生态学和传染病学是两个重要的研究领域.生态流行病学是理论生物学的 一个比较新的研究分支,通过处理生态学和传染病学的问题来处理这种情 况. 它可以被看做是生态学中的捕食者-食饵(竞争)模型和传染病学中 的SI,SIS或SIRS模型的结合. 了解捕食者与食饵的动力学关系是种群 动力学的主要目的之一. 传染病在捕食者-食饵系统中也有很重要的 影响^{[1, 2, 3, 4]}. 捕食者-食饵的另一个重要关系是捕食者捕食食饵 的捕食率,也就是捕食者的功能性反应. 文献^[5]研究了一个 具有传染率,疾病发作率等等的捕食者功能性反应的寄主-寄生虫-捕食者 模型动力学行为. 根据下面的假设生态流行病模型 $$\left\{ \begin{array}{l} \frac{{\rm d}S}{{\rm d}t}=rS(1-\frac{S+I}{K})-\lambda IS-h(S)P,\\ [3mm] \frac{{\rm d}I}{{\rm d}t}=\lambda IS-\beta IP-\mu I,\\ [3mm] \frac{{\rm d}P}{{\rm d}t}=-\theta\beta IP-\delta P+\theta h(S)P. \end{array} \right. (1.1)$$ 系统(1.1)在初值条件 $S(0)>0; ~I(0)>0;~ P(0)>0$ 情况下进行分析. 系统的平衡点已经由文献^[5]给出,也是系统稳定性的条件.

众所周知,扩散在决定捕食者-食饵模型的动力学行为中扮演重 要的角色^{[6, 7, 8]}. 因为种群密度通常不是齐次分布,实际上捕食者和 食饵的自然发展策略是幸存. 扩散可以描述捕食者和食饵种群密度不同时 种群的移动也不相同这一个复杂情况. 文献^[]研究了一个具有疾 病和扩散的Holling I捕食者-食饵模型. 文献^[11]研究了一个具有疾 病和扩散的比率依赖捕食者-食饵模型.

考虑到在一个固定的有界区域$\Omega$内,对于任意给定的时刻 $t$种群的分布不均匀,这里$\Omega\subset {\Bbb R}^{N},(N\geq1)$.而且, 根据下面的假设,$h(u_1)$是系统(1.1)的Holling II功能性反应项. 对系统(1.1)进行无量纲化 $u_1=\frac{S}{K}$,$u_2=\frac{I}{K}$, $u_3=\frac{P}{K}$,$\tau=\lambda Kt$,$b=\frac{r}{\lambda K}$, $d=\frac{\beta}{\lambda}$,$e=\frac{\mu}{\lambda K}$, $g=\frac{\delta}{\lambda K}$,$l=\frac{K}{a}$, $m=\frac{\alpha}{\lambda a}.$ 我们把$\tau$仍记为$t$,得到 下面反应扩散类型的PDE系统 $$\left\{\begin{array}{ll} u_{1t}-d_{1}\triangle u_{1}=bu_{1}[1-(u_{1}+u_{2})]-u_{1}u_{2}-\frac{mu_1u_3}{1+lu_1},\\ [3mm] u_{2t}-d_{2}\triangle u_{2}=u_{1}u_{2}-du_{2}u_{3}-eu_{2}, &x\in\Omega,t>0,\\ [3mm] u_{3t}-d_{3}\triangle u_{3}=\frac{mu_1u_3}{1+lu_1}-du_{2}u_{3}-gu_{3},\\ [3mm] \frac{\partial u_{1}}{\partial v}=\frac{\partial u_{2}}{\partial v}=\frac{\partial u_{3}}{\partial v}=0,&x\in\partial\Omega,t>0,\\ [2mm] u_{1}(x,0),u_{2}(x,0),u_{3}(x,0)\geq0, &x\in\Omega, \end{array} \right.(1.2)$$ 这里$\triangle$是Laplace算子,$\Omega$是${\Bbb R}^{N}$中边界光滑的区域, $\nu$是$\Omega$的单位外法向量. 正常数$d_{1},~d_{2},~d_{3}$ 是扩散系数,并且有初值$u_{i}(x,0)~(i=1,2,3)$是连续函数. 通 过文献^[5]可知系统(1.2)存在五个平衡点,分别为 $E_0=(0,0,0),$ $E_1=(1,0,0),$ $E_2=(e,\frac{b(1-e)}{b+1},0),$ $E_3=(\frac{g}{m-gl},0,\frac{b(m-gl-g)}{(m-gl)^2}),$ $E_{*}=(u_1^{*},u_2^{*},u_3^{*}),$ 其中 $$u_2^{*}=\frac{1}{d}\bigg[\frac{mu_1^{*}}{1+lu_1^{*}}-g\bigg], ~~~~ u_3^{*}=\frac{u_1^{*}-e}{d},$$ 其中$u_1^{*}$满足方程 $$bdl{u_1^{*}}^2-[bd(l-1)+(b+1)gl-m(b+2)]u_1^{*}-[me+(b+1)g+bd]=0. (1.3)$$ 平衡点$E_0$,$E_1$对于所有的参数值总是存在的; $E_2$存在当且仅当$e<1$; $E_3$存在当且仅当$m>g(l+1)$; 系统的唯一正平衡点$E_{*}$存在当且仅当下列条件成立 $$g(l+1)<m<\frac{gl(b+1)+bd(l-1)}{b+2},~~~~ u_1^{*}>\max\bigg\{e,\frac{g}{m-gl}\bigg\},~~e<1.(1.4)$$

本文的主要目的之一是研究系统(1.2)的正稳态解 的存在性和不存在性,也就是下面椭圆系统的非常值正解的存在性与不存在性 $$\left\{\begin{array}{ll} -d_{1}\triangle u_{1}=bu_{1}(1-u_{1})-(b+1)u_{1}u_{2}-\frac{mu_{1}u_{3}}{1+lu_1},\\ [3mm] -d_{2}\triangle u_{2}=u_{1}u_{2}-du_{2}u_{3}-eu_{2}, &x\in\Omega,\\ [3mm] -d_{3}\triangle u_{3}=\frac{mu_{1}u_{3}}{1+lu_1}-du_{2}u_{3}-gu_{3},\\ [3mm] \frac{\partial u_{1}}{\partial {\nu}}=\frac{\partial u_{2}}{\partial {\nu}}=\frac{\partial u_{3}}{\partial {\nu}}=0,&x\in\partial\Omega. \end{array} \right.(1.5)$$

文章剩余部分大体由以下几部分构成. 在第二部分, 通过线性化局部分析证明非负平衡点的局部渐进稳定性. 在第三部分, 对正稳态解的上下界进行先验估计. 在第四部分, 通过构造适当的Lyapunov函数证明了非负平衡点的全局渐进稳定性. 在第五、 六部分,我们研究了非常值正解的存在性和不存在性.

