"仓室"模型在传染病动力学的研究中起到非常重要的作用^[1].一个经典的SIR模型具有如下的形式

$ \begin{equation} \left\{ \begin{array}{ll} \frac{{\rm d}S}{{\rm d}t}=-\beta SI, \\[3mm] \frac{{\rm d}I}{{\rm d}t} =\beta SI-\gamma I, \\[3mm] \frac{{\rm d}R}{{\rm d}t} =\gamma I, \end{array} \right. \label{eq1.1} \end{equation} $

(1.1)

其中$S$, $I$, $R$分别表示易感者, 染病者与移出者的数量, $\!\!\beta$是传播系数, $\!\!\gamma$是恢复率.若初始时刻易感者的数量为$S(0)=S_0>0$, 则$R_0=\frac{\beta S_0}{\gamma}$可作为决定疾病是否消亡的一个阈值^[1], 称为基本再生数.

模型(1.1)中假设人口是均匀分布的并且传播过程是瞬时完成.但是, 空间均匀模型不能充分的解释疾病的扩散现象, 空间影响因素的引入使模型更加符合实际^[2-4].比如, Hosono和llyas^[3]考虑了下面模型的行波解问题

$ \begin{equation} \left\{ \begin{array}{ll} \frac{\partial S(x, t)}{\partial t} =\Delta S(x, t)-\beta S(x, t)I(x, t) , \\[3mm] \frac{\partial I(x, t)}{\partial t}=d\Delta I(x, t)+\beta S(x, t)I(x, t)-\gamma I(x, t) , \end{array} \right. \label{eq1.2} \end{equation} $

(1.2)

其中$d$是感染个体的扩散率.他们证明了当$S(x, 0)=S_0$为常数且$\frac{\beta S_0}{\gamma}>1$时, 对任意$c\geq c^*=2\sqrt{d(\beta S_0-\gamma)}$, 存在满足$S(-\infty)=S_0>S(+\infty)$且$ I(\pm\infty)=0$的行波解$(S(x+ct), I(x+ct))$; 当$\frac{\beta S_0}{\gamma}\leq1$时, 方程(1.2)不存在行波解.

在传染病模型中, 经常用到的发生率是线性发生率$\beta SI$, 其适用于宿主人口密度低的情况.但在疾病的实际传播过程中, 传播形式可能是非线性的^[5].另外, 时滞(如疾病潜伏期)的影响也是不可忽视的^[6]. Bai和Wu^[7]考虑了如下模型

$ \begin{equation} \left\{ \begin{array}{ll} \frac{\partial S(x, t)}{\partial t} =d_1\frac{\partial^2 S(x, t)}{\partial x^2}-f(S(x, t))g(I(x, t-\tau)) , \\[3mm] \frac{\partial I(x, t)}{\partial t}=d_2\frac{\partial^2 I(x, t)}{\partial x^2}+f(S(x, t))g(I(x, t-\tau))-\gamma I(x, t) , \\[3mm] \frac{\partial R(x, t)}{\partial t}=d_3\frac{\partial^2 R(x, t)}{\partial x^2}+\gamma I(x, t), \end{array} \right. \label{eq1.3} \end{equation} $

(1.3)

其中$d_i(i=1, 2, 3)$表示扩散系数.在对$f, g$的某些条件下, 他们研究了方程(1.3)的行波解的存在性和不存在性问题.

在模型(1.2)和(1.3)中, 扩散是通过Laplace算子来描述的.最近的一些研究发现, 非局部扩散在描述长距离传播方面比局部扩散更符合实际^[8].近年来, 非局部扩散问题引起了数学家与生物学家的广泛关注^[9-12].

综合考虑上述因素, 本文研究如下具有一般发生率和时滞的非局部扩散SIR传染病模型

$ \begin{equation} \left\{ \begin{array}{ll} \frac{\partial S(x, t)}{\partial t} =d_1(J*S(x, t)-S(x, t))-f(S(x, t))g(I(x, t-\tau)) , \\[3mm] \frac{\partial I(x, t)}{\partial t}=d_2(J*I(x, t)-I(x, t))+f(S(x, t))g(I(x, t-\tau))-\gamma I(x, t) , \\[3mm] \frac{\partial R(x, t)}{\partial t}=d_3(J*R(x, t)-R(x, t))+\gamma I(x, t), \end{array} \right. \label{eq1.4} \end{equation} $

(1.4)

其中$S(x, t), I(x, t), R(x, t)$分别表示$x$处和$t$时刻的易感者, 感染者及恢复者的密度. $d_i>0$ $(i=1, 2, 3)$表示扩散系数, $\gamma>0$表示恢复率. $\tau\geq0$表示疾病的潜伏时间. $J*S(x, t), $ $J*I(x, t), $ $J*R(x, t)$是关于空间$x$的标准卷积. $J*S(x, t)-S(x, t), $ $J*I(x, t)-I(x, t), $ $J*R(x, t)-R(x, t)$分别表示由易感者, 感染者, 恢复者的扩散导致的净增长率.本文假定

$\rm(H_1)$   $f, g\in C((0, +\infty), (0, +\infty))$, $f(0)=g(0)=0$, $f'(S)>0$, $\forall S\geq 0$, 且$g'(I)>0, $ $ g''(I)\leq 0$, $\forall I\geq 0$;

$\rm(H_2)$  $J\in C^1( {\mathbb{R}}), J(x)=J(-x)\geq0, \int_{{\mathbb{R}}}J(x){\rm d}x=1$且$J$具有紧支集.

在$\rm(H_1)$的假设下, $f(S)g(I)$包含多种形式的发生率.如当$f(S)=S, g(I)=I$时, 发生率为线性发生率^[12].当$f(S)=S, g(I)=\frac{I}{1+\alpha I}$时, 发生率表示为具有饱和效应的标准发生率^[13].

本文考察系统(1.4)的行波解问题.因为系统(1.4)的前两个方程中不包含变量$R$, 故只需考虑下面的子系统

$ \begin{equation} \left\{ \begin{array}{ll} \frac{\partial S(x, t)}{\partial t} =d_1(J*S(x, t)-S(x, t))-f(S(x, t))g(I(x, t-\tau)) , \\[3mm] \frac{\partial I(x, t)}{\partial t}=d_2(J*I(x, t)-I(x, t))+f(S(x, t))g(I(x, t-\tau))-\gamma I(x, t). \end{array} \right. \label{eq1.5} \end{equation} $

(1.5)

受文献[12-16]的启发, 当$R_0:=\frac{f(S_{-\infty})g'(0)}{r}>1, c>c^*$时, 通过对一个截断问题使用Schauder不动点定理以及取极限的方法证明了系统(1.5)的满足$S(-\infty)=S_{-\infty}, $ $I(-\infty)=0$的行波解$(S(x+ct), I(x+ct))$的存在性.然而, 有两个方面的因素导致难以得到行波解在$+\infty$处的极限.一方面非局部扩散导致$S(\xi)$单调性的缺失; 另一方面, 发生率的非线性性导致难以直接证明$I(\xi)$的有界性.通过对行波系统做细致的分析, 我们给出了行波解在$+\infty$处极限存在的充分性条件.进一步, 利用双边拉普拉斯变换^[17], 给出了当$0 <c <c^*, R_0>1$时, 系统行波解不存在性的证明.最后, 证明了当$R_0\leq1$时, 系统行波解的不存在性.

本文的安排如下:第二节研究行波解的存在性; 第三节讨论行波解的不存在性; 最后, 在第四节, 对本文的结果做了一个简单的总结.

本节考察模型(1.5)行波解的存在性.形如$(S(x+ct), I(x+ct))$的解称为行波解.令$\xi=x+ct$, 可得系统(1.5)的行波方程

$ \begin{array}{l} \left\{ \begin{array}{ll} cS'(\xi) =d_1(J*S(\xi)-S(\xi))-f(S(\xi))g(I(\xi-c\tau)) , \\ cI'(\xi)=d_2(J*I(\xi)-I(\xi))+f(S(\xi))g(I(\xi-c\tau))-\gamma I(\xi) . \end{array} \right. %\tag{2.1} \label{eq2.1} \end{array} $

(2.1)

假设初始无病平衡点为$(S_{-\infty}, 0)$, 其中$S_{-\infty}>0$.本节的主要结论如下

定理2.1  假定$\rm(H_1)$, $\rm(H_2)$且$R_0>1$.

(ⅰ)  对任意$c>c^*$, 系统(1.5)存在一个行波解$(S(\xi), I(\xi))$, 满足

$ S(-\infty)=S_{-\infty},~~~ I(-\infty)=0. $

(ⅱ)若$\limsup\limits_{\xi\rightarrow+\infty}I(\xi) <+\infty$, 则$I(+\infty)=0$, $S(+\infty) <S_{-\infty}$.

(ⅲ)若$c>\max\{c^*, \frac{3}{2}d_2\sigma_0\}$, 其中$\sigma_0:=S_{-\infty}\int_{-\infty}^{+\infty}J(y)\mid y\mid {\rm d}y$, 则$\limsup\limits_{\xi\rightarrow+\infty}I(\xi) <+\infty$.

