微生物的连续培养是近几十年来发展起来的一门数学与微生物学交叉的学科,是通过微分方程建立数学模型来刻画微生物连续培养的一门生物技术.恒化器(chemostat)是一种用于微生物连续培养的实验装置,被广泛地应用于微生物的生产、环境污染的控制、污水处理、生物制药以及废料处理等领域,因而chemostat模型受到了国内外许多专家学者的广泛关注^[1-20].在微生物的连续培养过程中微生物会受到污染环境中毒素的侵入,导致培养的微生物数量降低,因此在微生物培养中考虑毒素等因素的影响是非常有意义的.如文献[2, 15]分别考察了具有内部毒素和外加毒素的均匀chemostat模型正解的定性性质.文献[13]分析了具有毒素的非均匀chemostat模型平衡态正解的存在性和稳定性.文献[14, 16]讨论了具有质粒载体微生物与质粒自由微生物以及毒素的非均匀chemostat模型平衡态正解的存在性、稳定性、唯一性和多解性.有关污染环境中毒素对种群影响的研究可见文献[21-24]及其参考文献.

本文讨论如下具有毒素的非均匀chemostat食物链模型

(1.1) $ \begin{equation} \left\{\begin{array}{lll} S_t = \Delta S-auf(S, u), &x\in\Omega, &t>0, \\ u_t = \Delta u+auf(S, u)-bvg(u, v)-\gamma pu, &x\in\Omega, &t>0, \\ v_t = \Delta v+b(1-k)vg(u, v), &x\in\Omega, &t>0, \\ p_t = \Delta p+bkvg(u, v), &x\in\Omega, &t>0, \\ { } \frac{\partial S}{\partial n}+rS = S^{0}(x), \frac{\partial u}{\partial n}+ru = 0, \frac{\partial v}{\partial n}+rv = 0, \frac{\partial p}{\partial n}+rp = 0, &x\in\partial\Omega, &t>0, \\ S(0, x) = S_0(x)\geq0, \not\equiv0, u(0, x) = u_0(x)\geq0, \not\equiv0, &x\in\Omega, \\ v(0, x) = v_0(x)\geq0, \not\equiv0, p(0, x) = p_0(x)\geq0, \not\equiv0, &x\in\Omega, \\ \end{array}\right. \end{equation} $

其中$ S(x, t), u(x, t), v(x, t), p(x, t) $分别为反应器中营养物、两微生物和毒素的浓度. $ \Omega $是$ {{{\Bbb R}} }^{N} $中带有光滑边界$ \partial\Omega $的有界区域. $ f(S, u) = \frac{S}{(1+h_{1}S)(1+k_{1}u)}, g(u, v) = \frac{u}{(1+h_{2}u)(1+k_{2}v)} $是Crowley-Martin反应函数,刻画了物种的生长规律, $ h_i, k_{i}(i = 1, 2) $都是正常数. $ a>0, b>0 $分别是$ u $和$ v $的最大生长率. $ \gamma>0 $为毒素$ p $对$ u $生长的致命作用. $ 0<k<1 $表示用于生产毒素的消耗部分. $ S^{0}(x)\geq0, \not\equiv0 $是营养物的输入浓度.

本文主要研究系统$ (1.1) $的平衡态系统

(1.2) $ \begin{equation} \left\{\begin{array}{lll} \Delta S-auf(S, u) = 0, &x\in\Omega, \\ \Delta u+auf(S, u)-bvg(u, v)-\gamma pu = 0, &x\in\Omega, \\ \Delta v+b(1-k)vg(u, v) = 0, &x\in\Omega, \\ \Delta p+bkvg(u, v) = 0, &x\in\Omega, \\ { } \frac{\partial S}{\partial n}+rS = S^{0}(x), \frac{\partial u}{\partial n}+ru = 0, \frac{\partial v}{\partial n}+rv = 0, \frac{\partial p}{\partial n}+rp = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