文献^[5]中, 作者研究了系统(1.1)中非负平衡点的稳定性.本文研究系统(1.2)非负 平衡点的稳定性. 令$0=\mu_{1}<\mu_{2}<\mu_{3}<\cdots$是 $-\Delta$在$\Omega$中的特征根,并且具有齐次Neumann边界条件,令 $$X=\bigg\{(u_1,u_2,u_3)\in[C^{1}(\bar\Omega)]^{3}|\frac{\partial u_1}{\partial\nu}=\frac{\partial u_2}{\partial\nu}=\frac{\partial u_3}{\partial\nu}=0,~x\in\partial\Omega\bigg\},$$ 且$X=\bigoplus\limits_{i=1}^{\infty}X_{i}$,这里 $X_i$是$\mu_{i}$对应的特征空间.~

对系统(1.2)的右端求导可得 $$\begin{eqnarray*} &&G_{u}(\cdot)=: \left( \begin{array}{ccc} a_{11}&~~a_{12}~~&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{array} \right) \\ &=&\left( \begin{array}{ccc} b-2bu_{1}-(b+1)u_{2}-\frac{mu_{3}}{1+lu_{1}}+\frac{mu_{1}u_{3}}{(1+lu_{1})^{2}} &-(b+1)u_{1}& -\frac{mu_{1}}{1+lu_{1}}\\ [2mm] u_{2}&~u_{1}-du_{3}-e~&-du_{2}\\[2mm] \frac{mu_{3}}{1+lu_{1}}-\frac{mu_{1}u_{3}}{(1+lu_{1})^{2}}& -du_{3}& \frac{mu_{1}}{1+lu_{1}}-du_{2}-g\end{array} \right). \end{eqnarray*}$$

定理2.1    系统(1.2)的平衡点具有以下性质

(i)~ $E_0$是不稳定的;

(ii)~ 当$e>1,m<g(1+l)$成立时,$E_1$是渐进稳定的;

(iii)~ 当$m<\frac{g(1+le)}{e}$成立时,$E_2$是渐进稳定的;

(iv)~ 当$\frac{g(1+el)}{e}<m<\frac{gl(1+l)}{l-1}, \frac{l-1}{2l}<e<1,$ 或 $g(1+l)<m<\frac{gl(1+l)}{l-1}, e>1$成立时,$E_3$是渐进稳定的.

证    下面仅对(ii)进行证明,其余的证明类似.

(ii)~ 令 $D={\rm diag}(d_{1},d_{2},d_{3}),L=D\Delta+G_{u}(E_1).$ 系统(1.2)在$E_1$的线性化系统为 $u_{t}=Lu.$ 对于任意的$i\geqslant 1$,$X_i$在算子$L$下是不变的. 这意味着$\lambda$是$L$ 在$X_i$上的特征值当且仅当$\lambda$是矩阵 $-\mu_iD+f_u(u_1,u_2,u_3)$的特征值. $-\mu_iD+f_u(u_1,u_2,u_3)$的特征多项式为 $$\Psi_i(\lambda)=\lambda^3+B_{1i}\lambda^2+B_{2i}\lambda+B_{3i},$$ 这里 $$B_{1i}=\mu_i(d_1+d_2+d_3)+A_1,$$ $$B_{2i}=\mu_i^2(d_1d_2+d_1d_3+d_1d_3) -\mu_i[d_1(a_{22}+a_{33})+d_3(a_{11}+a_{22})+d_2(a_{11}+a_{33})]+A_2,$$ $$\begin{eqnarray*} B_{3i}&=&\mu_i^3d_1d_2d_3-\mu_i^2(d_1d_2a_{33}+d_2d_3a_{11} +d_1d_3a_{22})\\ &&+ \mu_i(d_1a_{22}a_{33}+d_{2}a_{11}a_{33}+d_3a_{11}a_{22})+A_{3}, \end{eqnarray*}$$ $$A_1=-(a_{11}+a_{22}+a_{33}),$$ $$A_2=a_{11}a_{22}+a_{22}a_{33}+a_{11}a_{33},$$ $$A_3=-\det\{G_u(E_1)\},$$ 其中 $$\left\{ \begin{array}{l} a_{11}=-b<0,~~a_{12}=-(b+1)<0, ~~a_{13}=-\frac{m}{1+l}<0,\\[2mm] a_{21}=0,~~a_{22}=1-e<0,~~a_{23}=0,\\ [2mm] a_{31}=0,~~a_{32}=0,~~a_{33}=\frac{m}{1+l}-g<0,\end{array} \right. (2.1)$$ 所以有$A_{1}>0,A_{2}>0,A_{3}>0,B_{1i}>0,B_{3i}>0,$ $$B_{1i}B_{2i}-B_{3i}=M_1\mu_i^3+M_2\mu_i^2+M_3\mu_i+A_1A_2-A_3,$$ 其中 $$M_1=(d_2+d_3)(d_1d_2+d_1d_3+d_2d_3)+d_1(d_1d_2+d_1d_3)>0,$$ $$\begin{eqnarray*} M_2&=&-[a_{11}(d_1d_2+d_2d_3)+a_{22}(d_1d_2+d_2d_3)+a_{33}(d_1d_3+d_2d_3)]\\ && -(d_1+d_2+d_3)[d_1(a_{22}+a_{33})+d_2(a_{11}+a_{33})+d_3(a_{11}+a_{22})]>0, \end{eqnarray*}$$ $$\begin{eqnarray*} M_3&=&2(d_1+d_2+d_3)(a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33})\\ &&+(d_2+d_3)a_{11}^{2}+(d_1+d_3)a_{22}^{2}+(d_1+d_2)a_{33}^{2}>0, \end{eqnarray*}$$ $$A_1A_2-A_3>0.$$

于是,对于任意的$i\geq 0,B_{1i}B_{2i}-B_{3i}>0,$则由Routh-Hurwitz判据可知 $\Psi_i(\lambda)=0$的三个根$\lambda_{i,1},\lambda_{i,2}, \lambda_{i,3}$都具有负实部.

下面证明存在一个正常数$\delta$使得 $${\rm Re}\{\lambda_{i,1}\},{\rm Re}\{\lambda_{i,2}\},{\rm Re}\{\lambda_{i,3}\}\leqslant-\delta, \forall i\geq 1. (2.2)$$ 因此,令$\lambda=\mu_{i}\zeta$,则 $$\Psi_i(\lambda)=\mu_{i}^3\zeta^3+B_{1i}\mu_{i}^2\zeta^2+B_{2i}\mu_{i}\zeta+B_{3i}\triangleq\tilde{\Psi_i}(\zeta).$$ 因为当$i\rightarrow\infty$时,$\mu_{i}\rightarrow\infty,$故 $$\lim_{i\rightarrow+\infty}\{\tilde{\Psi_i}(\zeta)/\mu_{i}^3\} =\zeta^3+(d_1+d_2+d_3)\zeta^2+(d_1d_2+d_2d_3+d_3d_1)\zeta+d_1d_2d_3\triangleq\bar{\Psi}(\zeta).$$ 应用Routh-Hurwitz判据可得$\bar{\Psi}(\zeta)=0$的三个根都具有负实部. 因此,存在一个正常数$\bar{\delta}$使得 $${\rm Re}\{\zeta_1\}, ~{\rm Re}\{\zeta_2\},~{\rm Re}\{\zeta_3\}\leqslant-\bar{\delta}.$$ 由连续性, 我们有$\exists i_{0},$使得$\tilde{\Psi_i}(\zeta)=0$的三个根$\zeta_{i,1},\zeta_{i,2}, \zeta_{i,3}$满足 ${\rm Re}\{\zeta_{i,1}\},$ ${\rm Re}\{\zeta_{i,2}\},$ ${\rm Re}\{\zeta_{i,3}\}\leqslant-\frac{\bar{\delta}}{2},$ $\forall i\geqslant i_{0}.$ 从而有 ${\rm Re}\{\lambda_{i,1}\}, ~{\rm Re}\{\lambda_{i,2}\}, ~{\rm Re}\{\lambda_{i,3}\}\leqslant-\frac{\mu_{i}\bar{\delta}}{2},$$~\forall i\geqslant i_{0}.$ 令$-\tilde{\delta}= \max_{1\leqslant i\leqslant i_{0}}\{{\rm Re}\{\lambda_{i,1}\},$ $~{\rm Re}\{\lambda_{i,2}\},~{\rm Re}\{\lambda_{i,3}\}\},$ 则$\tilde{\delta}>0,$当$\delta=\min\{\tilde{\delta},~\frac{\bar{\delta}}{2}\}$ 时, (2.2)式成立.因此, $L$的包含特征值的谱位于$\{{\rm Re}\lambda\leqslant-\delta\},$ 则$E_1$的局部稳定性得证.