注2.1  由定理2.1 (ⅲ)可知对任意的$c>\max\{c^*, \frac{3}{2}d_2\sigma_0\}$, $I(\xi)$是有界的.从而当$R_0>1, $ $c^*>\frac{3}{2}d_2\sigma_0$时, 对$\forall c>c^*$, 系统(1.5)存在满足$S(-\infty)=S_{-\infty}, S(+\infty) <S_{-\infty}, I(\pm\infty)=0$的行波解.我们猜测当$c^* <c <\frac{3}{2}d_2\sigma_0$时, $I(\xi)$仍是有界的.特别地, 当$c=c^*$时, 利用逼近技术^[18]可得系统(1.5)的行波解, 但具体的渐近边界条件很难得出, 有待进一步的研究.

对模型(2.1)的第二个方程在$(S_{-\infty}, 0)$处线性化, 则有

$ cI'(\xi)=d_2(J*I(\xi)-I(\xi))+f(S_{-\infty})g'(0)I(\xi-c\tau)-\gamma I(\xi). $

得特征方程

$ \begin{array}{l} \Delta(\lambda, c)=d_2\int_{-\infty}^{+\infty}J(y)e^{-\lambda y}{\rm d}y-d_2-c\lambda+f(S_{-\infty})g'(0)e^{-\lambda c \tau}-\gamma=0. %\tag{2.2} \label{eq2.2} \end{array} $

(2.2)

易验证, 下面的结论成立.

引理2.1  假设$R_0:=\frac{f(S_{-\infty})g'(0)}{r}>1$.则存在$c^*>0$和$\lambda^*>0$, 使得(2.2)式有

$ \frac{\partial \Delta(\lambda, c)}{\partial \lambda}\Big|_{(\lambda^*, c^*)}=0 \mbox{ 且 }\Delta(\lambda^*, c^*)=0 $

成立, 并且有下列结论

(ⅰ)当$c>c^*$时, 方程$\Delta(\lambda, c)=0$有两个实根$\lambda_1(c)$, $\lambda_2(c)$且$0 <\lambda_1(c) <\lambda^* <\lambda_2(c) <+\infty$.当$\lambda\in(\lambda_1(c), \lambda_2(c))$时, $\Delta(\lambda, c) <0$; 当$\lambda\in(0, \lambda_1(c))\cup(\lambda_2(c), +\infty)$, 有$\Delta(\lambda, c)>0$.

(ⅱ)当$0 <c <c^*$时, 对所有$\lambda>0$, 都有$\Delta(\lambda, c)>0$.

证  由$R_0=\frac{f(S_{-\infty})g'(0)}{r}>1$及核函数$J$的性质, 简单的计算可得:对任意的$c>0$, 有

$ \Delta(0, c)= f(S_{-\infty})g'(0) -\gamma>0, \ \Delta(+\infty, c)=+\infty, $

$ \frac{\partial^2 \Delta(\lambda, c)}{\partial \lambda^2}=d_2\int_{-\infty}^{+\infty}J(y)e^{-\lambda y}y^2{\rm d}y+(c\tau)^2 f(S_{-\infty})g'(0)e^{-\lambda c \tau}>0. $

另外, 对给定的$\lambda>0$, 有

$ \Delta(\lambda, 0)= d_2\int_{-\infty}^{+\infty}J(y)e^{-\lambda y}{\rm d}y-d_2 +f(S_{-\infty})g'(0) -\gamma>0 , $

$ \frac{\partial \Delta(\lambda, c)}{\partial c}= -\lambda-\lambda \tau f(S_{-\infty})g'(0) <0, \ \Delta(\lambda, +\infty)=-\infty. $

由以上性质，易见结论成立.

在下面的讨论中, 总是假设$R_0 >1$和$c>c^*$.定义函数

$ S_+(\xi)=S_{-\infty}, S_-(\xi)= \max\{S_{-\infty}-\sigma e^{\alpha \xi} , 0 \}, $

$ I_+(\xi)= e^{\lambda_1\xi}, I_-(\xi)= \max\{e^{\lambda_1\xi}(1-Me^{\eta\xi}), 0 \}, $

其中$\sigma, \alpha, \eta$和$M$都是待定的正常数.

引理2.2  函数$I_+(\xi)= e^{\lambda_1\xi}$满足

$ \begin{array}{l} cI_+'(\xi)\geq d_2(J*I_+(\xi)-I_+(\xi))+f(S_{-\infty})g(I_+(\xi-c\tau))-\gamma I_+(\xi). %\tag{2.3} \label{eq2.3} \end{array} $

(2.3)

由条件$g''(u)\leq0$, $\forall u\geq0$与引理2.1可得不等式(2.3)成立.

引理2.3  对充分小的$0 <\alpha <\lambda_1$和充分大的$\sigma>S_{-\infty}$, 函数$S_-(\xi) $满足

$ \begin{array}{l} cS'_-(\xi) \leq d_1(J*S_-(\xi)-S_-(\xi))-f(S_-(\xi))g(I_+(\xi-c\tau)), \ \forall \xi\neq \frac{1}{\alpha}\ln\frac{S_{-\infty}}{\sigma}. %\tag{2.4} \label{eq2.4} \end{array} $

(2.4)

证  若$\xi>\xi_1:=\frac{1}{\alpha}\ln\frac{S_{-\infty}}{\sigma}$, 则$S_-(\xi)=0$.故(2.4)式成立.

若$\xi <\xi_1$, 则$S_-(\xi)=S_{-\infty}-\sigma e^{\alpha \xi}>0$.从而, 对$\sigma=\frac{1}{\alpha}$且$\alpha>0$充分小, 有

$ \begin{array}{l} cS'_-(\xi) -d_1(J*S_-(\xi)-S_-(\xi))+f(S_-(\xi))g(I_+(\xi-c\tau)) \\ \leq e^{\alpha\xi}\Big[-c\alpha\sigma-d_1\sigma \int_{-\infty}^{+\infty}J(y)(1-e^{-\alpha y}){\rm d}y+f(S_{-\infty})g'(0)e^{\lambda_1(\xi-c\tau)-\alpha\xi}\Big]\\ \leq e^{\alpha\xi}\Big[ -c\alpha\sigma-d_1\sigma [1-\int_{-\infty}^{+\infty}J(y)e^{-\alpha y}{\rm d}y]+f(S_{-\infty})g'(0)(\frac{S_{-\infty}}{\sigma})^{\frac{\lambda_1-\alpha}{\alpha}}\Big] \\ \leq 0. \end{array} $

即(2.4)式成立.证毕.

引理2.4  对$0 <\eta <\min \{\lambda_2-\lambda_1, \lambda_1\}$与充分大的常数$M$, 则函数$ I_-(\xi)$满足

$ \begin{array}{l} cI'_-(\xi)\leq d_2(J*I_-(\xi)-I_-(\xi))+f(S_-(\xi))g(I_-(\xi-c\tau))-\gamma I_-(\xi), \ \forall \xi\neq \frac{1}{\eta}\ln\frac{1}{M}. %\tag{2.5} \label{eq2.5} \end{array} $

(2.5)

证  若$\xi>\xi_2:=\frac{1}{\eta}\ln\frac{1}{M}$, 则$I_-(\xi)=0$, (2.5)式显然成立.

若$\xi <\xi_2$, 有$I_-(\xi)=e^{\lambda_1\xi}(1-Me^{\eta\xi})$.令$M$相当大, 使得$\xi_2 <\xi_1$成立.故当$\xi <\xi_2$时, 有$S_-(\xi)=S_{-\infty}-\sigma e^{\alpha \xi}$.此时(2.5)式等价于

$ \begin{array}{l} f(S_{-\infty})g'(0)I_-(\xi-c\tau)-f(S_-(\xi))g(I_-(\xi-c\tau))\leq-Me^{(\lambda_1+\eta)\xi} \Delta(\lambda_1+\eta, c). \label{2.05} \end{array} $

(2.6)

利用条件$g''(u)\leq0$, $\forall u\geq0$, 不难证明

$ \begin{array}{l} f(S_{-\infty})g'(0)I_-(\xi-c\tau)-f(S_-(\xi))g(I_-(\xi-c\tau))\leq I^2_-(\xi-c\tau). \end{array} $

由于$I^2_-(\xi-c\tau) < e^{2\lambda_1\xi}$, 且对足够大的$M$, 有

$ I^2_-(\xi-c\tau)e^{-\eta\xi} < e^{2\lambda_1\xi-\eta\xi} <-Me^{\lambda_1\xi} \Delta(\lambda_1+\eta, c). $

故(2.6)式成立.证毕.