令$ Q(x) = (1-k)p-kv $,则由系统$ (1.2) $可知$ Q(x) $满足

$ \Delta Q = 0, x\in\Omega, \ \ \ \ \frac{\partial Q}{\partial n}+rQ = 0, x\in\partial\Omega. $

由线性椭圆方程的性质和最大值原理可知, $ Q\equiv0 $.即$ p = \frac{k}{1-k}v\triangleq\beta v. $于是系统$ (1.2) $简化为

(1.3) $ \begin{equation} \left\{\begin{array}{lll} \Delta S-auf(S, u) = 0, &x\in\Omega, \\ \Delta u+auf(S, u)-bvg(u, v)-\gamma\beta uv = 0, &x\in\Omega, \\ \Delta v+b(1-k)vg(u, v) = 0, &x\in\Omega, \\ { } \frac{\partial S}{\partial n}+rS = S^{0}(x), \frac{\partial u}{\partial n}+ru = 0, \frac{\partial v}{\partial n}+rv = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

纵观近几十年来对chemostat模型的研究成果,基本都是均匀chemostat模型或者非均匀chemostat竞争模型,对具有毒素的非均匀chemostat食物链模型的研究很少见,诸如毒素对微生物的影响分析都是需要解决的新问题.当$ k = 0 $且没有毒素$ p $时,文献[8]考察了此时系统$ (1.3) $中$ f(S, u), g(u, v) $是Beddington-DeAngelis反应函数的非均匀chemostat模型,得到了正平衡解的存在性、稳定性和解的渐近行为.文献[19]利用不动点指数理论、分歧理论和扰动理论讨论了此时系统$ (1.3) $正解的存在性、稳定性和唯一性.本文在文献[19]的基础上研究系统$ (1.3) $,主要分析毒素对系统$ (1.3) $的影响,得到正解的存在性、稳定性和唯一性.值得一提的是,与文献[19]的系统相比较,由于毒素的引入使得系统$ (1.3) $的动力学行为的研究变得更加复杂.

首先给出一些预备知识和半平凡解的性质.

引理 2.1^[25]  设$ q_{1}(x), q_{2}(x)\in C(\overline{\Omega}) $且$ q_{1}(x)\geq0, q_{2}(x)>0 $.令$ \eta_{i}(q_{1}, q_{2})(i\geq1) $是如下问题的特征值

$ -\Delta\psi+q_{1}\psi = \eta_{i} q_{2}\psi, x\in\Omega, \ \ \ \ \frac{\partial \psi}{\partial n}+r\psi = 0, x\in\partial\Omega, $

则$ 0<\eta_{1}(q_{1}, q_{2})<\eta_{2}(q_{1}, q_{2})\leq\cdots\rightarrow\infty $. $ \eta_{i}(q_{1}, q_{2}) $关于$ q_{1}, q_{2} $连续且若$ q_{1}\leq q'_{1} $,则$ \eta_{i}(q_{1}, q_{2})\leq\eta_{i}(q'_{1}, q_{2}) $;若$ q_{2}\leq q'_{2} $,则$ \eta_{i}(q_{1}, q_{2})\geq\eta_{i}(q_{1}, q'_{2}) $.而且$ \eta_{1}(q_{1}, q_{2}) $是简单的.

由椭圆方程的极值原理可知如下问题

$ \Delta z = 0, \ x\in \Omega, \ \ \frac{\partial z}{\partial n}+rz = S^{0}(x), \ x\in\partial\Omega $

存在唯一正解,记为$ z. $而且根据极值原理可知$ z $在$ \overline{\Omega} $上有界.显然, $ (z, 0, 0) $是系统$ (1.3) $的平凡解.

令$ \lambda_{1} $为如下特征值问题的主特征值

$ \Delta\varphi+\lambda_{1} f(z, 0)\varphi = 0, x\in\Omega, \ \ \ \ \frac{\partial \varphi}{\partial n}+r\varphi = 0, x\in\partial\Omega, $

相应的特征函数为$ \varphi_{1} $且$ \|\varphi_{1}\|_{2} = 1 $.

