对捕食模型的研究已经有很多结果, 如Pang和Wang^[1]研究了带有比率依赖响应函数的三种群捕食模型, 证明了对任意的扩散系数, 其唯一的正常数解对于反应扩散方程组而言是全局渐近稳定的, 从而该方程组没有非常数正平衡解. 结论表明, 对于该捕食模型, 一般扩散不能导致平衡态模式. 但是对于强耦合模型存在非常数正解, 这说明交错扩散可以导致平衡态模式, 更多关于比率依赖响应函数的捕食模型的研究见文献[2–5]. 衣凤岐等研究了Lengyel-Epstein模型, 给出了该模型的图灵不稳定性和分支^[6], 这表明扩散导致了图灵不稳定性. Du和Hsu^[7]研究了具有Leslie-Gower功能反应函数的扩散捕食模型, 结果表明在适当的条件下, 该系统不存在非常数正稳态解, 而当物种集中在空间栖息地的某个区域时, 该系统存在非常数正稳态解. 适当的选取系数函数, 食饵和捕食者的种群在子区域中趋于灭亡, 同时得到了异质环境中的特殊模式解. 在文献[8]中, 作者研究了具有食饵趋化项的捕食模型的正常数平衡态是局部和全局渐近稳定的. 林支桂等^[9]研究了一类带Michaelis-Menten函数响应系统平衡解的局部渐近稳定性和全局渐近稳定性. 刘蒙等^[10]研究了如下带有Beddington-DeAngelis响应函数的对食饵具有阶段结构的捕食模型

(1.1) $ \begin{equation} \left\{\begin{array}{ll} \frac{{\rm d}u_1}{{\rm d} t}=r u_2-c_1 u_1-bu_1- au^2_1, \\ \frac{{\rm d}u_2}{{\rm d}t}=b u_1-c_2 u_2-a_{11}u^2_2-\frac{a_{12}u_2u_3}{1+mu_2+nu_3}, \\ \frac{{\rm d}u_3}{{\rm d} t}=-c_3u_3+\frac{a_{21}u_2u_3}{1+mu_2+nu_3}-a_{22}u^2_3, \end{array}\right. \end{equation} $

其中$ a_{21}, a_{12}, {a_{22}}, m, n $是正常数, $ u_1(t) $和$ u_2(t) $分别表示幼年食饵和成年食饵在$ t $时刻的种群密度, $ u_3(t) $表示捕食者在$ t $时刻的种群密度, 捕食者仅以成年食饵为食, 正常数$ r, c_1, a $和$ b $分别为幼年食饵的出生率, 死亡率, 过度拥挤率和幼年食饵转化为成年食饵的转化率; 正常数$ c_2, a_{11} $和$ a_{12} $分别为成年食饵的死亡率, 过度拥挤率和捕食者的捕获率; 正常数$ c_3 $表示捕食者的死亡率; $ a_{21} /a_{12} $表示营养物质转化为捕食者繁殖的速率, $ m $为捕食者对所捕获猎物的平均处理时间, $ nu_3 $反应了捕食者之间的相互干扰.

为了研究方便, 令$ a=a_{11}=a_{22}=1 $, 并用$ p $代替$ c_1+b $, 显然$ p>b $, 则模型$ \rm (1.1) $可写为

$ \left\{\begin{array}{ll} \frac{{\rm d}u_1}{{\rm d} t}=r u_2-pu_1- u^2_1, \\ \frac{{\rm d}u_2}{{\rm d}t}=b u_1-c_2 u_2-u^2_2-\frac{a_{12}u_2u_3}{1+mu_2+nu_3}, \\ \frac{{\rm d}u_3}{{\rm d} t}=-c_3u_3+\frac{a_{21}u_2u_3}{1+mu_2+nu_3}-u^2_3. \end{array}\right. $

下面我们研究方程组

(1.2) $ \begin{equation} \left\{\begin{array}{ll} r x_2-p x_1- x^2_1=0, \\ b x_1-c_2 x_2-x^2_2-\frac{a_{12}x_2x_3}{1+mx_2+nx_3}=0, \\ -c_3x_3+\frac{a_{21}x_2x_3}{1+mx_2+nx_3}-x^2_3=0 \end{array}\right. \end{equation} $

的正常数解的存在性. 由方程组(1.2)的第一个方程, 得到

(1.3) $ \begin{equation} x_1=\frac{1}{2}(-p\pm\sqrt{p^2+4rx_2}). \end{equation} $

再由方程组(1.2)的第三个方程, 得到

(1.4) $ \begin{equation} x_3=\frac{1}{2n}\left[-(1+mx_2+c_3n)\pm\sqrt{(1+mx_2+c_3n)^2-4n(c_3+mc_3x_2-a_{21}x_2)}\right]. \end{equation} $

