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

(1.1) $ \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} $

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

(1.2) $ \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} $

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

(1.3) $ \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} $

及边界条件

(1.4) $ \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} $

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

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

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

$ \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.

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) $ \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} \end{array}\right. \end{equation} $

且满足边界条件

(2.2) $ \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} $

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

(2.3) $ \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} $

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

(2.4) $ \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} $

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.

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

(2.5) $ \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} $

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.

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

(2.6) $ \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} $

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.

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

(3.1) $ \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} $

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

(3.2) $ \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} $

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

(3.3) $ \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} $

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

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

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

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

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

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

下面证明

(3.7) $ \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} $

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

(3.8) $ \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} $

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

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

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

(3.10) $ \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} $

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

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

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

(3.12) $ \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} $

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

(3.13) $ \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} $

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

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

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

(3.15) $ \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} $

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

(3.16) $ \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} $

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

(3.17) $ \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} $

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.

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.

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

(4.1) $ \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} $

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

(4.2) $ \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} $

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

(4.3) $ \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} $

同理可得

(4.4) $ \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} $

将(4.3)式和(4.4)式相加,有

(4.5) $ \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} $

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

(4.6) $ \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} $

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

(4.7) $ \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} $

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

(4.8) $ \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} $

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.

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

(4.9) $ \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} $

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.