引理 2.2^[19]   问题

(2.1) $ \begin{equation} \Delta u+auf(z-u, u) = 0, x\in\Omega, \ \ \ \ \frac{\partial u}{\partial n}+ru = 0, x\in\partial\Omega \end{equation} $

当$ a\leq\lambda_{1} $时, (2.1)式只有零解.当$ a>\lambda_{1} $时, (2.1)式有惟一正解记为$ \Theta $且$ \Theta $满足以下条件

(i) $ \Theta<z; $

(ii) 当$ a>\lambda_{1} $时, $ \Theta $连续可微且当$ a $增大时$ \Theta $也逐点递增;

(iii) 若记(2.1)式在$ \Theta $处的线性化算子为$ L $,则

$ L = -\Delta-a[f(z-\Theta, \Theta)-\Theta f^{'}_{1}(z-\Theta, \Theta)+\Theta f^{'}_{2}(z-\Theta, \Theta)] $

的所有特征值都大于0.

于是,当$ a>\lambda_{1} $时系统$ (1.3) $存在半平凡解$ (z-\Theta, \Theta, 0). $下面给出平凡解和半平凡解的稳定性.

引理 2.3   (i) 若$ a>\lambda_{1} $,则平凡解$ (z, 0, 0) $不稳定;若$ a<\lambda_{1} $,则平凡解$ (z, 0, 0) $稳定.

(ii) 设$ a>\lambda_{1} $.若$ b>\frac{\lambda^{*}_{1}}{1-k}, $则半平凡解$ (z-\Theta, \Theta, 0) $不稳定;若$ b<\frac{\lambda^{*}_{1}}{1-k} $,则半平凡解$ (z-\Theta, \Theta, 0) $稳定.其中$ \lambda^{*}_{1} $为如下特征值问题的主特征值

(2.2) $ \begin{equation} \Delta\psi+\lambda^{*}_{1} g(\Theta, 0)\psi = 0, x\in\Omega, \ \ \ \ \frac{\partial \psi}{\partial n}+r\psi = 0, x\in\partial\Omega, \end{equation} $

相应的特征函数为$ \psi_{1} $且$ \|\psi_{1}\|_{2} = 1 $.

证   因为(i)和(ii)的证明一样,故证明(ii).考虑如下特征值问题

(2.3) $ \begin{equation} \left\{\begin{array}{lll} \Delta \xi-a\Theta f'_{1}(z-\Theta, \Theta)\xi-a[f(z-\Theta, \Theta)+\Theta f'_{2}(z-\Theta, \Theta)]\zeta+\mu \xi = 0, &x\in\Omega, \\ \Delta \zeta+a\Theta f'_{1}(z-\Theta, \Theta)\xi+a[f(z-\Theta, \Theta)+\Theta f'_{2}(z-\Theta, \Theta)]\zeta-bg(\Theta, 0)\eta\\ -\gamma\beta\Theta\eta+\mu\zeta = 0, &x\in\Omega, \\ \Delta \eta+b(1-k)g(\Theta, 0)\eta+\mu\eta = 0, &x\in\Omega, \\ { } \frac{\partial \xi}{\partial n}+r\xi = 0, \frac{\partial \zeta}{\partial n}+r\zeta = 0, \frac{\partial \eta}{\partial n}+r\eta = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