注 正平衡点$E_*$的稳定性需要进一步研究, 这里只给出关于$G_u(E_*)$的一些结果. 通过直接计算有 $$a_{11}=\frac{u_1^{*}}{d(1+lu_1^{*})^2}[-bdl^2{u_1^{*}}^2+u_1^{*}(m-2bdl)-(bd+me)].$$ 而且,$a_{11}<0$等价于 $$-bdl^2{u_1^{*}}^2+u_1^{*}(m-2bdl)-(bd+me)<0. (2.3)$$ $$\left\{ \begin{array}{l} a_{11}<0,~~a_{12}=-(b+1)u_1^{*}<0,~~a_{13}=-\frac{mu_1^{*}}{1+lu_1^{*}}<0,\\ [3mm] a_{21}=u_2^{*}>0,~~a_{22}=u_1^{*}-d\frac{u_1^{*}-e}{d}-e=0,~~a_{23}=-du_2^{*}<0,\\ [3mm] a_{31}=\frac{m(u_1^{*}-e)[1+(l-1)u_1^{*}]}{d(1+lu_1^{*})^2}>0,~~a_{32}=-du_3^{*}<0,\\ [3mm] a_{33}=\frac{mu_1^{*}}{1+lu_1^{*}}-d\frac{1}{d}[\frac{mu_1^{*}}{1+lu_1^{*}}-g]-g=0. \end{array} \right.(2.4)$$ 同时有 $\det G_{u}(E_*)=-a_{11}a_{23}a_{32}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}>0.$

这部分的主要目的是给出系统(1.5)正解的上下界的先验估计. 为了这个目的,我们首先引用下面已知的一些结果:

引理3.1    ^[12] 令$\omega\in C^{2}(\Omega)\cap C^{1}(\bar{\Omega})$是$\triangle \omega(x)+c(x)\omega(x)=0$的解,其中$c\in C(\bar{\Omega})$, $\omega$满足齐次Neumann边界条件,则存在一个正常数$C^{\ast}$, 其中$C^{\ast}$只依赖于$\alpha$,当$\|c\|_{\infty}\leq\alpha$时, 有$\max{\Omega}\bar{\omega}\leq C^{\ast}\min_{\bar{\Omega}}\omega$.

引理3.2    ^[13] 令$g(x,\omega)\in C(\Omega\times R^{1}),b_{j}(x)\in C(\bar\Omega),j=1,2,\cdots N,$

(I)~ 如果$\omega(x)\in C^{2}(\Omega)\cap C^{1}(\bar{\Omega})$在 $\Omega$上满足$\triangle \omega(x)+\sum\limits_{j=1}^N b_{j}(x)\omega_{x_{j}}+g(x,\omega(x)) \geq0$,在$\partial\Omega$上满足$\partial\omega/\partial\nu\leq0$ 且$\omega(x_{0})=\max_{\bar{\Omega}}\omega$,则$g(x_{0}, \omega(x_{0}))\geq0$.

(II)~ 如果$\omega(x)\in C^{2}(\Omega)\cap C^{1}(\bar{\Omega})$ 在$\Omega$上满足$\triangle \omega(x)+\sum\limits_{j=1}^N b_{j}(x) \omega_{x_{j}}+g(x,\omega(x))\leq0$,在$\partial\Omega$上满足 $\partial\omega/\partial\nu\geq0$且$\omega(x_{0})= \min_{\bar{\Omega}}\omega$,则$g(x_{0},\omega(x_{0}))\leq0$.

为了简便,定义常数$(d_{1},d_{2},d_{3})=d_*$,$(b,~d_*,~e,g, m_1)=\Lambda$.下面给出系统(1.5)正稳态解的有界性.

定理3.1    (上界) 如果$(u_1,u_2,u_3)$ 是系统(1.5)的正解, 则 $$\max_{\bar{\Omega}}u_{1}(x)\leqslant1, \max_{\bar{\Omega}}u_{2}(x)\leqslant\frac{4ed_{1}+bd_{2}} {4ed_{2}(b+1)}, \max_{\bar{\Omega}}u_{3}(x)\leqslant\frac{4gd_{1}+bd_{3}}{4gd_{3}}. (3.1)$$

证    首先直接对系统(1.5)用最大值原理有 $-d_{1}\triangle u_{1}=bu_{1}(1-u_{1})-(b+1)u_{1}u_{2} -\frac{mu_{1}u_{3}}{1+lu_1}\leqslant{bu_{1}(1-u_{1})}$ 使得$\max_{\bar{\Omega}}u_{1}(x)\leqslant1.$

令$\omega_{1}=d_{1}u_{1}+(b+1)d_{2}u_{2},$ 则 $$\left\{\begin{array}{ll} -\triangle\omega_{1}=bu_{1}(1-u_{1})- \frac{mu_{1}u_{3}}{1+lu_1}-d(b+1)u_{2}u_{3}-e(b+1)u_{2}, ~~&x\in\Omega,\\[3mm] \frac{\partial \omega_{1}}{\partial{\nu}}=0, &x\in\partial\Omega.\end{array} \right.(3.2)$$ 如果$\omega_{1}(x_{0})=\max_{\Omega}\omega_{1}(x)$,由引理3.2可得 $$e(b+1)u_{2}(x_{0})\leqslant{bu_{1}(1-u_{1})-\frac{mu_{1}u_{3}}{1+lu_1}-d(b+1)u_{2}u_{3}}\leqslant{\frac{b}{4}},$$ 进而可得 $$(b+1)d_{2}\max_{\bar{\Omega}}u_{2}(x)\leqslant{\max_{\Omega}\omega_{1}(x)}= d_{1}u_{1}(x_{0})+(b+1)d_{2}u_{2}(x_{0})\leqslant{d_{1}+{\frac{bd_{2}}{4e}}}=\frac{4ed_{1}+bd_{2}}{4e},$$ 因此 $$\max_{\bar{\Omega}}u_{2}(x)\leqslant\frac{4ed_{1}+bd_{2}}{4ed_{2}(b+1)}.$$ 令$\omega_{2}=d_{1}u_{1}+d_{3}u_{3},$ 我们有 $$\left\{\begin{array}{ll} -\triangle\omega _{2}=bu_{1}(1-u_{1})-(b+1)u_{1}u_{2}-du_{2}u_{3}-gu_{3}, ~~&x\in\Omega,\\[2mm] \frac{\partial \omega_{2}}{\partial {\nu}}=0, &x\in\partial\Omega. \end{array} \right. (3.3)$$ 如果$\omega_{2}(x_{0})=\max_{\Omega}\omega_{2}(x)$, 由引理3.2可得 $$gu_{3}(x_{0})\leqslant{bu_{1}(1-u_{1})-(b+1)u_{1}u_{2}-du_{2}u_{3}}\leqslant{\frac{b}{4}} ,$$ 进而可得 $$d_{3}\max_{\bar{\Omega}}u_{3}(x)\leqslant{\max_{\Omega}\omega_{2}(x)}= d_{1}u_{1}(x_{0})+d_{3}u_{3}(x_{0})\leqslant{d_{1}+{\frac{bd_{3}}{4g}}}=\frac{4gd_{1}+bd_{3}}{4g},$$ 因此 $$\max_{\bar{\Omega}}u_{3}(x)\leqslant\frac{4gd_{1}+bd_{3}}{4gd_{3}}~.$$ 证毕.