为了寻求方程组(1.2)的正常数解, 我们在(1.3)和(1.4)式中取正号, 并且要求$ a_{21}x_2>c_3+mc_3x_2 $, 当$ a_{21}>mc_3 $时, 此即

$ x_2>\frac{c_3}{a_{21}-mc_3}:=x_0. $

设

$\begin{array}{c}\varphi(x)=\frac{1}{2}(-p+\sqrt{p^2+4rx}), \\ \psi(x)=\frac{1}{2n}\left[-(1+mx+c_3n)+\sqrt{(1+mx+c_3n)^2-4n(c_3+mc_3x-a_{21}x)}\right]. \end{array} $

把(1.3)式以及(1.4)式代入方程组(1.2)的第二个方程, 我们得到

$ x^2_2+c_2 x_2+\frac{a_{12}x_2\psi(x_2)}{1+mx_2+n\psi(x_2)}-b \varphi(x_2)=0. $

令

$ \Phi(x)=x^2+c_2 x+\frac{a_{12}x\psi(x)}{1+mx+n\psi(x)}-b \varphi(x). $

如果$ a_{21}>mc_3 $且$ x>x_0 $, 通过直接计算, 有

$ \varphi'(x)=\frac{r}{\sqrt{p^2+4rx}}<\frac{r}{p}, $

$ \begin{eqnarray*} \psi'(x)=\frac{1}{2n} \left[-m+\frac{m(1+mx+c_3n)-2n(mc_3-a_{21})}{\sqrt{(1+mx+c_3n)^2-4n(c_3+mc_3x-a_{21}x)}}\right] >-\frac{m}{2n}. \end{eqnarray*} $

$ \begin{eqnarray*} \Phi'(x)&=&2x+c_2+a_{12}\frac{\psi(x)(1+n \psi(x))+x\psi'(x)(1+mx)}{(1+mx+n\psi(x))^2}-b \varphi'(x)\\ &\geq& 2x+c_2-a_{12}\frac{\frac{m}{2n}x(1+mx)}{(1+mx+n\psi(x))^2}- \frac{br}{p}\\ &> &2x_0+ c_2-\frac{a_{12}}{2n}- \frac{br}{p}. \end{eqnarray*} $

当$ 2x_0+c_2\geq\frac{a_{12}}{2n}+ \frac{br}{p} $时, $ \Phi'(x)>0 $. 为了保证$ \Phi(x)=0 $有比$ x_0 $大的解, 并且注意到

$ \lim\limits_{x\to +\infty} \Phi(x)=+\infty, $

且$ \Phi(x) $在$ [x_0, +\infty) $上连续, 所以我们可以要求$ \Phi(x_0)<0 $. 显然$ \psi(x_0)=0 $, 且

$ \Phi(x_0)=x^2_0 +c_2 x_0-b \varphi(x_0) =x^2_0 +c_2 x_0+ \frac{bp}{2}-\frac{b}{2}\sqrt{p^2+4rx_0}. $

因此$ \Phi(x_0)<0 $等价于

$ x_0(x_0+c_2)^2+bp(x_0+c_2)<b^2r. $

从上面的分析可以看出, 如果

(1.5) $ \begin{equation} a_{21}>mc_3, \ \ 2x_0+c_2\geq\frac{a_{12}}{2n}+\frac{br}{p}, \ \ x_0(x_0+c_2)^2+bp(x_0+c_2)<b^2r, \end{equation} $

则$ \Phi(x)=0 $在$ (x_0, +\infty) $内只有一个正解, 我们用$ \tilde{u}_2 $表示这个正解, 令

$ \tilde{u}_1=\varphi(\tilde{u}_2), \;\;{ } \tilde{u}_3=\psi(\tilde{u}_2), \;\;{ } \tilde {\bf{u}}=(\tilde{u}_1, \tilde{u}_2, \tilde{u}_3), $

则$ (x_1, x_2, x_3)=\tilde {\bf{u}} $是方程组(1.2)的正解.

例如当$ a_{21}=9, m=c_3=1, b=5, r=1, c_2=\frac{1}{8}, p= 16, \frac{a_{12}}{n}=\frac{1}{16} $时, 条件(1.5)满足.

