence of traveling waves for discrete quasilinear monostable dynamics

Math Ann, 2003, 326, 123- 146

DOI:10.1007/s00208-003-0414-0 [本文引用: 1]

[9]

Ferguson

N M

, Cummings

, Fraser

, et al.

Strategies for mitigating an influenza pandemic

Nature, 2006, 442, 448- 452

DOI:10.1038/nature04795 [本文引用: 1]

[10]

Hosono

, IIyas

Traveling waves of a simple diffusive epidemic model

Math Model Meth Appl Sci, 1995, 5, 935- 966

DOI:10.1142/S0218202595000504 [本文引用: 1]

[11]

Lin

, Andreasen

, Casagrandi

, Levin

S A

Traveling waves in a model of influenza a drift

J Theoret Biol, 2003, 222, 437- 445

DOI:10.1016/S0022-5193(03)00056-0 [本文引用: 1]

[12]

Lipsitch

, Cohen

, Muary

, Levin

B R

Antiviral resistantce and the control of pandemic influenza

PLoS Med, 2007, 4, 0111- 0120

DOI:10.1371/journal.pmed.0040111 [本文引用: 2]

[13]

, Li

W T

, Yang

F Y

Traveling waves for a nonlocal dispersal SIR model with delay and external supplies

Appl Math Compute, 2014, 247, 723- 740

URL [本文引用: 2]

[14]

Merler

, Ajelli

The role of population heterogeneity and human mobility in the spread of pandemic influenza

Proc R Soc B, 2010, 277, 557- 565

DOI:10.1098/rspb.2009.1605 [本文引用: 1]

[15]

Murray

J D

Mathematical Biology. Berlin: Springer, 2002

[本文引用: 1]

[16]

Nuno

, Feng

, Martcheva

, Castillo-Chavez

Dynamics of two-strain influenza with isolation and partial crossimmunity

SIAM J Appl Math, 2005, 65, 964- 982

DOI:10.1137/S003613990343882X [本文引用: 1]

[17]

Ruan S G. Spatial-temporal Dynamics in Nonlocal Epidemiological Models//Takeuchi Y, et al. Mathematics for Life Science and Medicine. New York: Springer, 2007: 97-122

[本文引用: 1]

[18]

Ruan S, Wu J. Modeling Spatial Spread of Communicable Diseases Involving Animal Hosts//Cantrell S, et al. Spatial Ecology. New York: Chapman and Hall/CRC, 2009: 293-316

[本文引用: 1]

[19]

Thieme

H R

Asymptotic estimates of the the solutions of nonlinear integral equations and asymptotic speeds for the spread of populations

J Renie Angew Math, 1979, 306, 94- 121

URL [本文引用: 1]

[20]

Thieme

H R

, Zhao

X Q

Asymptotic speeds of spread and traveling waves for integral equations and delayed reaction-diffusion models

J Differential Equations, 2003, 195, 430- 470

DOI:10.1016/S0022-0396(03)00175-X

[21]

C C

Existence of traveling waves with the critical speed for a discrete diffusive epidemic model

J Differential Equations, 2017, 262, 272- 282

DOI:10.1016/j.jde.2016.09.022 [本文引用: 1]

[22]

Wang

J B

, Li

W T

, Yang

F Y

Traveling waves in a nonlocal dispersal SIR model with nonlocal delayed transmission

Commun Nonlinear Sci Numer Simulat, 2015, 27, 136- 152

DOI:10.1016/j.cnsns.2015.03.005 [本文引用: 1]

[23]

Wang

X S

, Wang

, Wu

Traveling waves of diffusive predator-prey systems:disease outbreak propagation

Discrete Contin Dyn Syst B, 2002, 32, 3303- 3324

URL

[24]

Wang

Z C

, Wu

, Liu

Traveling waves of the spread avian influenza

Proc Amer Math Soc, 2012, 140, 3931- 3946

DOI:10.1090/S0002-9939-2012-11246-8

[25]

, Zhao

X Q

Asymptotic speed of spread and traveling waves for a nonlocal epidemic model

Discrete Contin Dyn Syst (Ser B), 2005, 5 (4): 1043- 1056

DOI:10.3934/dcdsb.2005.5.1043

[26]