定理3.2     (下界) 设$\Lambda,\underline{d}_{1}, \underline{d}_{2},\underline{d}_{3}$是正数,如果 $$\frac{A}{b}<1-e,~(b+1)A^2-4bl[mB-b+A(b+1)]<0,~(3.4)$$ 其中 $A=\frac{4ed_{1}+bd_{2}}{4ed_{2}(b+1)} ,$ $B=\frac{4gd_{1}+bd_{3}}{4gd_{3}}, ~(d_{1},d_{2},d_{3})\in{{[\underline{d}_{1},{\infty})} \times{[\underline{d}_{2},{\infty})}\times{[\underline{d}_{3}, {\infty})}}.$ 那么,存在一个正常数$\underline{C}=\underline{C}(\Lambda,\underline{d}_{1},\underline{d}_{2},\underline{d}_{3})$,使得系统(1.5)的正解$u_1,u_2,u_3$满足 $$\min_{\bar{\Omega}}u_{i}>\underline{C} ,~i=1,2,3.$$

证     令 $$c_{1}(x)=d_{1}^{-1} \bigg[b(1-u_{1})-(b+1)u_{2}-\frac{mu_{3}}{1+lu_1}\bigg],$$ $$c_{2}(x)=d_{2}^{-1}[u_{1}-du_{3}-e] ,$$ $$c_{3}(x)=d_{3}^{-1}\bigg[\frac{mu_{1}}{1+lu_1}-du_{2}-g\bigg].$$ 由(3.1)式可知,当$d_{1},d_{2},d_{3}>\overline{d}$时, 存在一个常数$\overline{C}(\overline{d}, \Lambda)$使得$\|c_{1}\|_{\infty},\|c_{2}\|_{\infty}, \|c_{3}\|_{\infty}\leqslant\overline{C}$. 由于$u_{1},u_{2},u_{3}$ 满足$\triangle{u_{i}}+c_{i}(x)u_{i}=0,\frac{\partial u_{i}}{\partial {\nu}}=0,(i=1,2,3)$. 根据引理3.1可知存在一个正常数$C_{*}=C_{*}(\Lambda,\overline{d})$ 使得$\max_{\bar{\Omega}}u_{i}\leqslant{C_{*}\min_{\bar{\Omega}}u_{i}}, ~i=1,2,3.$

应用反证法, 假设存在一个序列$(d_{1i},d_{2i},d_{3i}),$ 使得系统(1.5)对应的正解$(u_{1i},u_{2i},u_{3i})$满足$\max_{\bar{\Omega}}u_{1i}\rightarrow0$ 或者$\max_{\bar{\Omega}}u_{2i}\rightarrow0$ 或者$\max_{\bar{\Omega}}u_{3i}\rightarrow0,i\rightarrow{\infty}.$\\ 由引理3.2可得$u_{1i}\leqslant1$. 分部积分可得 $$\left\{\begin{array}{l} \int_{\Omega}bu_{1i}(1-u_{1i})-(b+1)u_{1i}u_{2i}-\frac{mu_{1i}u_{3i}}{1+lu_{1i}}{\rm d}x=0,\\ [3mm] \int_{\Omega}u_{1i}u_{2i}-du_{2i}u_{3i}-eu_{2i}{\rm d}x=0,\\ [3mm] \int_{\Omega}\frac{mu_{1i}u_{3i}}{1+lu_{1i}}-du_{2i}u_{3i}-gu_{3i}=0, ~~~i=1,2,3.\end{array} \right.(3.5)$$ 由椭圆方程的标准正则定理可知存在${(u_{1i},u_{2i},u_{3i})}$ 的一个子序列,这里仍记为${(u_{1i},u_{2i},u_{3i})}$, 以及三个非负函数$u_{1},u_{2},u_{3}\in{C^{2}(\overline{\Omega})},$ 使得在$[{C^{2}(\overline{\Omega})}]^{3}$中, ${(u_{1i},u_{2i},u_{3i})\rightarrow{(u_{1},u_{2},u_{3})}}, i\rightarrow{\infty}$. 由(3.5)式可得$u_{1}\equiv0$ 或$u_{2}\equiv0$ 或$u_{3}\equiv0.$

进一步,令$(d_{1i},d_{2i},d_{3i})\rightarrow ({\overline{d}_{1}, \overline{d}_{2},\overline{d}_{3}}) \in{{[\underline{d}_{1},{\infty})}\times{[\underline{d}_{2}, {\infty})}\times{[\underline{d}_{3},{\infty})}}$, 其中$\overline{d}_{1},\overline{d}_{2},\overline{d}_{3}$满足(3.4)式. 由(3.5)式可得,当$i\rightarrow{\infty}$时, $$\left\{\begin{array}{l} \int_{\Omega}bu_{1}(1-u_{1})-(b+1)u_{1}u_{2}-\frac{mu_{1}u_{3}}{1+lu_1}{\rm d}x=0,\\ [3mm] \int_{\Omega}u_{1}u_{2}-du_{2}u_{3}-eu_{2}{\rm d}x=0,\\ [3mm] \int_{\Omega}\frac{mu_{1}u_{3}}{1+lu_1}-du_{2}u_{3}-gu_{3}=0.\end{array} \right.(3.6)$$

我们考虑以下三种情况.

第一种情况: $u_{1}\equiv0,~u_2>0,u_3>0.$

由于当$i\rightarrow{\infty}$时,$u_{1i}\rightarrow{u_{1}}$, 故$u_{1i}-du_{3i}-e<0$, $$0=d_{2i}\int_{\partial{\Omega}}\partial{\nu}u_{2i}{\rm d}S =-d_{2i}\int_{\Omega}\Delta{u_{2i}}{\rm d}x=\int_{\Omega}u_{2i}(u_{1i}-du_{3i}-e){\rm d}x<0{ , }i\gg1,$$ 产生矛盾.