假设$ \Omega $是$ {{\bf R}}^N $中的一个具有光滑边界的有界区域, 如果物种的分布密度不均匀, 则可用下面的反应扩散方程组的初边值问题来描述:

(1.6) $ \begin{equation} \left\{\begin{array}{ll} \frac{\partial u_1}{\partial t}-d_1\Delta u_1=r u_2-p u_1- u^2_1, \ \ &x\in\Omega, \ t>0, \\ \frac{\partial u_2}{\partial t}-d_2\Delta u_2=b u_1-c_2 u_2-u^2_2-\frac{a_{12}u_2u_3}{1+mu_2+nu_3}, \ \ &x\in\Omega, \ t>0, \\ \frac{\partial u_3}{\partial t}-d_3\Delta u_3=-c_3u_3+\frac{a_{21}u_2u_3}{1+mu_2+nu_3}-u^2_3, \ \ &x\in\Omega, \ t>0, \\ \frac{\partial u_1}{\partial\nu}=\frac{\partial u_2}{\partial\nu}= \frac{\partial u_3}{\partial\nu}=0, \ &x\in\partial\Omega, \ t>0, \\ u_i(x, 0)=u_{i, 0}(x)\geq 0, \not\equiv 0, i=1, 2, 3, \ &x\in\Omega, \end{array}\right. \end{equation} $

其中$ \nu $为$ \partial\Omega $上的单位外法向量; 正常数$ d_1, d_2, d_3 $表示扩散系数; $ u_{i, 0}\ (i=1, 2, 3) $是连续函数. 因此当(1.5)式成立时, $ (u_1, u_2, u_3)=\tilde {\bf{u}} $是初边值问题(1.6)唯一的正常数平衡解.

设$ 0=\mu_1<\mu_2<\mu_3<\cdots $是算子$ -\Delta $在$ \Omega $上带有齐次Neumann边界条件的特征值, $ E(\mu_i) $是对应于$ \mu_i $在$ C^1(\bar\Omega) $中的特征子空间. 设

$ {\bf{X}}=\{(u_1, u_2, u_3)\in [C^1(\bar \Omega)]^3 |\partial_\nu{u_1}=\partial_\nu{u_2}=\partial_\nu{u_3}=0 , x\in \partial\Omega\}, $

$ \{\phi_{ij}\, |\, j=1, \, \cdots , \, \hbox{ dim}\, E(\mu_i)\} $是$ E(\mu_i) $的一组正交基, 令$ {{\bf X}}_{ij}=\{{{\bf c}}\phi_{ij} | {{\bf c}} \in R^3\}, $则

$ {\bf X}_i= \oplus_{j=1}^{{\rm dim\, }E(\mu_i)}{\bf X}_{ij}, \ \ \ {\bf X}=\oplus_{i=1}^\infty {\bf X}_i. $

定理2.1   如果$ p>b $且条件(1.5)成立, 那么问题(1.6)的正常数平衡解$ \tilde{\bf{u}} $是局部渐近稳定的.

证   令$ D={\rm diag}(d_1, d_2, d_3) $, $ {{\bf u}}=(u_1, u_2, u_3)^T $,

$ {\bf G}({{\bf u}})=\left(\begin{array}{cccc} r u_2-pu_1- u^2_1\\ { } b u_1-c_2 u_2-u^2_2-\frac{a_{12}u_2u_3}{1+mu_2+nu_3} \\ { } -c_3u_3+\frac{a_{21}u_2u_3}{1+mu_2+nu_3}-u^2_3\end{array}\right). \nonumber $

于是

$ {\bf G}_{{\bf u}} (\tilde{{\bf u}})=\left(\begin{array}{cccc} c_{11} &\; \; c_{12} \; \; &c_{13}\\ c_{21} &c_{22} & c_{23} \\ c_{31} &c_{32} &c_{33} \end{array}\right), $

其中

$ \begin{eqnarray*} & & c_{11}= -p-2\tilde{u}_1<0, \ \ c_{12}= r>0, \ \ c_{13}=0, \ \ c_{21}=b>0, \\ & &c_{22}=-c_2-2\tilde{u}_2-\frac{a_{12}\tilde{u}_3(1+n\tilde{u}_3)}{(1+m\tilde{u}_2+n\tilde{u}_3)^2}<0, \\ & & c_{23}= -\frac{a_{12}\tilde{u}_2(1+m\tilde{u}_2)}{(1+m\tilde{u}_2+n\tilde{u}_3)^2}<0, c_{31}= 0, \ \ c_{32}=\frac{a_{21}\tilde{u}_3(1+n\tilde{u}_3)}{(1+m\tilde{u}_2+n\tilde{u}_3)^2}>0, \\ && c_{33}=-c_3+\frac{a_{21}\tilde{u}_2(1+m\tilde{u}_2)}{(1+m\tilde{u}_2+n\tilde{u}_3)^2}-2\tilde{u}_3 =-\frac{a_{21}n\tilde{u}_2\tilde{u}_3}{(1+m\tilde{u}_2+n\tilde{u}_3)^2}-\tilde{u}_3<0.\\ \end{eqnarray*} $