令$X>\max\{\frac{1}{\alpha}\ln\frac{\sigma}{S_{-\infty}}, \frac{1}{\eta}\ln M\}$, 定义集合

$ \Gamma_X=\left\{ (\phi(\cdot), \varphi(\cdot))\in C([-X, X], {\mathbb{R}}^2)\left| \begin{array}{lll} \phi(-X)=S_-(-X), \ \varphi(-X)=I_-(-X), \\ S_-(\xi)\leq\phi(\xi)\leq S_{-\infty}, \forall\xi\in[-X, X], \\ I_-(\xi)\leq\varphi(\xi)\leq I_+(\xi), \forall\xi\in[-X, X]. \end{array} \right. \right\}. $

对任意的$(\phi(\cdot), \varphi(\cdot))\in\Gamma_X$, 将其延拓到${\mathbb{R}}$上

$ \hat{\phi}(\xi)= \left\{ \begin{array}{lll} \phi(X), &\xi>X, \\ \phi(\xi), & |\xi|\leq X, \\ S_-(\xi), & \xi <-X, \end{array} \right. , \hat{\varphi}(\xi)= \left\{ \begin{array}{lll} \varphi(X), &\xi>X, \\ \varphi(\xi), & |\xi|\leq X, \\ I_-(\xi), & \xi <-X. \end{array} \right. $

考虑初值问题

$ \begin{equation} cS'(\xi) =d_1\int_{-\infty}^{+\infty}J(y)\hat{\phi}(\xi-y){\rm d}y-d_1S(\xi)- f(\phi(\xi))g(\hat{\varphi}(\xi-c\tau)) , \label{eq2.8} \end{equation} $

(2.7)

$ \begin{equation} cI'(\xi)=d_2\int_{-\infty}^{+\infty}J(y)\hat{\varphi}(\xi-y){\rm d}y-d_2I(\xi)+ f(\phi(\xi))g(\hat{\varphi}(\xi-c\tau))-\gamma I(\xi) , \label{eq2.9} \end{equation} $

(2.8)

其中

$ \begin{array}{l} S(-X)=S_-(-X), I(-X)=I_-(-X).\label{eq2.10} \end{array} $

(2.9)

根据常微分方程理论, 可知初值问题(2.7)-(2.9)存在一个解$(S_X(\xi), I_X(\xi))$满足$S_X(\xi)\in C^1([-X, X]), $ $I_X(\xi)\in C^1([-X, X])$.定义算子$F=(F_1, F_2):\Gamma_X\rightarrow C([-X, X])$为

$ S_X(\xi)=F_1[\phi, \varphi](\xi), I_X(\xi)=F_2[\phi, \varphi](\xi). $

根据引理2.2-2.4, 容易验证下面结论成立.

引理2.5  算子$F:\Gamma_X\rightarrow\Gamma_X$.

引理2.6  算子$F:\Gamma_X\rightarrow\Gamma_X$是全连续算子.

证  首先证明$F$是连续的.由(2.7)-(2.9)式可得

$ \begin{equation} S_X(\xi) =S_-(-X)\exp\bigg\{-\frac{d_1}{c}(\xi+X)\bigg\}+\frac{1}{c} \int_{-X}^{\xi}e^{-\frac{d_1}{c}(\xi-\eta)}f_{\phi, \varphi} (\eta){\rm d}\eta , \label{eq2.11} \end{equation} $

(2.10)

$ \begin{equation} I_X(\xi)= I_-(-X)\exp\bigg\{-\frac{d_2+\gamma}{c}(\xi+X)\bigg\}+\frac{1}{c} \int_{-X}^{\xi}e^{-\frac{d_2+\gamma}{c}(\xi-\eta)}g_{\phi, \varphi} (\eta){\rm d}\eta, \label{eq2.12} \end{equation} $

(2.11)

其中

$ f_{\phi, \varphi}(\xi)=d_1\int_{-\infty}^{+\infty}J(y)\hat{\phi}(\xi-y){\rm d}y- f(\phi(\xi))g(\hat{\varphi}(\xi-c\tau)), $

$ g_{\phi, \varphi}(\xi)=d_2\int_{-\infty}^{+\infty}J(y)\hat{\varphi}(\xi-y){\rm d}y+ f(\phi(\xi))g(\hat{\varphi}(\xi-c\tau)). $

令$(\phi_i(\cdot), \varphi_i(\cdot))\in \Gamma_X$且$F_1[\phi_i, \varphi_i](\xi)=S_{X, i}(\xi), F_2[\phi_i, \varphi_i](\xi)=I_{X, i}(\xi), i=1, 2$.因为

$ \begin{array}{l} \Big|\int_{-\infty}^{+\infty}J(y)(\hat{\phi}_1(\xi-y)-\hat{\phi}_2(\xi-y)){\rm d}y\Big| \\ \leq \int_{-X}^{X}J(\xi-y)|\phi_1(y)-\phi_2(y)|{\rm d}y+ \int_{X}^{+\infty}J(\xi-y)|\phi_1(X)-\phi_2(X)|{\rm d}y\\ \leq \max\limits_{y\in[-X, X]}|\phi_1(y)-\phi_2(y)| , \end{array} $

$ \begin{array}{l} \mid f(\phi_1(\xi))g(\hat{\varphi_1}(\xi-c\tau))-f(\phi_2(\xi))g(\hat{\varphi_2}(\xi-c\tau)) \mid\\ \leq f(\phi_1(\xi))\mid g(\hat{\varphi_1}(\xi-c\tau))-g(\hat{\varphi_2}(\xi-c\tau))\mid +g(\hat{\varphi_2}(\xi-c\tau))\mid f(\phi_1(\xi))-f(\phi_2(\xi))\mid \\ \leq f(S_{-\infty})g'(0)\max\limits_{y\in[-X, X]}|\varphi_1(y)-\varphi_2(y)|+Ng'(0)e^{\lambda_1X} \max\limits_{y\in[-X, X]}|\phi_1(y)-\phi_2(y)|, \end{array} $

其中$N:=\max f'(\xi)$, 可得

$ \begin{array}{l} \mid f_{\phi_1, \varphi_1}(\xi)- f_{\phi_2, \varphi_2}(\xi) \mid \\ %\leq\mid\int_{-\infty}^{+\infty}J(\xi-y)(\hat{\phi}_1(y)-\hat{\phi}_2(y)){\rm d}y\mid +\mid % f(\phi_1(\xi))g(\hat{\varphi}_1(\xi-c\tau)-f(\phi_2(\xi))g(\hat{\varphi}_2(\xi-c\tau) \mid \\ \leq (d_1+Ng'(0)e^{\lambda_1X})\max\limits_{y\in[-X, X]}|\phi_1(y)-\phi_2(y)| +f(S_{-\infty})g'(0)\max\limits_{y\in[-X, X]}|\varphi_1(y)-\varphi_2(y)|, \end{array} $

由$F$的定义及方程(2.10)可知$F_1$是连续的.同理可得, $F_2$是连续的.

由$S_X\in C^1([-X, X])$, $I_X\in C^1([-X, X])$并结合(2.7)和(2.8)式可知:对任意$(\phi(\cdot), \varphi(\cdot)) \in \Gamma_X$, $ S_X', I_X'$都是有界的.所以$F$是紧的.证毕.

由$\Gamma_X$的定义可知, $\Gamma_X$是一个闭凸集合.利用Schauder不动点定理可知, 存在一个不动点$(S_X(\cdot), I_X(\cdot))\in \Gamma_X$, 对任意$\xi\in (-X, X)$, 都有

$ (S_X(\xi), I_X(\xi))=F[S_X, I_X](\xi). $

为得到模型(1.5)的行波解.下面给出$S_X(\cdot), I_X(\cdot)$的一些估计.

定理2.2  给定任意$Y>0$, 存在常数$C(Y)>0$, 当$X>\max\{\frac{1}{\alpha}\ln\frac{\sigma}{S_{-\infty}}, \frac{1}{\eta}\ln M\}$和$X>Y+R_1$时, 有

$ \parallel S_X \parallel_{C^{1, 1}([-Y, Y])}\leq C(Y) \mbox{ , } \parallel I_X \parallel_{C^{1, 1}([-Y, Y])}\leq C(Y), $

其中$R_1$是$J$的支撑半径.

证  显然, $(S_X(\cdot), I_X(\cdot))$满足

$ \begin{equation} \label{eq2.13} cS_X'(\xi) =d_1\int_{-\infty}^{+\infty}J(y)\hat{S}_X(\xi-y){\rm d}y-d_1S_X(\xi)- f(S_X(\xi))g(\hat{I}_X(\xi-c\tau)) , \end{equation} $

(2.12)

$ \begin{equation} \label{eq2.14} cI_X'(\xi)=d_2\int_{-\infty}^{+\infty}J(y)\hat{I}_X(\xi-y){\rm d}y-d_2I_X(\xi)+ f(S_X(\xi))g(\hat{I}_X(\xi-c\tau))-\gamma I_X(\xi). \end{equation} $

(2.13)

对$ \forall \xi\in[-Y, Y]$, 有$S_X(\xi)\leq S_{-\infty}$, $I_X(\xi)\leq I_+(Y)=e^{\lambda_1Y}$, 得