第二种情况: $u_{2}\equiv0,u_{1}>0,u_3>0.$

$u_{1},u_{2},u_{3}$满足 $$-d_{1}\triangle u_{1}=u_{1}\bigg[b(1-u_{1})-(b+1)u_{2} -\frac{mu_{3}}{1+lu_1}\bigg],~\frac{\partial u_1}{\partial\nu}=0.$$ 令$u_{1}(x_{0})=\min_{\bar{\Omega}}u_{1}(x)$, 由引理3.2可得 $$b(1-u_{1}(x_{0}))-(b+1)u_{2}(x_{0}) -\frac{mu_{3}(x_{0})}{1+lu_1(x_0)}\leqslant0,$$ 于是 $$bu_{1}(x_{0})\geqslant{b-(b+1)u_{2}(x_{0}) -\frac{mu_{3}(x_{0})}{1+lu_1(x_0)}}\geqslant b-(b+1)A -\frac{mB}{1+lu_1(x_0)},$$ 则 $$\frac{blu_1(x_0)^2+(b+1)Au_1(x_0)+mB-[b-(b+1)A]} {1+lu_1(x_0)}\geqslant0.$$ 对$u_{2i}$的微分方程在$\Omega$上分部积分有 $$0=d_{2i}\int_{\partial{\Omega}}\partial{\nu}u_{2i}{\rm d}S =-d_{2i}\int_{\Omega}\Delta{u_{2i}}{\rm d}x=\int_{\Omega}u_{2i}(u_{1i} -du_{3i}-e){\rm d}x>0 ,~~i\gg1,$$ 产生矛盾.

三种情况: $u_{3}\equiv0,u_{1}>0,u_{2}>0.$

$u_{1},u_{2}$满足 $$-d_{1}\triangle u_{1}=u_{1}[b(1-u_{1})-(b+1)u_{2}],{~~}\frac{\partial {u_{1}}}{\partial{\nu}}=0.$$ 令$u_{1}(x_{0})=\min_{\bar{\Omega}}u_{1}(x)$,通过引理3.2有 $$b(1-u_{1}(x_{0}))-(b+1)u_{2}(x_{0})\geqslant{b-\frac{4ed_{1}+bd_{2}}{4ed_{2}}},$$ 则有 $$\min_{\bar{\Omega}}u_{1}(x)\geqslant{1-\frac{4ed_{1}+bd_{2}}{4bed_{2}}}.$$ 由于$\overline{d}_{1},\overline{d}_{2}$满足(3.4)式, 易得$u_{1i}-du_{3i}-e>0,i\gg1.$ 对$u_{2i}$的微分方程在$\Omega$上分部积分有 $$0=d_{2i}\int_{\partial{\Omega}}\partial{\nu}u_{2i}{\rm d}S =-d_{2i}\int_{\Omega}\Delta{u_{2i}}{\rm d}x=\int_{\Omega}u_{2i}(u_{1i} -du_{3i}-e){\rm d}x>0 ,~i\gg1,$$ 矛盾产生,证毕.

定理4.1     当$e>1,~m<g(1+l)$成立时, 系统(1.2)的常值解$(1,0,0)$是全局渐近稳定的.

证     为了研究系统(1.2)的稳定性,考虑下面辅助系统 $$\left\{ \begin{array}{l} \omega_{1t}-d_1\Delta\omega_1=b\omega_1(1-\omega_1)-(b+1)\omega_1\omega_2,\\ \omega_{2t}-d_2\Delta \omega_2=\omega_1\omega_2-e\omega_2,\\ [2mm] \frac{\partial \omega_1}{\partial\nu}=\frac{\partial\omega_2}{\partial\nu}=0,~~~~x\in\partial\Omega,\\ [2mm] \omega_1(x,0)\geqslant0,\omega_2(x,0)\geqslant0. \end{array} \right.(4.1)$$ 定义下面Lyapunov函数 $$V(t)=\int_{\Omega}(\omega_1-1-\ln \omega_1){\rm d}x+(b+1)\int_{\Omega}\omega_2{\rm d}x,$$ 则我们有 $$\begin{eqnarray*} V'(t)&=&\int_{\Omega}\frac{\omega_{1}-1}{\omega_{1}}\omega_{1t}{\rm d}x+(b+1)\int_{\Omega}\omega_{2t}{\rm d}x\\ &=&\int_{\Omega} \bigg\{\frac{d_{1}(\omega_{1}-1)\triangle\omega_{1}}{\omega_{1}}+(\omega_{1}-1)[b(1-\omega_1)-(b+1)\omega_2] \bigg\}{\rm d}x\\ &&+(b+1)\int_{\Omega}(d_{2}\triangle \omega_{2}+\omega_1\omega_2-e\omega_2){\rm d}x\\ &\leq&-b\int_{\Omega}(\omega_1-1)^2{\rm d}x-(e-1)(b+1)\int_{\Omega}\omega_2{\rm d}x\\ &\leqslant &0. \end{eqnarray*}$$ 由定理3.1有 $$\lim_{t\rightarrow+\infty}\|\omega_{1}-1\|^{2}=0, \quad \lim_{t\rightarrow+\infty}\|\omega_{2}-0\|^{2}=0.$$ 因此,取$\varepsilon>0$充分小,$\exists T,$ 使得当$t>T$时有$\omega_{1}(x,t)\leq1+\varepsilon, ~\omega_{2}(x,t)\leq0+\varepsilon=\varepsilon$. 通过比较原理, (1.2)式的任意正解满足$u_{1}(x,t)\leq \omega_{1}(x,t)\leq1+\varepsilon, ~ u_{2}(x,t)\leq \omega_{2}(x,t)\leq0+\varepsilon.$ 因此 $$\left\{ \begin{array}{ll} u_{3t}-d_{3}\triangle u_{3}= u_{3}\bigg(\frac{mu_1}{1+lu_1}-du_2-g\bigg)\leqslant u_3\bigg[\frac{m(1+\varepsilon)}{1+l(1+\varepsilon)}-g\bigg], ~~ &x\in\Omega,\\ [3mm] \frac{\partial u_{3}}{\partial\nu}=0 ,&x\in\partial\Omega,~t>T,\\ [3mm] u_{3}(x,T)>0 ,&x\in\Omega. \end{array} \right.$$ 令$\omega_3'(t)=\omega_3[\frac{m(1+\varepsilon)}{1+l(1+\varepsilon)}-g],$ 可得$\omega_3(t)=e^{c[\frac{m(1+\varepsilon)}{1+l(1+\varepsilon)}-g]t},$ 其中 $\frac{m(1+\varepsilon)}{1+l(1+\varepsilon)}-g<0$如果$m<g(1+l)$并且$\varepsilon$充分小. 故 $$\lim_{t\rightarrow+\infty} \omega_3(t)=0.$$ 因为$u_3(x,t)\leqslant \omega_3(t),$ 通过比较原理可得 $$\lim_{t\rightarrow+\infty} u_3(x,\cdot)=0.$$ 因此 $$\lim_{t\rightarrow+\infty} \|(u_1(x,t),u_2(x,t),u_3(x,t))-(1,0,0)\|=0.$$ 证毕.

下面定理的证明与定理4.1类似.

定理4.2     当$e>1,m<\frac{g(1+le)}{e}$时,(1.2)式的 平衡点$(e,\frac{b(1-e)}{b+1},0)$是全局渐近稳定的.

这部分,研究系统(1.5)非常值正解的不存在性. 定理5.1     令$d_{2}^{\ast},d_{3}^{\ast}$是 固定的常数, 并且满足$d_{2}^{\ast}\mu_1>1-e,~d_{3}^{\ast}\mu_1>\frac{m}{1+l}-g,$ 则存在一个正常数$d_{1}=d_{1}(\Lambda,d_{2}^{*},d_{3}^{*}),$ 使得当$d_{i}\geq d_{i}^{\ast}(i=1,2,3)$时, 系统(1.5)没有非常值正解.