为了叙述方便, 记

$ \begin{eqnarray*} &&A_1=-(c_{11}+c_{22}+c_{33})>0, \\ &&A_2 =c_{11}c_{22}+c_{11}c_{33}+c_{22}c_{33}-c_{23}c_{32}-c_{12}c_{21}, \\ &&A_3=-\det{{\bf G}_{{{\bf u}}}(\tilde{{\bf u}})} =-(c_{11}c_{22}-c_{12}c_{21})c_{33}+c_{11}c_{23}c_{32}. \end{eqnarray*} $

设$ {\cal L}= {\cal D}\Delta+{\bf G}_{{{\bf u}}}(\tilde{{\bf u}}) $, 则初边值问题(1.6)在$ \tilde{{\bf u}} $处的线性化为

$ {\bf u}_t={\cal L}{\bf u}. $

对每个$ i\geq 1 $, $ {\bf X}_i $在算子$ {\cal L} $下是不变的. $ \lambda $为$ {\cal L} $在$ {\bf X}_i $上的特征值当且仅当$ \lambda $为矩阵$ -\mu_i{\cal D}+{\bf G}_{{{\bf u}}}(\tilde{{\bf u}}) $的特征值, $ -\mu_i{\cal D}+{\bf G}_{{{\bf u}}}(\tilde{{\bf u}}) $的特征多项式为

$ \psi_i(\lambda)=\lambda^3+B_1\lambda^2+B_2\lambda+B_3, $

其中

$ \begin{eqnarray*} && B_1=\mu_i(d_1+d_2+d_3)+A_1>0, \\ &&B_2=\mu_i^2(d_1d_2+d_1d_3+d_2d_3)-\mu_i[d_1(c_{22}+c_{33})+d_2(c_{22}+c_{33})+d_3(c_{11}+c_{22})]+A_2, \\ && B_3=\mu_i^3d_1d_2d_3-\mu_i^2(c_{33}d_1d_2+c_{11}d_1d_3+c_{22} d_2d_3)\\ &&{\qquad}\ +\mu_i[d_3(c_{11}c_{22}-c_{12}c_{21})+d_2c_{11}c_{33}+d_1(c_{22}c_{33}-c_{23}c_{32})]+A_3. \end{eqnarray*} $

由(1.5)式的第二个不等式可知$ p(2x_0+c_2)>br $, 并注意到$ \tilde{u}_2>x_0 $, 于是有

$ \begin{eqnarray*} c_{11}c_{22}-c_{12}c_{21}&=&(-p-2\tilde{u}_1)(-c_2-2\tilde{u}_2-\frac{a_{12}\tilde{u}_3(1+n\tilde{u}_3)} {(1+m\tilde{u}_2+n\tilde{u}_3)^2})-br\\ &=&p(c_2+2\tilde{u}_2)-br+\frac{pa_{12}\tilde{u}_3(1+n\tilde{u}_3)} {(1+m\tilde{u}_2+n\tilde{u}_3)^2}+2\tilde{u}_1(c_2+2\tilde{u}_2+\frac{a_{12}\tilde{u}_3(1+n\tilde{u}_3)} {(1+m\tilde{u}_2+n\tilde{u}_3)^2})\\ &>&p(c_2+2x_0)-br+\frac{pa_{12}\tilde{u}_3(1+n\tilde{u}_3)} {(1+m\tilde{u}_2+n\tilde{u}_3)^2}+2\tilde{u}_1(c_2+2\tilde{u}_2+\frac{a_{12}\tilde{u}_3(1+n\tilde{u}_3)} {(1+m\tilde{u}_2+n\tilde{u}_3)^2})\\ &>&0. \end{eqnarray*} $

又因为$ c_{11}<0, c_{22}<0, c_{33}<0, c_{23}<0, c_{32}>0 $, 所以有$ A_2>0 $, $ A_3>0 $.

直接计算得到

$ B_1B_2-B_3=M_3\mu_i^3+M_2\mu_i^2+M_1\mu_i+A_1A_2-A_3, $

其中

$ \begin{eqnarray*} M_3&=&(d_1+d_2+d_3)(d_1d_2+d_2d_3+d_3d_1)-d_1d_2d_3>0, \\ M_2&=&-(c_{11}+c_{22})(d_1d_2+d_1d_3+d_2d_3+d^2_3)-(c_{11}+c_{33})(d_1d_2+d_2d_3+d_1d_3+d^2_2)\\ && -(c_{22}+c_{33})(d_1d_2+d_1d_3+d_2d_3+d^2_1)>0, \\ M_1&=&d_1[A_2+c_{23}c_{32}+c_{22}(c_{11}+c_{22}+c_{33})+c_{33}(c_{11}+c_{33})] \\ && +d_2[A_2+c_{11}(c_{11}+c_{22}+c_{33})+c_{33}(c_{22}+c_{33})] \\ && +d_3[A_2+c_{12}c_{21}+c_{11}(c_{11}+c_{22}+c_{33})+c_{22}(c_{22}+c_{33})], \end{eqnarray*} $

因为

$ \begin{eqnarray*} &&A_2+c_{23}c_{32}=(c_{11}c_{22}-c_{12}c_{21})+c_{11}c_{33}+c_{22}c_{33}>0, \\ &&A_2+c_{12}c_{21}=c_{11}c_{22}+c_{11}c_{33}+c_{22}c_{33}-c_{23}c_{32}>0, \end{eqnarray*} $

所以, 我们有$ M_1>0 $.