$ \begin{array}{l} \mid S'_X(\xi)\mid %&\leq\frac{d_1}{c}\mid\int_{-\infty}^{+\infty}J(y)\hat{S}_X(\xi-y){\rm d}y-S_X(\xi)\mid %+\frac{1}{c}\mid f(S_X(\xi))g(\hat{I}_X(\xi-c\tau)) \mid \\ \leq\frac{2d_1S_{-\infty}+f(S_{-\infty})g'(0)e^{\lambda_1Y}}{c}, \ \mid I'_X(\xi)\mid %&\leq\frac{d_2}{c}\mid\int_{-\infty}^{+\infty}J(y)\hat{I}_X(\xi-y){\rm d}y-I_X(\xi)\mid %+\frac{1}{c} f(S_X(\xi))g(\hat{I}_X(\xi-c\tau))+\frac{\gamma}{c}e^{\lambda_1Y} \\ &\leq\frac{2d_2+\gamma+f(S_{-\infty})g'(0)}{c}e^{\lambda_1Y}. \end{array} $

因此, 存在常数$L_1(Y)>0$, 使得对任意$\xi, \eta\in[-Y, Y]$, 有

$ \mid S_X(\xi)-S_X(\eta)\mid < L_1(Y)\mid\xi-\eta\mid , \mid I_X(\xi)-I_X(\eta)\mid < L_1(Y)\mid\xi-\eta\mid , $

结合方程(2.12), 得出

$ \begin{array}{l} c\mid S'_X(\xi)-S'_X(\eta)\mid \leq d_1\Big|\int_{-\infty}^{+\infty}(J(\xi-y)-J(\eta-y))\hat{S}_X(y){\rm d}y\Big|+d_1\mid S_X(\xi)-S_X(\eta)\mid \\ +\mid f(S_X(\xi))g(\hat{I}_X(\xi-c\tau))-f(S_X(\eta))g(\hat{I}_X(\eta-c\tau))\mid \\ =A_1+A_2+A_3. \end{array} $

由(H$_2)$可得, 核函数$J$是李普希兹连续的, 令$L$是李普希兹常数, 有

$ \begin{array}{l} A_1=d_1\Big| \int_{-\infty}^{+\infty}J(y)\hat{S}_X(\xi-y){\rm d}y-\int_{-\infty}^{+\infty}J(y)\hat{S}_X(\eta-y){\rm d}y\Big| \nonumber\\ =d_1\Big| \int_{\xi-R_1}^{\xi+R_1}J(\xi-y)\hat{S}_X(y){\rm d}y- \int_{\eta-R_1}^{\eta+R_1}J(\eta-y)\hat{S}_X(y){\rm d}y\Big| \nonumber\\ = d_1\Big| -\int_{\eta-R_1}^{\xi-R_1}J(\eta-y)\hat{S}_X(y){\rm d}y+\int_{\eta+R_1}^{\xi+R_1}J(\xi-y)\hat{S}_X(y){\rm d}y \nonumber\\ +\int_{\xi-R_1}^{\eta+R_1}(J(\xi-y)-J(\eta-y))\hat{S}_X(y){\rm d}y\Big| \nonumber\\ \leq d_1\Big(2S_{-\infty}\parallel J\parallel_{L\infty}\mid\xi-\eta\mid +\Big|\int_{\xi-R_1}^{\eta-R_1}(J(\xi-y)-J(\eta-y))\hat{S}_X(y){\rm d}y\Big| \nonumber\\ +\Big|\int_{\eta-R_1}^{\eta+R_1}(J(\xi-y)-J(\eta-y))\hat{S}_X(y){\rm d}y\Big|\Big)\nonumber \\ \leq d_1(4S_{-\infty}\parallel J\parallel_{L\infty}+2R_1LS_{-\infty})\mid\xi-\eta\mid, \label{eq2.15} \end{array} $

(2.14)

$ \begin{array}{l} A_3=\mid f(S_X(\xi))g(\hat{I}_X(\xi-c\tau))-f(S_X(\eta))g(\hat{I}_X(\eta-c\tau))\mid \nonumber\\ \leq f(S_{-\infty})g'(0)\mid \hat{I}_X(\xi-c\tau)- \hat{I}_X(\eta-c\tau)\mid +Ng'(0) \hat{I}_X(\eta-c\tau) \mid S_X(\xi)- S_X(\eta)\mid \nonumber\\ \leq f(S_{-\infty})g'(0)\max\limits_{\xi, \eta\in[-Y, Y]} \mid I_X(\xi)-I_X(\eta)\mid + Ng'(0)e^{\lambda_1Y} \mid S_X(\xi)-S_X(\eta)\mid \nonumber\\ \leq \Big[ f(S_{-\infty})g'(0)+ Ng'(0)e^{\lambda_1Y} \Big]L_1(Y)\mid\xi-\eta\mid, \label{eq2.16} \end{array} $

(2.15)

其中$N:=\max\limits_{\xi\in[-Y, Y]} f'(\xi)$.由不等式(2.14)和(2.15), 可知存在常数$C_1(Y)>0$, 对任意$\xi, \eta\in[-Y, Y]$, 有

$ \mid S'_X(\xi)-S'_X(\eta)\mid\leq C_1(Y)\mid \xi-\eta\mid. $

同理可得, 对任意$\xi, \eta\in[-Y, Y]$, 有

$ \mid I'_X(\xi)-I'_X(\eta)\mid\leq C_1(Y)\mid \xi-\eta\mid. $

所以, 对满足$R_1+Y <X$的任意$Y$(独立于$X$), 存在一个常数$C(Y)$, 使得

$ \parallel S_X \parallel_{C^{1, 1}([-Y, Y])}\leq C(Y) \mbox{ , } \parallel I_X \parallel_{C^{1, 1}([-Y, Y])}\leq C(Y) $

成立.证毕.

定理2.1(ⅰ)的证明  取满足$X_n>\max\{\frac{\ln M}{\eta}, \frac{1}{\alpha}\ln\frac{\sigma}{S_{-\infty}}\}$, $X_n>Y+R_1$, $\lim\limits_{n\rightarrow+\infty}X_n=+\infty$的序列$\{X_n\}^{\infty}_{n=1}$.对任意$n$, 存在$(S_{Xn}, I_{Xn})\in\Gamma_{Xn}$.由定理2.2知, 存在满足如下条件的子序列$\{X_{nk}\}$:当$k\rightarrow+\infty$时, $X_{nk}\rightarrow+\infty$, 且$ S_{nk}\rightarrow S \mbox{ , } I_{nk}\rightarrow I \mbox{ 在 } C^1_{loc}( {\mathbb{R}})\mbox{中成立}. $由于$J$是紧支集, 由勒贝格控制收敛定理, 得对任意$\xi\in {\mathbb{R}}$, 有

$ \lim\limits_{k\rightarrow+\infty}\int_{-\infty}^{+\infty}J(y)S_{X_{nk}}(\xi-y){\rm d}y= \int_{-\infty}^{+\infty}J(y)S(\xi-y){\rm d}y=J*S(\xi), $

$ \lim\limits_{k\rightarrow+\infty}\int_{-\infty}^{+\infty}J(y)I_{X_{nk}}(\xi-y){\rm d}y= \int_{-\infty}^{+\infty}J(y)I(\xi-y){\rm d}y=J*I(\xi). $

进一步，$(S, I)$满足系统(2.1)且$ S_-(\xi) <S(\xi) <S_{-\infty}, I_-(\xi) <I(\xi) <I_+(\xi).$由$S_-(\xi), I_-(\xi)$的定义可得: $S(-\infty)=S_{-\infty}, I(-\infty)=0$.

下面考察定理2.1(ⅰ)中所得到的行波解在$+\infty$处的渐近行为.先证明如下引理.

引理2.7   $\int_{-\infty}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi <+\infty$, 且$\liminf\limits_{\xi\rightarrow+\infty}S(\xi) <S_{-\infty}$.

证  由

$ \begin{array}{l} \int_{z}^{x}(J*S(\xi)-S(\xi)){\rm d}\xi =\int_{z}^{x}\int_{-\infty}^{+\infty}J(y)[S(\xi-y)-S(\xi)]{\rm d}y{\rm d}\xi \\ =-\int_{z}^{x}\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}S'(\xi-ty){\rm d}t{\rm d}y{\rm d}\xi \\ =\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}[S(z-ty)-S(x-ty)] {\rm d}t{\rm d}y \end{array} $

和$S(-\infty)=S_{-\infty}$, 可得:当$z\rightarrow-\infty$时, 对任意的$x\in{\mathbb{R}}$, 有

$ \begin{array}{l} \Big|\int_{-\infty}^{x}(J*S(\xi)-S(\xi)){\rm d}\xi\Big| =\Big|\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}[S_{-\infty}-S(x-ty)]{\rm d}t{\rm d}y\Big| \\ =\Big|\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}S(x+ty){\rm d}t{\rm d}y\Big| \leq \sigma_0, \end{array} $

其中$\sigma_0:=S_{-\infty}\int_{-\infty}^{+\infty}J(y)\mid y\mid{\rm d}y$.由(2.1)式的第一个方程可得

$ \begin{array}{l} \beta\int_{-\infty}^{x}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi =d_1\int_{-\infty}^{x}(J*S(\xi)-S(\xi)){\rm d}\xi+cS_{-\infty}-cS(x)\\ \leq d_1\sigma_0+cS_{-\infty}. \end{array} $

即证得$\int_{-\infty}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi <+\infty$.