证    $\forall u_i\in{{L}^{1}(\Omega)}$, 令$u_i=\frac{1}{|\Omega|}\int_{\Omega}{u_i}{\rm d}x.$ 首先, 在(1.5)式两端同时乘以${u}_{i}-\bar{u}_{i},$ 然后积分有 $$\sum_{i=1}^{3}\int_{\Omega}d_{i}|\nabla u_{i}|^{2}{\rm d}x = \sum_{i=1}^{3}\int_{\Omega}G_{i}(u_{i})(u_{i}-\bar{u}_{i}){\rm d}x = \sum_{i=1}^{3}\int_{\Omega} (G_{i}(u_{i})-G_{i}(\bar{u}_{i}))(u_{i}-\bar{u}_{i}){\rm d}x. (5.1)$$ 利用Young's不等式 $$|ab|\leqslant\frac{1}{p}|a|^{p}+\frac{1}{q}|b|^{q},~~ \frac{1}{p}+\frac{1}{q}=1,$$ 有 $$\begin{eqnarray*} && \sum_{i=1}^{3}\int_{\Omega} (G_{i}(u_{i})-G_{i}(\bar{u}_{i}))(u_{i}-\bar{u}_{i}){\rm d}x\\ &=&\int_{\Omega}\bigg[b-bu_{1}-b\bar{u}_{1}-(b+1)u_{2} -\frac{mu_{3}}{(1+lu_1)(1+l\bar{u}_1)}\bigg](u_{1}-\bar{u}_{1})^{2}{\rm d}x \\ && +\int_{\Omega}(u_{1}-du_{3}-e)(u_{2}-\bar{u}_{2})^{2}{\rm d}x +\int_{\Omega} \bigg(\frac{mu_{1}}{1+lu_1}-du_{2}-g\bigg)(u_{3}-\bar{u}_{3})^{2}{\rm d}x \\ &&+\int_{\Omega}[-(b+1)\bar{u}_{1}+\bar{u}_{2}](u_{1}-\bar{u}_{1})(u_{2}-\bar{u}_{2}){\rm d}x\\ &&+\int_{\Omega} \bigg(-\frac{m\bar{u}_{1}}{1+l\bar{u}_1}+\frac{m\bar{u}_{3}} {(1+lu_1)(1+l\bar{u}_1)}\bigg)(u_{1}-\bar{u}_{1}) (u_{3}-\bar{u}_{3}){\rm d}x \\ && -\int_{\Omega}(d\bar{u}_{2}+d\bar{u}_{3})(u_{2}-\bar{u}_{2}) (u_{3}-\bar{u}_{3}){\rm d}x\\ &\leqslant & \int_{\Omega}b(u_{1}-\bar{u}_{1})^{2}{\rm d}x +\int_{\Omega}(u_{1}-e)(u_{2}-\bar{u}_{2})^{2}{\rm d}x+\int_{\Omega} \bigg(\frac{mu_{1}}{1+lu_1}-g\bigg)(u_{3}-\bar{u}_{3})^{2}{\rm d}x\\ &&+\int_{\Omega}[-(b+1)\bar{u}_{1}+\bar{u}_{2}] (u_{1}-\bar{u}_{1})(u_{2}-\bar{u}_{2}){\rm d}x \\ && +\int_{\Omega} \bigg(-\frac{m\bar{u}_{1}}{1+l\bar{u}_1} +\frac{m\bar{u}_{3}}{(1+lu_1)(1+l\bar{u}_1)}\bigg) (u_{1}-\bar{u}_{1})(u_{3}-\bar{u}_{3}){\rm d}x\\ &&-\int_{\Omega}(d\bar{u}_{2}+d\bar{u}_{3}) (u_{2}-\bar{u}_{2})(u_{3}-\bar{u}_{3}){\rm d}x\\ &\leqslant & \int_{\Omega}\bigg[(b+C_{1}+C_{2})(u_{2}-\bar{u}_{2})^{2}+(1-e+\varepsilon_{1})(u_{2}-\bar{u}_{2})^{2} +\bigg(\frac{m}{1+l}-g+\varepsilon_{2}\bigg) (u_{3}-\bar{u}_{3})^{2}\bigg]{\rm d}x, \end{eqnarray*}$$ 其中$\varepsilon_{1},\varepsilon_{2}$是由Young's不等式引 起的任意小的正常数, $C_{1}=C_{1}(\Lambda,d_{2}^{*},d_{3}^{*},\varepsilon_{1}),$ $C_{2}=C_{2}(\Lambda,$ $d_{2}^{*},d_{3}^{*},\varepsilon_{2}).$

由Poincare不等式$\mu_{1}\int_{\Omega}(f-\bar{f})^{2}{\rm d}x\leq \int_{\Omega}|\nabla f|^{2}{\rm d}x$,我们有 $$\begin{eqnarray*} \mu_{1} \sum_{i=1}^{3}\int_{\Omega}d_{i}|\nabla u_{i}|^{2}{\rm d}x & \leqslant& \int_{\Omega}\bigg[(b+C_{1}+C_{2})(u_{2}-\bar{u}_{2})^{2}+(1-e+\varepsilon_{1})(u_{2}-\bar{u}_{2})^{2}\\ &&+\bigg(\frac{m}{1+l}-g+\varepsilon_{2}\bigg)(u_{3}-\bar{u}_{3})^{2}\bigg]{\rm d}x. \end{eqnarray*}$$ 选取足够小的$\varepsilon_{2},\varepsilon_{2}>0$使得$\mu_{1}d_{2}^{*}\geqslant1-e+\varepsilon_{1}, \mu_{1}d_{3}^{*}\geqslant \frac{m}{1+l}-g+\varepsilon_{2},$积分上面不等式可得$u_{2}\equiv\bar{u}_{2}\equiv$ 常量, $u_{3}\equiv\bar{u}_{3}\equiv$ 常量, 如果$d_{1}>d_{1}^*\triangleq\mu_{1}^{-1}(b+C_{1}+C_{2}),$ 则$u_{1}\equiv\bar{u}_{1}\equiv$ 常量. 证毕.

d 本部分我们研究系统(1.5)在$E_*=u^*=(u_1^*,u_2^*,u_3^*~)$的线性化. 令$X$如第二部分定义,且定义 $$\begin{array}{l} {\bf X}^{+}=\{{\bf u}\in{\bf X}|u_{i}>0~~\mbox{on}~~{\bar\Omega},i=1,2,3\},\\ {B}(C)=\{{\bf u}\in{\bf X}|C^{-1}<u_{i}<C~~\mbox{on}~~{\bar\Omega},i=1,2,3\}. \end{array} %(1.2)$$ 则系统(1.5)可以被记为 $$\left\{\begin{array}{ll} -{D}\Delta{\bf u}={\bf G}({\bf u}),~~&x\in\Omega,\\ \partial_{\nu}{\bf u}=0,~~~~&x\in\partial\Omega, \end{array} \right.(6.1)$$ $\mathbf u$是系统(1.5)的一个正解当且仅当 在${\bf X}^{+}$中 $$~{\bf F}({\bf u})\triangleq {\bf u}-({\bf I}-\Delta)^{-1}\{{D}^{-1}{\bf G}({\bf u})+{\bf u}\}=0 ,$$ 其中$({\bf I}-\Delta)^{-1}$是$({\bf I}-\Delta)$的逆. 由于${\bf F}(\cdot)$是恒同算子的紧摄动,如果在$\partial {B}$上${\bf F}(u)\neq 0$,对任意的${B}={B}(c)$ Leray-Schauder度deg$({\bf F}(\cdot),0,{B})$有定义. 进一步,我们有 $${\bf D}_{u}{\bf F}({\bf u}^{*}) ={\bf I}-({\bf I}-\Delta)^{-1}\{{D}^{-1}{\bf G}_{u}({\bf u}^{*})+{\bf I}\} ,$$ 如果${\bf D}_{u}{\bf F}({\bf u}^{*})$可逆,${\bf F}$在${\bf u}^{*}$的指数定义为 $\mbox{index}({\bf F}(\cdot),{\bf u}^{*})=(-1)^{\gamma}$,其中$\gamma$是${\bf D}_{u}{\bf F}({\bf u}^{*})$的负特征根的个数.