通过计算可以看出

$ \begin{eqnarray*} A_1A_2-A_3 &=& -(c_{22}+c_{33})[(c_{11}c_{22}-c_{12}c_{21})+c_{11}c_{33}+c_{22}c_{33}-c_{23}c_{32}] \\ && -c_{11}[(c_{11}c_{22}-c_{12}c_{21})+c_{11}c_{33}]-c_{12}c_{21}c_{33}>0, \end{eqnarray*} $

从而$ B_1B_2-B_3\geq A_1A_2-A_3>0 $. 由Routh-Hurwitz判别定理可知, 对每一个$ i\geq 1 $, $ \psi_i(\lambda)=0 $的三个根$ \lambda_{i1}, \, \lambda_{i2}, \, \lambda_{i3} $都有负实部. 令$ \lambda=\mu_i\xi $, 则

$ \psi_i(\lambda)={\mu_i}^3\xi^3+B_1{\mu_i}^2\xi^2+B_2{\mu_i}\xi+B_3\triangleq \tilde{\psi}_i(\xi). $

注意到当$ i\rightarrow \infty $时, $ \mu_i\rightarrow \infty $, 所以

$ \lim\limits_{i\to\infty}\{\tilde{\psi}_i(\xi)/{\mu_i}^3\}=\xi^3+(d_1+d_2+d_3)\xi^2+(d_1d_2+d_1d_3+d_2d_3)\xi+d_1d_2d_3 \triangleq \tilde{\psi}(\xi). $

因为$ \tilde{\psi}(\xi)=0 $有三个负实根: $ -d_1, -d_2, -d_3 $, 取$ \bar{\delta}=\min \{d_1, d_2, d_3\}, $由连续性知存在$ i_0 $, 使得对所有的$ i\geq i_0 $, $ \mu_i\geq 1 $, 方程$ \tilde{\psi}_i(\xi)=0 $的三个根$ \xi_{i1}, \xi_{i2}, \xi_{i3} $满足

$ {\rm Re}\xi_{i1}, {\rm Re}\xi_{i2}, {\rm Re}\xi_{i3}\leq -\frac{\bar{\delta}}{2}, $

从而

$ {\rm Re}\lambda_{i1}, {\rm Re}\lambda_{i2}, {\rm Re}\lambda_{i3}\leq -\mu_i\frac{\bar{\delta}}{2}\leq -\frac{\bar{\delta}}{2}, i\geq i_0. $

取

$ -\tilde{\delta}=\max\limits_{1\leq i\leq i_0}\{ {\rm Re}\lambda_{i1}, {\rm Re}\lambda_{i2}, {\rm Re}\lambda_{i3}\}. $

则$ \tilde{\delta}>0 $, 取$ \delta=\min\{\tilde{\delta}, \frac{\bar{\delta}}{2}\}>0 $, 则有

$ {\rm Re}\lambda_{i1}, {\rm Re}\lambda_{i2}, {\rm Re}\lambda_{i3}\leq -\delta, \;\;{ } i\geq 1. $

故算子$ {\cal L} $的谱位于左半平面$ \{ {\rm Re} \lambda\leq -\delta\} $, 由文献[11, 定理5.1.1]可得$ \tilde {\bf{u}} $的局部渐近稳定性. 证毕.

显然初边值问题(1.6)存在唯一的非负整体解. 由强极值原理可知, 当$ u_i(x, 0)\not\equiv0 $$ (i=1, 2, 3) $时, $ u_i(x, t)>0 $$ (i=1, 2, 3), x\in\bar{\Omega}\times (0, \infty) $. 利用最大值原理, 我们得到

$ \sup\limits_{\bar{\Omega}\times [0, \infty)} u_3(x, t)\leq \max\left\{\frac{a_{21}}{m}-c_3, \max\limits_{\bar{\Omega}}u_{3, 0}(x)\right\}\triangleq C_0. $