Yang

F Y

, Li

W T

, Wang

Z C

Travelling waves in a nonlocal dispersal Kermack-Mckendrick epidemic model

Discrete Contin Dyn Syst (Ser B), 2013, 18, 1969- 1993

DOI:10.3934/dcdsb.2013.18.1969 [本文引用: 3]

[27]

Yang

F Y

, Li

W T

, Wang

Z C

Traveling waves in a nonlocal dispersal SIR epidemic model

Nonlinear Anal, 2015, 23, 129- 147

DOI:10.1016/j.nonrwa.2014.12.001 [本文引用: 3]

[28]

Yang

F Y. Li W T

Traveling waves in a nonlocal dispersal SIR epidemic with critical wave speed

J Math Anal Appl, 2018, 458, 1131- 1146

DOI:10.1016/j.jmaa.2017.10.016 [本文引用: 1]

[29]

Zhang

G B

, Li

W T

, Wang

Z C

Spreading speed and traveling waves for nonlocal dispersal equations with degenerate monostable nonlinearity

J Differential Equations, 2012, 252, 5096- 5124

DOI:10.1016/j.jde.2012.01.014 [本文引用: 4]

[30]

Zhang

T R

, Wang

W D

Existence of traveling wave solutions for influenza model with treatment

J Math Anal Appl, 2014, 419, 469- 495

DOI:10.1016/j.jmaa.2014.04.068 [本文引用: 5]

A model for influenza with vaccination and antiviral treatment

2008

... 众所周知,流感等传染病在现代生活中十分常见,参看文献[1-2, 4, 9, 12, 14, 16].研究各类传染病模型的时空传播动力学具有重要意义.在过去几十年中,扩散传染病模型的行波解得到广泛的研究,参看文献[5-6, 10-11, 13, 19-27, 30].特别地,文献[13, 22, 26-27]等研究了非局部扩散流行病模型的行波解.关于传染病模型行波解的更多工作请参见专著[15],综述论文[17-18]及相关参考文献. ...

A vaccination model for transmission dynamics of influenza

2004

Uniqueness and existence of traveling waves for discrete quasilinear monostable dynamics

2003

... 此外,我们考虑了当

$d_u = d_h$

且

$c = c*$

时行波解存在性.受文献[3, 8, 21, 28]启发,通过对超临界波速的递减序列取极限来证明.然而,由于系统(1.2)中出现了治疗项,估计

$I_{u}$

和

$I_{h}$

的有界性及S在

$-\infty$

处的边界渐近行为比较困难.我们通过一精细的分析证明了

$k_{h}I_{u}+k_{u}\delta I_{h}$

的有界性,从而解决了上述困难. ...

2000

Travelling wave solution for an infection-age strutured model with diffusion

2009

... 并研究了其行波解的存在性和不存在性.其中

$S(t, x)$

$I_{u}(x, t)$

$I_{h}(x, t)$

和

$R(t, x)$

分别表示易感者,感染而未治愈者,治愈者和恢复者.

$0<\mu<1$

$0<\delta<1$

,且

$k_{u}<k_{h}$

$d_{i}(i = s, u, h, r)$

分别表示易感者,感染未治愈者,治愈者,恢复者的扩散率.文献[30]引入一个辅助系统,构造了一个有界锥,从而利用Schauder不动点定理来证明系统(1.1)行波解的存在性,参见文献[5-7].其次,双边拉普拉斯变换证明了系统(1.1)的行波解的不存在性. ...

... 定理1.1把文献[5]的行波解的存在性和不存在性从反应扩散模型推广到了非局部扩散情形.受文献[5, 30]的启发,我们利用Schauder不动点定理结合上下解方法和取极限的方法,首先证明了

$R_0>1$

$c>c^*$

$d_u = d_h$

或

$c>c_1^*$

$d_u\neq d_h$

情形下行波解的存在性;然后利用双边拉普拉斯变换,得到

$R_0>1$

$0<c<c^*$

$d_u = d_h$

或

$0<c<c_2^*$

$d_u\neq d_h$

情形下行波解的不存在性.最后,利用Fubini定理,证明了

$R_0<1$

时行波解的不存在性. ...