首先,对任意的整数$i>0$和整数$1\leq j\leq\dim E(\mu_{i}),{\bf X}_{ij}$在${\bf D}_{u}{\bf F}({\bf u}^{*})$下是不变的. 并且$\lambda$是${\bf D}_{u}{\bf F}({\bf u}^{*})$在${\bf X}_{ij}$上的特征值,当且仅当它是下面矩阵的特征值. $${\bf I}-\frac{1}{1+\mu_{i}}[{D}^{-1}{\bf G}_{u}({\bf u}^{*}) +{\bf I}]=\frac{1}{1+\mu_{i}}[\mu_{i}{\bf I}-{D}^{-1}{\bf G}_{u}({\bf u}^{*})].$$ 因此,${\bf D}_{u}{\bf F}({\bf u}^{*})$是可逆的,当且仅当对所有的$i\geq0$, 矩阵${\bf I}-\frac{1}{1+\mu_{i}}[{D}^{-1}{\bf G}_{u}({\bf u}^{*})+{\bf I}]$是非奇异的. 定义 $$H(\mu)=H({\bf u}^{*};\mu)\triangleq \det\{\mu{\bf I}-{D}^{-1}{\bf G}_{u}({\bf u}^{*})\} =\frac{1}{d_{1}d_{2}d_{3}}\det\{\mu {D}-{\bf G}_{u}({\bf u}^{*})\}. (6.2)$$ 如果$H(\mu_{i})\neq 0$,则对任意的$1\leq j\leq \dim E(\mu_{i})$,${\bf D}_{u}{\bf F}({\bf u}^{*})$在${\bf X}_{ij}$上的负特征根的个数为奇数等价于$H(\mu_{i})<0$. 因此我们可以得到下面的结果.

引理6.1     假设对$\forall i\geq0$,矩阵$\mu_{i}{\bf I}-{D}^{-1}{\bf G}_{u}({\bf u}^{*})$是非奇异的,则 $$\mbox{index}({\bf F}(\cdot),{\bf u}^{*})=(-1)^\sigma,$$ 其中 $$\sigma=\Sigma_{i\geq0,H(\mu_{i})<0}\dim E(\mu_{i}).$$

为了计算$({\bf F}(\cdot){\bf u}^{*})$的指数, 我们需要考虑$H(\mu_i)$的符号 $$\det\{{\mu {D}-{\bf G}_{u}({\bf u}^{*})}\} =A_{3}(d_{2})\mu^{3}+A_{2}(d_{2})\mu^{2}+A_{1}(d_{2})\mu-\det{{{\bf G}_{u}({\bf u}^{*})}}\triangleq {\cal A}(d_{2};\mu), (6.3)$$ 其中 $$A_{3}(d_{2})=d_{1}d_{2}d_{3},$$ $$A_{2}(d_{2})=-a_{11}d_{2}d_{3},$$ $$A_{1}(d_{2})=-a_{12}a_{21}d_{3}-a_{13}a_{31}d_{2}-a_{23}a_{32}d_{1},$$ $$\det\{G_{u}(u^{*})\}=-a_{11}a_{23}a_{32}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32},$$ 这里$a_{ij}$在第二部分中已给出.

首先我们考虑$\mathcal {A}$与$d_2$的关系. 令$\tilde{\mu}_{1}(d_{2}),\tilde{\mu}_{2}(d_{2}), \tilde{\mu}_{3}(d_{2})$ 是${\cal A}$ $(d_2;\mu)=0$的三个根, 且$Re{\tilde{\mu}_{1}(k)}\leq Re{\tilde{\mu}_{2}(k)}\leq Re{\tilde{\mu}_{3}(k)}.$ 则 $$\tilde{\mu}_{1}(d_{2})\tilde{\mu}_{2}(d_{2})\tilde{\mu}_{3}(d_{2})=\det\{{\bf G}_{u}({\bf u}^{*})\}.$$ 由于$\det\{{\bf G}_{u}({\bf u}^{*})\}<0,~~A_{3}(d_{2})>0$. 因此$\tilde{\mu}_{1}(d_{2}),\tilde{\mu}_{2}(d_{2}),\tilde{\mu}_{3}(d_{2})$之一是正实数,并且另外两个的乘积是负数. 考虑下面的极限 $$\lim_{d_2\rightarrow+\infty} \frac{A_{3}(d_{2})}{d_{2}}=d_{1}d_{3} ,~~ \lim_{d_2\rightarrow+\infty}\frac{A_{2}(d_{2})}{d_{2}}=-d_{3}a_{11} ,~~ \lim_{d_2\rightarrow+\infty}\frac{A_{1}(d_{2})}{d_{2}}=-a_{13}a_{31} ,$$ $$\lim_{d_2\rightarrow+\infty}\frac{{\cal A}(\mbox{d}_{2};\mu)}{d_{2}}=d_{1}d_{3}\mu^{3}-d_{3}a_{11}\mu^{2}-a_{13}a_{31}\mu =\mu(d_{1}d_{3}\mu^{2}-d_{3}a_{11}\mu-a_{13}a_{31}) ,$$

如果参数$\Lambda,d_{1},d_{3}$满足 $${d_{3}}^{2}{a_{11}}^{2}+4d_{1}d_{3}a_{13}a_{31}>0,$$ 我们可以得到下面的引理.

引理6.2nbsp;    如果(1.4),(2.4)式成立,且$C^{-1}<\frac{ru_{1}^{*}(\alpha-1)}{bR}$,则存在一个正数$D_{2},$ 使得当$d_{2}\geq D_{2}$时, ${\cal A}(\mbox{d}_{2};\mu)=\mbox{0}$的三个根$\tilde{\mu}_{1}(d_{3}),\tilde{\mu}_{2}(d_{3}),\tilde{\mu}_{3}(d_{3})$都是实的且满足 $$\begin{array}{l} \lim_{d_2\rightarrow+\infty}\tilde{\mu}_1(d_{2})=\frac{1}{2d_{1}d_{2}}[d_{3}a_{11}-\sqrt{{d_{3}}^{2}{a_{11}}^{2}+4d_{1}d_{3}a_{13}a_{31}}]<0,\\ [3mm] \lim_{d_2\rightarrow+\infty}\tilde{\mu}_2(d_{2})=\frac{1}{2d_{1}d_{2}}[d_{3}a_{11}+\sqrt{{d_{3}}^{2}{a_{11}}^{2}+4d_{1}d_{3}a_{13}a_{31}}]<0,\\ [3mm] \lim_{d_2\rightarrow+\infty}\tilde{\mu}_3(d_{2})=0=\tilde{\mu}. \end{array}(6.4)$$ 进一步,如果$-\infty<\tilde{\mu}_1(d_{2})<\tilde{\mu}_2(d_{2})<0<\tilde{\mu}_3(d_{2}),$ 则 $$\begin{array}{l} \mu\in(-\infty,\tilde{\mu}_{\mbox{1}}(\mbox{d}_{2})) \cup(\tilde {\mu}_{2}(\mbox{d}_{2}),\tilde {\mu}_{3}(\mbox{d}_{2})), ~~{\cal A}(\mbox{d}_{2};\mu)<\mbox{0},\\ \mu\in(\tilde{\mu}_{\mbox{1}}(\mbox{d}_{2}), \tilde{\mu}_{2}(\mbox{d}_{2}))\cup(\tilde{\mu}_{3}(\mbox{d}_{2}), +\infty), ~~{\cal A}(\mbox{d}_{2};\mu)>\mbox{0}. \end{array}(6.5)$$

当$d_{2}$足够大,则存在$\Lambda,d_{i},i=1,3$使得系统$(1.5)$至 少有一个非常值正解.