对问题(1.6)的三个方程在$ \Omega $上积分并相加, 再利用上面的估计式得到

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\int_\Omega \left(u_1+u_2+u_3\right){\rm d}x&\leq &\int_\Omega \left( r u_2-p u_1- u^2_1+b u_1-c_2 u_2-u^2_2+u_3(\frac{a_{21}}{m}-c_3) \right){\rm d}x\\ &=&\int_\Omega \left[ -p u_1-c_2 u_2- (u_1-\frac{b}{2})^2+\frac{b^2}{4}-(u_2-\frac{r}{2})^2+ \frac{r^2}{4}\right]{\rm d}x\\ &&+\int_\Omega \left[u_3(\frac{a_{21}}{m}-c_3) \right]{\rm d}x, \\ &\leq&\int_\Omega \left[ -p u_1-c_2 u_2+\frac{b^2}{4}+ \frac{r^2}{4}+u_3(\frac{a_{21}}{m}-c_3) \right]{\rm d}x\\ &\leq & -\min\{p, c_2\}\int_\Omega(u_1+u_2+u_3){\rm d}x+C_1, \end{eqnarray*} $

其中$ C_1 $是一个仅依赖于$ b, r, a_{21}, m, c_3, C_0, p, c_2 $以及$ \Omega $的正常数, 因此$ \|u_i(\cdot , t)\|_{L^1(\Omega)}\ ( i=1, 2, 3) $在$ [0, +\infty) $上有界. 由文献[11, 第3.5节的习题4]知, 存在正常数$ C_2 $, 使得$ \|u_i(\cdot , t)\|_\infty\leq C_2\ ( i=1, 2, 3) , t\geq 0 $. 再由文献[12, 定理A2], 存在常数$ C>0 $, 使得

(3.1) $ \begin{equation} \|u_i(\cdot , t)\|_{C^{2, \alpha}(\bar{\Omega})}\leq C\ ( i=1, 2, 3), t\geq 1. \end{equation} $

引理3.1^[13]   设$ a, b $是两个正常数, 假设$ \phi, \varphi\in C^1([a, \infty)), \varphi(t)\geq 0 $, 且$ \phi $下方有界, 如果对某个正常数$ K $, 有$ \phi'(t)\leq b\varphi (t) $, 且$ \varphi'(t)\leq K, t\in [a, \infty) $, 则$ \lim\limits_{t\to \infty}\varphi(t)=0 $.

定理3.1   如果$ p>b $, 条件(1.5)成立, 且$ \frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3}<1 $, 那么初边值问题(1.6)的正常数平衡解$ \tilde{\bf{u}} $是全局渐近稳定的, 此时问题(1.6)没有非常数正平衡解.

证   定义

$ \begin{eqnarray*} E(t)&=&\int_{\Omega}\left[\left(u_{1}-\tilde{u}_{1}-\tilde{u}_{1}\ln\frac{{u}_{1}} {\tilde{u}_{1}}\right) +\frac{r\tilde{u}_2}{b\tilde{u}_1}\left(u_{2}-\tilde{u}_{2}-\tilde{u}_{2}\ln\frac{u_{2}}{\tilde{u_{2}}}\right)\right]{\rm d}x\\ & &+\int_{\Omega}\left[\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\left(u_{3}-\tilde{u}_{3} -\tilde{u}_{3}\ln\frac{u_{3}}{\tilde{u}_{3}}\right) \right]{\rm d}x. \end{eqnarray*} $