... ]的行波解的存在性和不存在性从反应扩散模型推广到了非局部扩散情形.受文献[5, 30]的启发,我们利用Schauder不动点定理结合上下解方法和取极限的方法,首先证明了

$R_0>1$

$c>c^*$

$d_u = d_h$

或

$c>c_1^*$

$d_u\neq d_h$

情形下行波解的存在性;然后利用双边拉普拉斯变换,得到

$R_0>1$

$0<c<c^*$

$d_u = d_h$

或

$0<c<c_2^*$

$d_u\neq d_h$

情形下行波解的不存在性.最后,利用Fubini定理,证明了

$R_0<1$

时行波解的不存在性. ...

Travelling wave solutions in Mltigroup Age-Structured Epidemic Models

2010

Travelling wave solution for an infection-age strutured epidemic model with external supplies

2011

... 并研究了其行波解的存在性和不存在性.其中

$S(t, x)$

$I_{u}(x, t)$

$I_{h}(x, t)$

和

$R(t, x)$

分别表示易感者,感染而未治愈者,治愈者和恢复者.

$0<\mu<1$

$0<\delta<1$

,且

$k_{u}<k_{h}$

$d_{i}(i = s, u, h, r)$

Uniqueness and existence of traveling waves for discrete quasilinear monostable dynamics

2003

... 此外,我们考虑了当

$d_u = d_h$

且

$c = c*$

时行波解存在性.受文献[3, 8, 21, 28]启发,通过对超临界波速的递减序列取极限来证明.然而,由于系统(1.2)中出现了治疗项,估计

$I_{u}$

和

$I_{h}$

的有界性及S在

$-\infty$

处的边界渐近行为比较困难.我们通过一精细的分析证明了

$k_{h}I_{u}+k_{u}\delta I_{h}$

的有界性,从而解决了上述困难. ...

Strategies for mitigating an influenza pandemic

2006

Traveling waves of a simple diffusive epidemic model

1995

Traveling waves in a model of influenza a drift

2003

Antiviral resistantce and the control of pandemic influenza

2007

... 2014年, Zhang和Wang^[30]将扩散引入到带治疗项的流感模型^[12]中,即以下反应扩散方程 ...

Traveling waves for a nonlocal dispersal SIR model with delay and external supplies

2014

... ].特别地,文献[13, 22, 26-27]等研究了非局部扩散流行病模型的行波解.关于传染病模型行波解的更多工作请参见专著[15],综述论文[17-18]及相关参考文献. ...

The role of population heterogeneity and human mobility in the spread of pandemic influenza

2010

2002

Dynamics of two-strain influenza with isolation and partial crossimmunity

2005

Asymptotic estimates of the the solutions of nonlinear integral equations and asymptotic speeds for the spread of populations

1979

Asymptotic speeds of spread and traveling waves for integral equations and delayed reaction-diffusion models

2003

Existence of traveling waves with the critical speed for a discrete diffusive epidemic model

2017

... 此外,我们考虑了当

$d_u = d_h$

且

$c = c*$

时行波解存在性.受文献[3, 8, 21, 28]启发,通过对超临界波速的递减序列取极限来证明.然而,由于系统(1.2)中出现了治疗项,估计

$I_{u}$

和

$I_{h}$

的有界性及S在

$-\infty$

处的边界渐近行为比较困难.我们通过一精细的分析证明了

$k_{h}I_{u}+k_{u}\delta I_{h}$

的有界性,从而解决了上述困难. ...

Traveling waves in a nonlocal dispersal SIR model with nonlocal delayed transmission

2015

Traveling waves of diffusive predator-prey systems:disease outbreak propagation

2002

Traveling waves of the spread avian influenza

2012

Asymptotic speed of spread and traveling waves for a nonlocal epidemic model

2005

Travelling waves in a nonlocal dispersal Kermack-Mckendrick epidemic model

2013

... 又

$0<\overline{I}_{u}(\xi)\leq-K\Delta_{h}(\lambda_{1}, c)$

,则

$\int^{+\infty}_{-\infty}I_{u}(\xi){\rm d}\xi<+\infty$

.显然,

$I_{u}^{'}(\xi)$

有界,则

$I_{u}(\pm\infty) = 0$

.同理可得

$I_{h}(\pm\infty) = 0$

$\liminf\limits_{\xi\rightarrow+\infty} S(\xi) = \limsup\limits_{\xi\rightarrow+\infty} S(\xi)<S^{0}$

的证明类似于文献[26,定理2.8],这里略去证明. ...