使用反证法证明: $\liminf\limits_{\xi\rightarrow+\infty}S(\xi) <S_{-\infty}$.假设$\liminf\limits_{\xi\rightarrow+\infty}S(\xi)\geq S_{-\infty}$, 则$\lim\limits _{\xi\rightarrow+\infty}S(\xi)=S_{-\infty}$.所以当$x\rightarrow+\infty$时, 有

$ \int_{-\infty}^{x}(J*S(\xi)-S(\xi)){\rm d}\xi=\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}S(x+ty){\rm d}t{\rm d}y \rightarrow0. $

即得$\int_{-\infty}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi=0$.与$\int_{-\infty}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi>0$相矛盾.故假设不成立.证毕.

定理2.1(ⅱ)的证明  分两种情形证明该结论.

情形1   $\liminf\limits_{\xi\rightarrow+\infty}S(\xi)>0$.此时我们证明$I(+\infty)=0$，且存在常数$\sigma>0$, 使得$S(+\infty)=\sigma <S_{-\infty}$.

由于$\liminf\limits_{\xi\rightarrow+\infty}S(\xi)>0$, 则存在$\xi_0>0, \delta_0>0$, 使得对任意$\xi>\xi_0$, 有$S(\xi)>\delta_0$.又因为对$\forall S\geq0$, 都有$f'(S)>0$.则$f(S(\xi))>f(\delta_0)$.又有$\int_{-\infty}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi <+\infty$, 则

$ f(\delta_0)\int_{\xi_0}^{+\infty}g(I(\xi-c\tau)){\rm d}\xi < \int_{\xi_0}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi <+\infty. $

另外, 因为$S(\xi)\leq S_{-\infty}, S(-\infty)=S_{-\infty}$, 可知存在$\xi'_0>0, \delta'_0>0$, 使得对任意$\xi <\xi'_0$, 有$S_{-\infty}-\delta'_0 <S(\xi)$成立.从而$f(S_{-\infty}-\delta'_0) <f(S(\xi))$.即

$ f(S_{-\infty}-\delta'_0)\int_{-\infty}^{\xi'_0}g(I(\xi-c\tau)){\rm d}\xi <\int_{-\infty}^{\xi'_0}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi 7 +\infty. $

综上可得$\int_{-\infty}^{+\infty}g(I(\xi-c\tau)){\rm d}\xi <+\infty$, 即$\int_{-\infty}^{+\infty}g(I(\xi)){\rm d}\xi <+\infty$.

由$\limsup\limits_{\xi\rightarrow+\infty}I(\xi) <+\infty$知, 存在$M>0$使得$0\leq I(\xi)\leq M$, $\forall \xi\in{\mathbb{R}}$.由$g$的性质可知: $0 <g'(M)\leq g'(I(\xi))\leq g'(0)$, $\forall \xi\in{\mathbb{R}}$.利用中值定理可得: $g(I(\xi))=g'(\theta I(\xi))I(\xi)\geq g'(M)I(\xi)$, $\forall \xi\in{\mathbb{R}}$, 其中$\theta\in(0, 1)$.所以

$ g'(M)\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi\leq\int_{-\infty}^{+\infty}g(I(\xi)){\rm d}\xi <+\infty. $

即$\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi <+\infty$.又由$I'(\xi)$有界得$I(+\infty)=0$.

下面证明:当$\xi\rightarrow+\infty$时, $S(\xi)$的极限存在.只需证$\liminf\limits_{\xi\rightarrow+\infty}S(\xi)=\limsup\limits_{\xi\rightarrow+\infty}S(\xi).$假设$\liminf\limits_{\xi\rightarrow+\infty}S(\xi) <\limsup\limits_{\xi\rightarrow+\infty}S(\xi)$成立.则存在点列$\{\xi_n\}^{+\infty}_{n=1}$和$\{\eta_n\}^{+\infty}_{n=1}$, 满足$\lim\limits_{n\rightarrow+\infty}\xi_n=+\infty, \lim\limits_{n\rightarrow+\infty}\eta_n=+\infty$, 且

$ \lim\limits_{n\rightarrow+\infty}S(\xi_n)=\limsup\limits_{\xi\rightarrow+\infty}S(\xi):=\sigma_1\leq S_{-\infty}, \ S'(\xi_n)=0, $

$ \lim\limits_{n\rightarrow+\infty}S(\eta_n)=\liminf\limits_{\xi\rightarrow+\infty}S(\xi):=\sigma_2 <\sigma_1 , \ S'(\eta_n)=0, $

$ \lim\limits_{n\rightarrow+\infty}J*S(\xi_n)\leq\sigma_1, \ \lim\limits_{n\rightarrow+\infty}J*S(\eta_n)\geq\sigma_2>0. $

由于

$ cS'(\xi_n) =d_1(J*S(\xi_n)-S(\xi_n))-f(S(\xi_n))g(I(\xi_n-c\tau)), $

则$\lim\limits_{n\rightarrow+\infty}J*S(\xi_n)=\sigma_1$.令$S_n(y)=S(\xi_n+y)$.取很小的正数$\epsilon$, 令$\Omega_\epsilon=\Omega\cap\{\lim\limits_{n\rightarrow+\infty}S(\xi_n) < \sigma_1-\epsilon\}$, 其中$\Omega:= {\rm supp} J$.可得

$ \begin{array}{l} \sigma_1=\lim\limits_{n\rightarrow+\infty}\int_{\Omega}J(y)S_n(y){\rm d}y\\ \leq \limsup\limits_{n\rightarrow+\infty}\int_{\Omega\setminus\Omega_\epsilon} J(y)S_n(y){\rm d}y+\limsup\limits_{n\rightarrow+\infty}\int_{\Omega_\epsilon}J(y)S_n(y){\rm d}y\\ \leq \sigma_1\int_{\Omega\setminus\Omega_\epsilon} J(y){\rm d}y+(\sigma_1-\epsilon)\int_{\Omega_\epsilon}J(y){\rm d}y\\ = \sigma_1-\epsilon\int_{\Omega_\epsilon}J(y){\rm d}y, \end{array} $

则$m(\epsilon)=0$, 其中$m$表示测度.所以$S_n(y)\rightarrow\sigma_1$在$\Omega$上几乎处处成立.又因为$\{S_n\}$是等度连续的, 所以$S_n(y)\rightarrow\sigma_1$在$\Omega$上处处成立.即:对任意$y\in\Omega$, 都有$S_n(y)\rightarrow\sigma_1$.

因为$J\in C^1$, 故必存在$R'\geq\delta'>0$, 使得

$ [R'-\delta', R'+\delta']\cup[-R'-\delta', -R'+\delta']\subseteq\Omega\mbox{ 成立 }. $

令${\xi_n}^{\pm}=\xi_n\pm R'$, 则对任意$y\in\Omega$, 当$n\rightarrow+\infty$时, 有$S(\xi^\pm_n +y)\rightarrow\sigma_1$.特别的:对任意$y\in[-\delta', \delta']$, 当$n\rightarrow+\infty$时, 有$S(\xi_n+ y)\rightarrow\sigma_1$.

重复上面的过程可得:对任意$y\in[-R_1, R_1]$, 当$n\rightarrow+\infty$时, 有$S(\xi_n+ y)\rightarrow\sigma_1$, 其中$R_1$表示$\Omega$的半径.另一方面, 因为$I(+\infty)=0, g(0)=0$且

$ 0=cS'(\eta_n) =d_1(J*S(\eta_n)-S(\eta_n))-f(S(\eta_n))g(I(\eta_n-c\tau)), $

则$\lim\limits_{n\rightarrow+\infty}J*S(\eta_n)=\sigma_2$.同理可得:对任意$y\in[-R_1, R_1]$, 当$n\rightarrow+\infty$时, 有$S(\xi_n+ y)\rightarrow\sigma_2$.

当$n\rightarrow+\infty$时, 有$\int_{\eta_n}^{\xi_n}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi=0$, 对方程(2.1)的第一个式子从$\eta_n$到$\xi_n$积分, 可得

$ \begin{array}{l} 0 < c(\sigma_1-\sigma_2)\\ =c\lim\limits_{n\rightarrow+\infty}[S(\xi_n)-S(\eta_n)]\\ =\lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}d_1[J*S(\xi)-S(\xi)]{\rm d}\xi- \lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi \\ =\lim\limits_{n\rightarrow+\infty}\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}[S(\eta_n-ty)-S(\xi_n-ty)] {\rm d}t{\rm d}y\\ -\lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi\\ =0, \end{array} $

产生矛盾, 所以$\liminf\limits_{\xi\rightarrow+\infty}S(\xi)=\limsup\limits_{\xi\rightarrow+\infty}S(\xi):=\sigma <S_{-\infty}.$

情形2   $\liminf\limits_{\xi\rightarrow+\infty}S(\xi)=0$.此时我们证明$S(+\infty)=I(+\infty)=0$.首先证明$S(+\infty)=0$.假设$\limsup\limits_{\xi\rightarrow+\infty}S(\xi)>0$.则存在$\{\xi_n\}^{+\infty}_{n=1}$和$\{\eta_n\}^{+\infty}_{n=1}$满足$\lim\limits_{n\rightarrow+\infty}\xi_n=+\infty, \lim\limits_{n\rightarrow+\infty}\eta_n=+\infty$,