则当$ \frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3}<1 $时, 有

$ \begin{eqnarray*} E'(t)&=&\int_{\Omega}\left(\frac{u_{1}-\tilde{u}_{1}}{u_{1}} \frac{\partial u_{1}}{\partial t}+\frac{r\tilde{u}_2}{b\tilde{u}_1}\frac{u_{2}-\tilde{u}_{2}}{u_{2}} \frac{\partial u_{2}}{\partial t} +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\frac{u_{3}-\tilde{u}_{3}}{u_{3}}\frac{\partial u_{3}}{\partial t}\right){\rm d}x\\ & =&-\int_{\Omega}\left(\frac{d_1 \tilde{u}_1}{u_1^2}|\nabla u_1|^2+\frac{r\tilde{u}_2}{b\tilde{u}_1}\frac{d_2\tilde{u}_2} {u_2^2}|\nabla u_2|^2 +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\frac{d_3\tilde{u}_3}{u_3^2}|\nabla u_3|^2\right){\rm d} x\\ &&+\int_{\Omega}(u_{1}-\tilde{u}_{1})\left(\frac{ru_{2}}{u_{1}}-p-u_1\right){\rm d}x \\ && +\frac{r\tilde{u}_2}{b\tilde{u}_1}\int_{\Omega}(u_{2}-\tilde{u}_{2})\left(\frac{bu_{1}}{u_{2}}-c_2-u_2-\frac{a_{12}u_3}{1+mu_2+nu_3} \right){\rm d}x\\ && +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\int_{\Omega}(u_{3}-\tilde{u}_{3})\left( -c_3+\frac{a_{21}u_2}{1+mu_2+nu_3}-u_3\right){\rm d}x \\ &=&-\int_{\Omega}\left(\frac{d_1 \tilde{u}_1}{u_1^2}|\nabla u_1|^2+\frac{r\tilde{u}_2}{b\tilde{u}_1}\frac{d_2\tilde{u}_2} {u_2^2}|\nabla u_2|^2 +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\frac{d_3\tilde{u}_3}{u_3^2}|\nabla u_3|^2\right){\rm d} x\\ && -\int_{\Omega}(u_1-\tilde{u}_1)^2{\rm d}x-\frac{r\tilde{u}_2}{b\tilde{u}_1}\int_{\Omega}(u_{2}-\tilde{u}_{2})^2{\rm d}x -\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\int_{\Omega}(u_{3}-\tilde{u}_{3})^2{\rm d}x\\ && -\int_{\Omega}\left(\sqrt{\frac{u_2r}{u_1\tilde{u}_1}} (u_{1}-\tilde{u}_{1}) -\sqrt{\frac{u_1r}{u_2\tilde{u}_1}}(u_{2}-\tilde{u}_{2})\right)^2{\rm d}x \\ && +\int_{\Omega}\frac{r\tilde{u}_2}{b\tilde{u}_1}\frac{a_{12}m\tilde{u}_3}{(1+mu_2+nu_3)(1+m\tilde{u}_2+n\tilde{u}_3)}(u_2-\tilde{u}_2)^2{\rm d}x \\ &&-\int_{\Omega}\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{b\tilde{u}_1(1+n\tilde{u}_3)}\frac{n\tilde{u}_2}{(1+mu_2+nu_3)(1+m\tilde{u}_2+n\tilde{u}_3)}(u_3-\tilde{u}_3)^2{\rm d}x \\ &\leq& -\int_{\Omega}\left(\frac{d_1 \tilde{u}_1}{u_1^2}|\nabla u_1|^2+\frac{r\tilde{u}_2}{b\tilde{u}_1}\frac{d_2\tilde{u}_2} {u_2^2}|\nabla u_2|^2 +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\frac{d_3\tilde{u}_3}{u_3^2}|\nabla u_3|^2\right){\rm d} x\\ && -\int_{\Omega}(u_1-\tilde{u}_1)^2{\rm d}x -\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\int_{\Omega}(u_{3}-\tilde{u}_{3})^2{\rm d}x\\ &&-\int_{\Omega}\left(\sqrt{\frac{u_2r}{u_1\tilde{u}_1}} (u_{1}-\tilde{u}_{1}){\rm d}x -\sqrt{\frac{u_1r}{u_2\tilde{u}_1}}(u_{2}-\tilde{u}_{2})\right)^2{\rm d}x\\ && -\int_{\Omega}\frac{r\tilde{u}_2}{b\tilde{u}_1}(1-\frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3})(u_2-\tilde{u}_2)^2{\rm d}x\\ && -\int_{\Omega}\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{b\tilde{u}_1(1+n\tilde{u}_3)} \frac{n\tilde{u}_2}{(1+mu_2+nu_3)(1+m\tilde{u}_2+n\tilde{u}_3)}(u_3-\tilde{u}_3)^2{\rm d}x \\ &\leq &-\int_{\Omega}\left(\frac{d_1 \tilde{u}_1}{u_1^2}|\nabla u_1|^2+\frac{r\tilde{u}_2}{b\tilde{u}_1}\frac{d_2\tilde{u}_2} {u_2^2}|\nabla u_2|^2 +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}\frac{d_3\tilde{u}_3}{u_3^2}|\nabla u_3|^2\right){\rm d} x\\ && -\int_{\Omega}\bigg[(u_1-\tilde{u}_1)^2+\frac{r\tilde{u}_2}{b\tilde{u}_1}(1-\frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3})(u_2-\tilde{u}_2)^2 \\ && +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}(u_{3}-\tilde{u}_{3})^2 \bigg]{\rm d}x. \end{eqnarray*} $

由于$ |{{\bf u}}|^2\leq C $, 因此有

$ \begin{eqnarray*} E'(t) &\leq& -\frac{1}{C}\int_{\Omega}\left(|\nabla u_1|^2+|\nabla u_2|^2 +|\nabla u_3|^2\right){\rm d} x\\ && -\int_{\Omega}\bigg[(u_1-\tilde{u}_1)^2 +\frac{r\tilde{u}_2}{b\tilde{u}_1}(1-\frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3})(u_2-\tilde{u}_2)^2 \\ && +\frac{a_{12}r\tilde{u}_2(1+m\tilde{u}_2)}{a_{21}b\tilde{u}_1(1+n\tilde{u}_3)}(u_{3}-\tilde{u}_{3})^2 \bigg]{\rm d}x\\ & \triangleq& -\psi_1(t)-\psi_2(t). \end{eqnarray*} $