... 这与(4.7)式的第二个等式矛盾.从而,

$\Delta_{u}(\lambda_{1}^{*}, c)<0$

且

$\Delta_{h}(\lambda_{1}^{*}, c)<0$

.由引理1.2,

$H(\lambda_{1}^{*})<0$

.令

$\overline{S}(\xi) = S^{0}-S(\xi)$

.由文献[26,定理3.1],存在正常数

$\omega$

使得当

$\xi\rightarrow\pm\infty$

时,

$\overline{S}(\xi){\rm e}^{-\omega\xi}\rightarrow 0$

,且

$G(\lambda) = L[g(.)](\lambda) = L[\overline{S}(\xi)(I_{u}(\xi)+\delta I_{h}(\xi)](\lambda)$

在

$[0, \lambda_{1}^{*}+\omega)$

上有定义.从而,由引理1.2,

$G(\lambda_{1}^{*})<+\infty.$

进一步地 ...

Traveling waves in a nonlocal dispersal SIR epidemic model

2015

... -27]等研究了非局部扩散流行病模型的行波解.关于传染病模型行波解的更多工作请参见专著[15],综述论文[17-18]及相关参考文献. ...

... 引理2.4的证明类似于文献[27,定理2.6],故略去. ...

Traveling waves in a nonlocal dispersal SIR epidemic with critical wave speed

2018

... 此外,我们考虑了当

$d_u = d_h$

且

$c = c*$

时行波解存在性.受文献[3, 8, 21, 28]启发,通过对超临界波速的递减序列取极限来证明.然而,由于系统(1.2)中出现了治疗项,估计

$I_{u}$

和

$I_{h}$

的有界性及S在

$-\infty$

处的边界渐近行为比较困难.我们通过一精细的分析证明了

$k_{h}I_{u}+k_{u}\delta I_{h}$

的有界性,从而解决了上述困难. ...

Spreading speed and traveling waves for nonlocal dispersal equations with degenerate monostable nonlinearity

2012

... 下述引理是文献[29,命题3.7]的特殊情形. ...

... 利用(3.11)式和文献[29,命题3.6],存在正常数

$C_{1}$

和

$C_{2}$

使得 ...

... 同理,由文献[29,命题3.6],由(3.13)式的第一个方程可得 ...

... 的两个正根.再由文献[29,命题3.6],易得存在正常数

$C_{i}(i = 2, 3, 4, 5)$

,使得 ...

Existence of traveling wave solutions for influenza model with treatment

2014

... 2014年, Zhang和Wang^[30]将扩散引入到带治疗项的流感模型^[12]中,即以下反应扩散方程 ...

... 并研究了其行波解的存在性和不存在性.其中

$S(t, x)$

$I_{u}(x, t)$

$I_{h}(x, t)$

和

$R(t, x)$

分别表示易感者,感染而未治愈者,治愈者和恢复者.

$0<\mu<1$

$0<\delta<1$

,且

$k_{u}<k_{h}$

$d_{i}(i = s, u, h, r)$

$R_0>1$

$c>c^*$

$d_u = d_h$

或

$c>c_1^*$

$d_u\neq d_h$

情形下行波解的存在性;然后利用双边拉普拉斯变换,得到

$R_0>1$

$0<c<c^*$

$d_u = d_h$

或

$0<c<c_2^*$

$d_u\neq d_h$

情形下行波解的不存在性.最后,利用Fubini定理,证明了

$R_0<1$

时行波解的不存在性. ...

... 以上两个引理的证明与文献[30]中的引理2.3、引理2.4相似,此处略去证明. ...

$c=c^{*}$ 情形下行波解的存在性

$R_{0}>1$ 情形下的不存在性

$R_{0}<1$ 情形下的不存在性

〈

〉