且

$ \lim\limits_{n\rightarrow+\infty}S(\xi_n)=\limsup\limits_{\xi\rightarrow+\infty}S(\xi):=\sigma_1>0 , S'(\xi_n)=0, $

$ \lim\limits_{n\rightarrow+\infty}S(\eta_n)=\liminf\limits_{\xi\rightarrow+\infty}S(\xi):=\sigma_2=0 , S'(\eta_n)=0. $

类似于情形1的证明，可以得出对任意$y\in[-R_1, R_1], $都有$ \lim_{n\rightarrow+\infty}S(\xi_n+y)=\sigma_1.$又因为$\limsup\limits_{\xi\rightarrow+\infty}I(\xi) <+\infty$, 则对任意$y\in[-R_1, R_1], $都有$ \lim_{n\rightarrow+\infty}S(\eta_n+y)=0.$对方程(2.1)的第一个式子从$\eta_n$到$\xi_n$积分, 可得

$ \begin{array}{l} 0 < c\sigma_1\\ =c\lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}S'(x){\rm d}x\\ =\lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}d_1[J*S(\xi)-S(\xi)]{\rm d}\xi- \lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi \\ =d_1\lim\limits_{n\rightarrow+\infty}\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}[S(\eta_n-ty)-S(\xi_n-ty)] {\rm d}t{\rm d}y\\ -\lim\limits_{n\rightarrow+\infty}\int_{\eta_n}^{\xi_n}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi =0, \end{array} $

产生矛盾, 所以$S(+\infty)=0$.

下面证明: $I(+\infty) =0$.如果$\limsup\limits_{\xi\rightarrow+\infty}I(\xi)>0$, 则

$ \limsup\limits_{\xi\rightarrow+\infty}I(\xi)=\liminf\limits_{\xi\rightarrow+\infty}I(\xi):=\sigma>0, \mbox{ 或 }\limsup\limits_{\xi\rightarrow+\infty}I(\xi)>\liminf\limits_{\xi\rightarrow+\infty}I(\xi)\geq0. $

若第一个式子成立, 则

$ \begin{array}{l} 0=c\lim\limits_{\xi\rightarrow+\infty}\int_{\xi}^{\xi+1}I'(x){\rm d}x\\ =d_2\lim\limits_{\xi\rightarrow+\infty}\int_{\xi}^{\xi+1}[J*I(x)-I(x)]{\rm d}x+ \lim\limits_{\xi\rightarrow+\infty}\int_{\xi}^{\xi+1}(f(S(x))g(I(x-c\tau))-\gamma I(x)){\rm d}x\\ =d_2 \lim\limits_{\xi\rightarrow+\infty}\int_{-\infty}^{+\infty}J(y)y\int_{0}^{1}[I(\xi-ty)-I(\xi+1-ty)] {\rm d}t{\rm d}y\\ +\lim\limits_{\xi\rightarrow+\infty}\int_{\xi}^{\xi+1}f(S(x))g(I(x-c\tau)){\rm d}x- \gamma\lim\limits_{\xi\rightarrow+\infty}\int_{\xi}^{\xi+1}I(x){\rm d}x\\ =-\gamma\sigma, \end{array} $

产生矛盾.如果第二个式子成立, 则存在点列$\zeta_n\rightarrow+\infty (n\rightarrow+\infty)$, 满足$\lim\limits_{n\rightarrow+\infty}I(\zeta_n)=\limsup\limits_{\xi\rightarrow+\infty}I(\xi):=\sigma>0$且$I'(\zeta_n)=0$.因为$\lim\limits_{n\rightarrow+\infty}S(\zeta_n)=0$, 则

$ \begin{array}{l} 0 \leq \liminf\limits_{n\rightarrow+\infty}\Big[cI'(\zeta_n)-d_2(J*I(\zeta_n) -I(\zeta_n))\Big] \\ =\liminf\limits_{n\rightarrow+\infty}[f(S(\zeta_n))g (I(\zeta_n-c\tau))-\gamma I(\zeta_n)] \\ =-\gamma\sigma, \end{array} $

产生矛盾.综上可得$I(+\infty)=0$.

定理2.1(ⅲ)的证明  因为$I_X(\xi)\in C^1([-X, X])$, 所以存在$\xi_0 \in[-X, X]$, 满足

$ I_X(\xi_0)=\max\limits_{\xi\in [-X, X]}I_X(\xi). $

将(2.13)式从$-X$到$\xi_0$积分可得

$ \begin{array}{l} cI_X(\xi_0)=d_2\int_{-X}^{\xi_0}\int_{-\infty}^{+\infty}J(y)(\hat{I}_X(\xi-y)- I_X(\xi)){\rm d}y{\rm d}\xi+\int_{-X}^{\xi_0}f(S_X(\xi))g(\hat{I}_X(\xi-c\tau)){\rm d}\xi \\ -\gamma \int_{-X}^{\xi_0}I_X(\xi){\rm d}\xi+cI_X(-X). \end{array} $

由于

$ \begin{array}{l} \int_{-X}^{\xi_0}\int_{-\infty}^{+\infty}J(y)(\hat{I}_ X(\xi-y)- I_X(\xi)){\rm d}y{\rm d}\xi \\ =\int_{-\infty}^{+\infty}J(y)\int_{-X-y}^{\xi_0-y}(\hat{I}_X(z)-I_X(z+y) ){\rm d}z{\rm d}y \\ =\Big\{\int_{-\infty}^{0}J(y)\int_{-X-y}^{X}(I_X(z)-I_X(z+y)){\rm d}z{\rm d}y \Big\} \\ +\Big\{\int_{0}^{+\infty}J(y)\int_{-X}^{\xi_0-y}(\hat{I}_X(z)- I_X(z+y)){\rm d}z{\rm d}y\Big\} \\ +\Big\{\int_{-\infty}^{0}J(y)\int_{X}^{\xi_0-y}(\hat{I}_X(z)- I_X(z+y)) {\rm d}z{\rm d}y \\ +\int_{0}^{+\infty}J(y)\int_{-X-y}^{ -X}(I_-(z)-I_X(z+y)){\rm d}z{\rm d}y\Big\} \\ =A_{I_1}+A_{I_2}+A_{I_3} , \end{array} $

且$0\leq I_X(\xi)\leq I_X(\xi_0)$, 所以

$ \begin{array}{l} A_{I_1}=\int_{-\infty}^{0}J(y)\int_{-X-y}^{X}(I_X(z)-I_X(z+y)){\rm d}z{\rm d}y\\ =\int_{-\infty}^{0}J(y)\int_{-X-y}^{X}(-y)\int_{0}^{1}I'_X(z+ty){\rm d}t{\rm d}z{\rm d}y\\ =\int_{-\infty}^{0}J(y)(-y)\int_{0}^{1}[I_X(X+ty)-I_X(-X-y+ty)]{\rm d}t{\rm d}y\\ \leq I_X(\xi_0)\int_{0}^{+\infty}J(y)y{\rm d}y, \end{array} $

$ \begin{array}{l} A_{I_2} =\int_{0}^{\xi_0+X}J(y)\int_{-X}^{\xi_0-y}(I_X(z)-I_X(z+y)){\rm d}z{\rm d}y \\ -\int_{\xi_0+X}^{+\infty}J(y)\int_{\xi_0-y}^{-X}(I_-(z)-I_X(z+y)){\rm d}z{\rm d}y\\ \leq I_X(\xi_0)\int_{0}^{\xi_0+X}J(y)y{\rm d}y+ \int_{\xi_0+X}^{+\infty}J(y)\int_{-X-y}^{-X}I_X(z+y){\rm d}z{\rm d}y\\ \leq I_X(\xi_0)\int_{0}^{+\infty}J(y)y{\rm d}y , \end{array} $

同理得

$ A_{I_3}\leq I_X(\xi_0)\int_{0}^{+\infty}J(y)y{\rm d}y+\int_{-\infty}^{\frac{1}{\eta}\ln\frac{1}{M}}I_-(y){\rm d}y. $

综上所述, 可得

$ \int_{-X}^{\xi_0}\int_{-\infty}^{+\infty}J(y)(\hat{I}_X(\xi-y)- I_X(\xi)){\rm d}y{\rm d}\xi \leq 3I_X(\xi_0)\int_{0}^{+\infty}J(y)y{\rm d}y+\int_{-\infty}^{\frac{1}{\eta}\ln\frac{1}{M}}I_-(y){\rm d}y. $

另一方面, 将(2.12)式从$-X$到$X$积分可得

$ \begin{array}{l} \int_{-X}^{X}f(S_X(\xi))g(I_X(\xi-c\tau)){\rm d}\xi =d_1\int_{-X}^{X}\int_{-\infty}^{+\infty}J(y) (\hat{S}_X(\xi-y)- S_X(\xi)){\rm d}y{\rm d}\xi\\ -cS_X(X)+cS_X(-X). \end{array} $

因为对任意的$\xi\in[-X, X]$, 都有$0\leq S_X(\xi)\leq S_{-\infty}$.同理可得

$ d_1\int_{-X}^{X}\int_{-\infty}^{+\infty}J(y)(\hat{S}_X(\xi-y)- S_X(\xi)){\rm d}y{\rm d}\xi\leq4d_1S_{-\infty}\int_{0}^{+\infty}J(y)y{\rm d}y. $

所以

$ \int_{-X}^{X}f(S_X(\xi))g(I_X(\xi-c\tau)){\rm d}\xi\leq4d_1S_{-\infty}\int_{0}^{+\infty}J(y)y{\rm d}y+ cS_{-\infty}. $

又因为$c>\frac{3}{2}d_2\sigma_0$, 所以

$ \begin{array}{l} I_X(\xi_0)& \leq&(c-\frac{3}{2}d_2\sigma_0)^{-1} \Big(d_2\int_{-\infty}^{\frac{1}{\eta}\ln\frac{1}{M}}I_-(y){\rm d}y+ \int_{-X}^{X}f(S_X(\xi))g(I_X(\xi-c\tau)){\rm d}\xi+cI_-(-X)\Big)\\ \leq C_0, \end{array} $

其中$C_0$是与$X$无关的常数.