由(1.6)和(3.7)式可知, $ \psi'_1(t), \psi'_2(t) $在$ [1, \infty) $上有界, 对上式应用引理3.1得到

$ \lim\limits_{t\rightarrow \infty}\psi_1(t)=0, \lim\limits_{t\rightarrow \infty}\psi_2(t)=0. $

因此

(3.2) $ \begin{equation} \lim\limits_{t\rightarrow \infty}\int_{\Omega}\left(|\nabla u_1|^2+|\nabla u_2|^2 +|\nabla u_3|^2\right){\rm d} x=0, \end{equation} $

(3.3) $ \begin{equation} \lim\limits_{t\rightarrow \infty}\int_{\Omega}(u_i-\tilde{u}_i)^2{\rm d}x=0{ } \;\;\;(i=1, 2, 3). \end{equation} $

由(3.8)式和Poincáre不等式可得

(3.4) $ \begin{equation} \lim\limits_{t\rightarrow \infty}\int_{\Omega}(u_i-\bar{u}_i)^2{\rm d}x=0{ } \;\;\;(i=1, 2, 3), \end{equation} $

其中$ \bar{u}_i=\frac{1}{|\Omega|}\int_{\Omega}u_i{\rm d}x, i=1, 2, 3 $, 由于

$ \begin{eqnarray*} |\Omega|(\bar{u}_i(t)-\tilde{u}_i(t))^2&=&\int_{\Omega}(\bar{u}_i-\tilde{u}_i)^2{\rm d}x=\int_{\Omega}(\bar{u}_i-u_i+u_i-\tilde{u}_i)^2{\rm d}x \\ & \leq& 2\int_{\Omega}(\bar{u}_i-u_i)^2{\rm d}x+2\int_{\Omega}(u_i-\tilde{u}_i)^2{\rm d}x. \end{eqnarray*} $

再结合(3.9)和(3.10)式可得

(3.5) $ \begin{equation} \lim\limits_{t\rightarrow \infty}\bar{u}_i=\tilde{u}_i{ } \;\;\;(i=1, 2, 3). \end{equation} $

由(3.1)式可知, 存在序列$ \{t_m\}^\infty_{m=1} $的子列仍记为$ \{t_m\}^\infty_{m=1} $及非负函数$ w_1, w_2, w_3\in C^2(\bar{\Omega}) $, 使得

$ \lim\limits_{m\rightarrow \infty}\|u_i(\cdot, t_m)-w_i(\cdot)\|_{C^2(\bar{\Omega})} =0, { } \;\;\;i=1, 2, 3. $

因此, $ w_i\equiv \tilde{u}_i\ (i=1, 2, 3) $, 于是

$ \lim\limits_{m\rightarrow \infty}\|u_i(\cdot, t_m)-\tilde{u}_i\|_{C^2(\bar{\Omega})} =0, { } \;\;\;i=1, 2, 3. $

再由$ \tilde{{\bf u}} $的局部渐近稳定性可得$ \tilde{{\bf u}} $是全局渐近稳定的. 定理3.1得证.

注   由于

$ b \tilde{u}_1-c_2 \tilde{u}_2-u^2_2-\frac{a_{12}\tilde{u}_2\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3}=0, $

所以

$ \frac{a_{12}\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3}=\frac{b\tilde{u}_{1}}{\tilde{u}_{2}}-c_2-\tilde{u}_2, $

又因为$ \frac{-p+\sqrt{p^2+4rx}}{2x} $在$ (0, \infty) $内关于$ x $非增且$ \tilde{u}_2>x_0 $, 从而

$ \begin{eqnarray*} 1-\frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3}&=&1-m(\frac{b\tilde{u}_{1}}{\tilde{u}_{2}}-c_2-\tilde{u}_2)\\ &=&1+m(c_2+\tilde{u}_2-\frac{b(-p+\sqrt{p^2+4r\tilde{u}_2})}{2\tilde{u}_{2}})\\ &\geq& 1+m(c_2+x_0-\frac{b(-p+\sqrt{p^2+4rx_0})}{2x_0})\\ &=&1+\frac{m}{x_0}\Phi(x_0). \end{eqnarray*} $

由此可见, 当$ \Phi(x_0)>-\frac{x_0}{m} $时, 有

$ 1-\frac{a_{12}m\tilde{u}_3}{1+m\tilde{u}_2+n\tilde{u}_3}>0, $

所以, 当$ p>b $, 条件(1.5)成立, 且$ \Phi(x_0)>-\frac{x_0}{m} $时, 初边值问题(1.6) 的正常数平衡解$ \tilde{\bf{u}} $是全局渐近稳定的.