定理6.1    假设参数$\Lambda,d_{1},d_{3}$固定, $\tilde{\mu}\in(\mu_{0},\mu_{1})$且$\sigma_{0}= \dim_{i\geqslant0,H(\mu_{i})<0}E(\mu_{i})$是奇数, 则存在一个正常数$D_{2}$,使得当$d_{2}\geq D_{2}$时, 系统(1.5)至少有一个非常值正解.

证    通过引理6.2可知,存在一个正常数$D_{2}$使 得当$d_{2}\geq D_{2}$时,(6.5)式成立并且 $$~~~~0=\mu_{0}<\tilde{\mu}_{3}(d_{2})<\mu_{1}.(6.6)$$ 要证明$\forall d_{2}\geq D_{2},$系统(1.5)至少有一个非常值正解.

我们将基于拓扑度的同伦不变性,利用反证法来证明. 假设存在$d_{2}=\tilde{d_{2}}\geq D_{2}$使得结论不成立. 固定$d_{2}=\tilde{d_{2}},~{d_{2}}^{*}=\frac{2-e}{\mu_{1}},$

由定理$5.1,$ 存在一个正常数$D_{1}=D_{1}(\Lambda,d_{2}^{*},d_{3}^{*}),$ $\hat{d_{2}}\geqslant{d_{2}}^{*}, ~\hat{d_{3}}\geqslant\max\{{d_{3}}^{*},d_{3}\}, ~\hat{d_{1}}\geqslant\max\{D_{1},~d_{1}\}.$ 对于$t\in^{[0, 1]},$定义${D}(t)={\rm diag}(d_{1}(t),~d_{2}(t),~d_{3}(t))$其中$d_{i}(t)=td_{i}+(1-t)\hat{d_{i}},i=1,2,3. $考虑问题$$\left\{\begin{array}{ll} -{D}(t)\Delta{\bf u}={\bf G}({\bf u}), ~~&x\in\Omega,\\ \partial_{\nu}{\bf u}=0, &x\in\partial\Omega. \end{array}\right.(6.7)$$则${\bf u}$是系统(1.5)的一个非常值解当且仅当${\bf u}$是方程(6.7)当$t=1$时的一个解. 显然对任意的$0\leq t\leq 1$,${\bf u}^{*}$是方程(6.7)的唯一正常值解.

对任意的$0\leq t\leq 1$,${\bf u}$是方程(6.7) 的正解当且仅当 在${\bf X}^{+}$上 $${\bf F}(t;{\bf u})\triangleq{\bf u}-({\bf I}-\Delta)^{-1}\{{D}^{-1}(t){\bf G}({\bf u})+{\bf u}\}=0 .$$ 显然${\bf F}(1;{\bf u})={\bf F}({\bf u}),$ 由定理5.1可知${\bf F}(0;{\bf u})={\bf F}({\bf u})=0$在${\bf X}^{+}$中只有正解$\mathbf u^{*}$,直接计算可得 $${\bf D}_{u}{\bf F}(t;{\bf u}^{*})= {\bf I}-({\bf I}-\Delta)^{-1}\{{D}^{-1}(t){\bf G}_{u}({\bf u}^{*})+{\bf I}\}.$$ 特别地, $${\bf D}_{u}{\bf F}(0;{\bf u}^{*})= {\bf I}-({\bf I}-\Delta)^{-1}\{\tilde{{D}}^{-1}{\bf G}_{u}({\bf u}^{*})+{\bf I}\},$$ $${\bf D}_{u}{\bf F}(1;{\bf u}^{*})= {\bf I}-({\bf I}-\Delta)^{-1}\{{D}^{-1}{\bf G}_{u}({\bf u}^{*})+{\bf I}\}={\bf D}_{u}F({\bf u}^{*}).$$ 其中$\hat{D}={\rm diag}(\hat{d_{1}},\hat{d_{2}},\hat{d_{3}})$. 由(6.2), (6.3)式可知 $$H(\mu)=\frac{1}{d_{1}d_{2}d_{3}} {\cal A}(\mbox{d}_{2};\mu),(6.8)$$ 由(6.5), (6.6)式通过直接计算有 $$\left\{\begin{array}{l} H(\mu_{0})=H(0)<0,\\ H(\mu_{i})>0,~~i\geq 1. \end{array}\right.(6.9)$$ 因此, 对任意的$i\geq 0$,$0$不是矩阵 $\mu_{i}{\bf I}-{D}^{-1}{\bf G}_{u}({\bf u}^{*})$的特征值, 且有 $$\Sigma_{i\geq0,H(\mu_{i})<0}\dim E(\mu_{i})=\sum_{i=1}^n\dim E(\mu_{i})=\sigma_{n}~~~~\mbox{是奇数.}$$ 由引理6.1有 $$\mbox{index}({\bf F}(1;\cdot),{\bf u}^{*})=(-1)^{\gamma}=(-1)^{\sigma_{n}}=-1. (6.10)$$

类似地我们可以证明 $$\mbox{index}({\bf F}(0;\cdot),{\bf u}^{*})=(-1)^{0}=1. (6.11)$$ 通过定理3.1, 存在一个正常数$C$使得对所有的$0\leq t\leq 1$, (6.7)的正解满足$C^{-1}<u_{1},u_{2},u_{3}<C$. 因此,在$\partial{\cal B}\mbox{(c)}$上,对所有的$C^{-1}<u_{1},u_{2},u_{3}<C$,${\bf F}(t;{\bf u})\neq 0$. 由拓扑度的同伦不变性可知 $$~~~~\mbox{deg}({\bf F}(1;\cdot),0,{\cal B}\mbox{(c)})=\mbox{deg}({\bf F}(0;\cdot),0,{\cal B}\mbox{(c)}). (6.12)$$ 另一方面,通过假设,方程 ${\bf F}(1;{\bf u})=0$ 和 ${\bf F}(0;{\bf u})=0$在 ${\cal B}\mbox{(c)}$都只有一个正解,因此, 通过(6.10)和(6.11) 式 $$\mbox{deg}({\bf F}(0;\cdot),0,{\cal B}\mbox{(c)})=\mbox{index}({\bf F}(0;\cdot),{\bf u}^{*})=(-1)^{0}=1,$$ $$\mbox{deg}({\bf F}(1;\cdot),0,{\cal B}\mbox{(c)})=\mbox{index}({\bf F}(1;\cdot),{\bf u}^{*})=(-1)^{0}=-1.$$ 这与(6.12)式矛盾, 证毕.