本节考察当$0<c<c^*, R_0>1$或$c>0, R_0\leq1$时, 系统(1.5)的行波解的不存在性.

定理3.1  若$R_0>1$.则对任意$0 <c <c^*$, 系统(1.5)不存在满足边界条件$S(-\infty)=S_{-\infty}, $ $I(\pm\infty)=0, $ $S(+\infty) <S_{-\infty}$的非平凡行波解$(S(\xi), I(\xi))$.

证  用反证法证明.假设系统(1.5)存在非平凡行波解$(S(\xi), I(\xi))$.根据$g'(0)$可知, 对任意$\epsilon \in(0, g'(0))$, 存在正数$\delta_0$使得$\frac{g(I)}{I}\geq g'(0)-\epsilon, {\ } 0 <I <\delta_0$.又因为当$\xi\rightarrow -\infty$时, $f(S(\xi))\rightarrow f(S_{-\infty}), I(\xi)\rightarrow 0$.故存在$\xi_* <0$, 对任意$\xi <\xi_*$, 有

$ f(S(\xi))>\frac{f(S_{-\infty})g'(0)+\gamma}{2g'(0)} \mbox{ , }I(\xi) <\delta_0. $

所以当$\xi <\xi_*$时, 有

$ \begin{array}{l} cI'(\xi)%&=d_2(J*I(\xi)-I(\xi))+f(S(\xi))g(I(\xi-c\tau)-\gamma I(\xi)\\ \geq d_2(J*I(\xi)-I(\xi))+\frac{f(S_{-\infty})g'(0)+\gamma}{2g'(0)}(g'(0)-\epsilon)I(\xi-c\tau)-\gamma I(\xi). \end{array} $

上面不等式对任意$\epsilon$都成立, 则当$\xi <\xi_*$时

$ \begin{equation} cI'(\xi)\geq d_2(J*I(\xi)-I(\xi))+\frac{f(S_{-\infty})g'(0)+\gamma}{2}[I(\xi-c\tau)-I(\xi)]+ \frac{f(S_{-\infty})g'(0)-\gamma}{2}I(\xi). \end{equation} $

(3.1)

定义$K(\xi):=\int_{-\infty}^\xi I(x){\rm d}x$.在$-\infty$到$\xi (\xi <\xi_*)$区间上对(3.1)式积分可得

$ \begin{array}{l} cI(\xi)\geq d_2 \int_{-\infty}^\xi(J*I(t)-I(t)){\rm d}t-\frac{f(S_{-\infty})g'(0)+\gamma}{2} \int_{\xi-c\tau}^\xi I(x){\rm d}x +\frac{f(S_{-\infty})g'(0)-\gamma}{2}K(\xi). \end{array} $

因为

$ \int_{-\infty}^\xi J*I(t){\rm d}t=\int_{-\infty}^{+\infty}J(x)\int_{-\infty}^{\xi-x}I(t){\rm d}t{\rm d}x=J*K(\xi), $

则

$ \begin{equation} d_2 (J*K(\xi)-K(\xi))+\frac{f(S_{-\infty})g'(0)-\gamma}{2}K(\xi)\leq cI(\xi)+\frac{f(S_{-\infty})g'(0)+\gamma}{2}\int_{\xi-c\tau}^\xi I(x){\rm d}x. \end{equation} $

(3.2)

又因为

$ \begin{array}{l} \int_{-\infty}^\xi(J*K(t)-K(t)){\rm d}t &= &\int_{-\infty}^\xi\int_{-\infty}^{+\infty}J(x)(K(t-x)-K(t)){\rm d}x{\rm d}t\\ &= &\int_{-\infty}^\xi \int_{-\infty}^{+\infty}(-x)J(x)\int_{0}^{1}K'(t-\theta x){\rm d}\theta {\rm d}x{\rm d}t \\ &= &\int_{-\infty}^{+\infty}(-x)J(x)\int_{0}^{1}K(\xi-\theta x){\rm d}\theta {\rm d}x, \end{array} $

$ \begin{array}{l} \int_{-\infty}^\xi\int_{x-c\tau}^x I(s){\rm d}s{\rm d}x =\int_{-\infty}^{\xi-c\tau}\int_{s}^{s+c\tau}I(s){\rm d}x{\rm d}s +\int_{\xi-c\tau}^{\xi}\int_{s}^{\xi}I(s){\rm d}x{\rm d}s \\ \leq c\tau\int_{-\infty}^\xi I(s){\rm d}s=c\tau K(\xi). \end{array} $

所以对(3.2)式从$-\infty$到$\xi'$积分, 得出关于变量$\xi'$的不等式, 我们用$\xi$代替$\xi'$.可得

$ \frac{f(S_{-\infty})g'(0)-\gamma}{2}\int_{-\infty}^\xi K(s){\rm d}s \leq m_1 K(\xi)+d_2\int_{-\infty}^{+\infty}xJ(x)\int_{0}^{1}K(\xi-\theta x){\rm d}\theta {\rm d}x, $

其中$m_1=c+\frac{f(S_{-\infty})g'(0)+\gamma}{2}c\tau$.因为$xK(\xi-\theta x)$在$\theta \in [0, 1]$上是非增函数, 则$xK(\xi-\theta x)\leq xK(\xi)$.所以

$ \frac{f(S_{-\infty})g'(0)-\gamma}{2}\int_{-\infty}^\xi K(s){\rm d}s \leq m_1 K(\xi). $

又因为$K(\xi)$关于$\xi$是非减函数, 则对所有$\eta>0, \xi <\xi_*$, 有

$ \frac{f(S_{-\infty})g'(0)-\gamma}{2}\eta K(\xi-\eta)\leq m_1 K(\xi). $

所以存在充分大的正数$\eta_0$, 使得$ K(\xi-\eta_0)\leq \frac{1}{2}K(\xi), \forall \xi <\xi_* $成立.令$\mu_0=\frac{\ln2}{\eta_0}>0$和$L(x)=K(x)e^{-\mu_0 x}$.则对所有$\xi <\xi_*$, $L(\xi-\eta_0)=K(\xi-\eta_0)e^{-\mu_0 (\xi-\eta_0)} <K(\xi)e^{-\mu_0 \xi}=L(\xi)$都成立.故当$\xi\rightarrow -\infty$时, $L(\xi)$是有界的.由于

$ J*K(x)e^{-\mu_0 x}=\int_{-\infty}^{+\infty}J(x-y)e^{-\mu_0 (x-y)}K(y)e^{-\mu_0 y}{\rm d}y \leq \sup\limits_{y\in {\mathbb{R}}}L(y)\int_{-\infty}^{+\infty}J(z)e^{-\mu_0 z}{\rm d}z, $

且

$ cI'(\xi)\leq d_2(J*I(\xi)-I(\xi))+f(S_{-\infty})g'(0)I(\xi-c\tau)-\gamma I(\xi), $

则当$\xi\rightarrow -\infty$时, $I(\xi)e^{-\mu_0 \xi}$和$I'(\xi)e^{-\mu_0 \xi}$都是有界的.又因为$I(\xi)$是有界的, 则

$ \sup\limits_{\xi\in {\mathbb{R}}}\{I(\xi)e^{-\mu_0 \xi}\} <+\infty \mbox{ , } \sup\limits_{\xi\in {\mathbb{R}}}\{I'(\xi)e^{-\mu_0 \xi}\} <+\infty . $

令$\lambda\in {\Bbb C}$且$0 <{\rm Re}\lambda <\mu_0$, 定义$I$的双边拉普拉斯变换$\mathit{£}(\lambda)=\int_{-\infty}^{+\infty}I(\xi)e^{-\lambda \xi}{\rm d}\xi.$因为

$ \int_{-\infty}^{+\infty}J*I(\xi)e^{-\lambda \xi}{\rm d}\xi=\mathit{£}(\lambda)\int_{-\infty}^{+\infty}J(x)e^{-\lambda x}{\rm d}x, $

$ \begin{array}{l} d_2(J*I(\xi)-I(\xi))-cI'(\xi)+f(S_{-\infty})g'(0)I(\xi-c\tau)-\gamma I(\xi)\\ =f(S_{-\infty})g'(0)I(\xi-c\tau)-f(S_{-\infty})g(I(\xi-c\tau)), \end{array} $

则

$ \begin{equation} \Delta(\lambda, c){\mathit{£}}(\lambda)=\int_{-\infty}^{+\infty}e^{-\lambda \xi} \Big[f(S_{-\infty})g'(0)I(\xi-c\tau)-f(S_{-\infty})g(I(\xi-c\tau))\Big]{\rm d}\xi. \end{equation} $

(3.3)

考虑$F(\xi):=f(S_{-\infty})g'(0)I(\xi-c\tau)-f(S_{-\infty})g(I(\xi-c\tau)).$由$I(-\infty)=0, S(-\infty)=S_{-\infty}$结合引理2.4可知, 当$\xi\rightarrow-\infty$时, 有$f(S_{-\infty})g'(0)I(\xi-c\tau)-f(S_{-\infty})g(I(\xi-c\tau))\leq I^2(\xi-c\tau).$令$\xi\rightarrow-\infty$, 有

$ \begin{array}{l} e^{-2\mu_0\xi}F(\xi) \leq e^{-2\mu_0\xi}I^2(\xi-c\tau) =\Big(e^{-\mu_0(\xi-c\tau)}I(\xi-c\tau)\Big)^2e^{-2\mu_0c\tau}\\ \leq \Big[\sup\limits_{\xi\in {\mathbb{R}}}\{e^{-\mu_0(\xi-c\tau)}I(\xi-c\tau)\}\Big]^2 <+\infty. \end{array} $

又因为当$\xi\rightarrow+\infty$时, $e^{-2\mu_0\xi}F(\xi)$是有界的.所以$\sup_{\xi\in {\mathbb{R}}}\{e^{-2\mu_0\xi}F(\xi)\} <+\infty. $由双边拉普拉斯变换的性质^[17]可知, $\mathit{£}(\lambda)$定义域有两种情况:存在实数$\alpha>0$, 使得$\lambda\in {\Bbb C}, 0 <{\rm Re}\lambda <\alpha$并且$\lambda=\alpha$是$\mathit{£}(\lambda)$的奇异点, 或$\lambda\in {\Bbb C}, {\rm Re}\lambda>0$.由(3.3)式右边的积分在$\lambda\in {\Bbb C}, $ $0 <{\rm Re}\lambda <2\mu_0$上有定义, 且当$0 <c <c^*$时, 对所有$\lambda>0$, 都有$\Delta(\lambda, c)>0$, 可得$\mathit{£}(\lambda)$的定义域为$\lambda\in {\Bbb C}, $ Re$\lambda>0$. (3.3)式整理为

$ \int_{-\infty}^{+\infty}e^{-\lambda \xi}\Big[\Delta(\lambda, c)I(\xi)-F(\xi)\Big]{\rm d}\xi=0. $

事实上, 对$\forall0 <c <c^*$, 当$\lambda\rightarrow +\infty$时, 有$\Delta(\lambda, c)\rightarrow +\infty$.这与上式矛盾, 故假设不成立.

定理3.1得证.

定理3.2   (ⅰ)若$R_0 <1$.则对任意$c>0$, 系统(1.5)不存在满足边界条件

$ \begin{equation} S(-\infty)=S_{-\infty}, \ I(\pm\infty)=0, \ S(+\infty) <S_{-\infty} \end{equation} $

(3.4)

的非平凡行波解$(S(\xi), I(\xi))$.

(ⅱ)若$R_0=1$.则对任意$c>0$, 系统(1.5)不存在满足(3.4)式与$ \sup_{{\mathbb{R}}} S(\xi)\leq S_{-\infty}$的非平凡行波解$(S(\xi), I(\xi))$.

证   (ⅰ)反证法.假设系统(1.5)存在满足(3.4)的非平凡行波解$(S(\xi), I(\xi))$.当$R_0=\frac{f(S_{-\infty})g'(0)}{r} <1$时, 对(2.1)式的第二个式子在${\mathbb{R}}$上积分, 得

$ \begin{array}{l} c\int_{-\infty}^{+\infty}I'(\xi){\rm d}\xi =d_2\int_{-\infty}^{+\infty}(J*I(\xi)-I(\xi)){\rm d}\xi+ \int_{-\infty}^{+\infty}f(S(\xi))g(I(\xi-c\tau)){\rm d}\xi\\ -\gamma\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi. \end{array} $

因为$I(\pm\infty)=0$, 则

$ \gamma\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi\leq d_2\int_{-\infty}^{+\infty}J*I(\xi){\rm d}\xi- d_2\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi+f(S_{-\infty})g'(0)\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi. $

即

$ \int_{-\infty}^{+\infty}I(\xi){\rm d}\xi \leq\frac{d_2}{\gamma+d_2-f(S_{-\infty})g'(0)}\int_{-\infty}^{+\infty}J*I(\xi){\rm d}\xi <\int_{-\infty}^{+\infty}I(\xi){\rm d}\xi, $

产生矛盾.

(ⅱ)同样采用反证法.假设系统(1.5)存在满足(3.4)式与$ \sup_{{\mathbb{R}}} S(\xi)\leq S_{-\infty}$的非平凡行波解$(S(\xi), I(\xi))$.当$R_0=1$即$f(S_{-\infty})g'(0)=r$时, 由(2.1)式的第二个方程可得

$ cI'(\xi)=d_2(J*I(\xi)-I(\xi))+f(S(\xi))g(I(\xi-c\tau))-f(S_{-\infty})g'(0)I(\xi). $

因为$I(\pm\infty)=0$, 将上式在${\mathbb{R}}$上积分, 得

$ \int_{-\infty}^{+\infty}\Big[f(S(\xi))g(I(\xi-c\tau))-f(S_{-\infty})g'(0)I(\xi-c\tau)\Big]{\rm d}\xi=0. $

由$f(0)=0, $ $f'(\cdot)>0$及$S(\cdot)\leq S_{-\infty}$, 可得$0\leq f(S(\xi))\leq f(S_{-\infty})$.又因为$0\leq g(I(\xi))\leq g'(0)I(\xi)$, 从而

$ f(S(\xi))g(I(\xi-c\tau))\leq f(S_{-\infty})g'(0)I(\xi-c\tau). $

由$f, g, S, I$的连续性可得

$ f(S(\xi))g(I(\xi-c\tau))- f(S_{-\infty})g'(0)I(\xi-c\tau)=0. $

因为行波解是非平凡的，存在$\xi_1\in{\mathbb{R}}$使得$I(\xi_1-c\tau)>0$.因为$f(S_{-\infty})>0$, 由上式有

$ f(S(\xi_1))=f(S_{-\infty}) \mbox{ , } g(I(\xi_1-c\tau))=g'(0)I(\xi_1-c\tau). $

由$f$的严格单调性知$S(\xi_1)=S_{-\infty}$.又由$S(\cdot)\leq S_{-\infty}$，得$S'(\xi_1)=0.$进一步可得

$ \begin{array}{l} 0=cS'(\xi_1)=d_1(J*S(\xi_1)-S(\xi_1))-f(S(\xi_1))g(I(\xi_1-c\tau))\\ \leq -f(S(\xi_1))g(I(\xi_1-c\tau)) <0, \end{array} $

产生矛盾.证毕.

本文考察了一类具有非线性发生率与时滞的非局部扩散SIR模型的行波解.由定理2.1, 3.1, 3.2可知, 当$R_0>1$且传播速度较大时, 系统(1.5)存在从无病平衡点到无病平衡点的行波解.当$0 <c <c^*, R_0>1$或$c>0, R_0\leq1$时, 系统(1.5)不存在行波解.

下面考虑时滞对最小波速的影响.由

$ \frac{\partial \Delta(\lambda, c)}{\partial \lambda}\Big|_{(\lambda^*, c^*)}=0 \mbox{ 和 }\Delta(\lambda^*, c^*)=0, $

简单的计算可得

$ \frac{{\rm d} c^*}{{\rm d}\tau}=-\frac{c^*f(S_{-\infty})g'(0)}{e^{\lambda^* c^* \tau}+\tau f(S_{-\infty})g'(0)} <0, \ \frac{{\rm d} c^*}{{\rm d}d_2}=\frac{\int_{-\infty}^{+\infty}J(y)e^{- \lambda y}{\rm d}y-1}{\lambda^* (1+ \tau f(S_{-\infty})g'(0)e^{-\lambda^* c^* \tau})}>0, $

所以, 疾病的潜伏期$\tau$的增大会导致$c^*$的减小, 而$d_2$的增大会导致$c^*$的增大.因此可得时滞可以降低疾病的传播速度, 染病个体的移动可以加快疾病的传播速度.

与行波解的最小波速密切相关的一个问题是渐近传播速度，我们猜测$c^*$是系统(1.5)的渐近传播速度.但由于该系统不满足比较原理，因此证明$c^*$是其渐近传播速度比较困难的, 这也是我们拟进一步研究的问题.