令$ \omega = \xi+\zeta, $则系统$ (2.3) $成为

(2.4) $ \begin{equation} \left\{\begin{array}{lll} \Delta \xi+a[f(z-\Theta, \Theta)-\Theta f'_{1}(z-\Theta, \Theta)+\Theta f'_{2}(z-\Theta, \Theta)]\xi\\ -a[f(z-\Theta, \Theta)+\Theta f'_{2}(z-\Theta, \Theta)]\omega+\mu\xi = 0, &x\in\Omega, \\ \Delta \omega-[bg(\Theta, 0)+\gamma\beta\Theta]\eta+\mu\omega = 0, &x\in\Omega, \\ \Delta \eta+b(1-k)g(\Theta, 0)\eta+\mu\eta = 0, &x\in\Omega, \\ { } \frac{\partial \xi}{\partial n}+r\xi = 0, \frac{\partial \omega}{\partial n}+r\omega = 0, \frac{\partial \eta}{\partial n}+r\eta = 0, &x\in\partial\Omega.\\ \end{array}\right. \end{equation} $

由Riesz-Schauder定理和特征值理论可知系统(2.4)的特征值由算子:$ L, -\Delta $和$ L^{*} = -\Delta-b(1-k)g(\Theta, 0) $的特征值构成.显然, $ L, -\Delta $的所有特征值大于0.若$ b>\frac{\lambda^{*}_{1}}{1-k}, $则$ L^{*} $有负特征值,于是$ (z-\Theta, \Theta, 0) $不稳定.同理,若$ b<\frac{\lambda^{*}_{1}}{1-k}, $则$ L^{*} $的所有特征值大于0,平凡解$ (z-\Theta, \Theta, 0) $稳定.

其次给出系统$ (1.3) $正解的先验估计.

引理 2.4  如果$ (S, u, v) $是系统$ (1.3) $的非负解且$ u\not\equiv0, v\not\equiv 0 $,则

(i) $ 0<S<z, \ 0<u<\Theta, \ 0<v<z(1-k); $

(ii) $ a>\lambda_{1}, \ b>\frac{\lambda^{*}_{1}}{1-k}. $

证   首先由强最大值原理可知$ S>0, u>0, v>0. $令$ P = z-S-u-\frac{v}{1-k} $,则

$ -\Delta P = \gamma\beta uv\geq0, \ x\in \Omega, \ \ \ \frac{\partial P}{\partial n}+rP = 0, \ x\in\partial\Omega. $

又由强最大值原理可知$ P\geq0. $则$ 0<S<z, S< z-u, v<z(1-k) $且$ S< z-\frac{v}{1-k} $.关于$ a>\lambda_{1} $的证明与文献[1,引理4.1]类似.根据系统$ (1.3) $的第二个方程得

$ \Delta u+auf(S, u)-bvg(u, v)-\gamma\beta uv\leq\Delta u+auf(S, u)\leq\Delta u+auf(z-u, u). $

故$ u $是方程(2.1)的下解,而$ z $是其上解,由上下解方法、$ \Theta $的唯一性及极值原理可知$ u<\Theta. $$ b>\frac{\lambda^{*}_{1}}{1-k} $的证明同$ a>\lambda_{1} $.

设$ z' = z-S. $则系统$ (1.3) $成为

(2.5) $ \begin{equation} \left\{\begin{array}{lll} \Delta z'+auf(z-z', u) = 0, &x\in\Omega, \\ \Delta u+auf(z-z', u)-bvg(u, v)-\gamma\beta uv = 0, &x\in\Omega, \\ \Delta v+b(1-k)vg(u, v) = 0, &x\in\Omega, \\ { } \frac{\partial z'}{\partial n}+rz' = 0, \frac{\partial u}{\partial n}+ru = 0, \frac{\partial v}{\partial n}+rv = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

显然(0, 0, 0)是系统(2.5)的平凡解,当$ a>\lambda_{1} $时, $ (\Theta, \Theta, 0) $是系统(2.5)的唯一半平凡解.

记

$ C_{0}(\overline{\Omega}) = \{u\in C(\overline{\Omega}): \frac{\partial u}{\partial n}+ru = 0, x\in\partial\Omega\}, X = C_{0}(\overline{\Omega})\times C_{0}(\overline{\Omega})\times C_{0}(\overline{\Omega}), $

$ W = \{( z', u, v)\in X: z'\geq0, u\geq0, v\geq0, x\in\overline{\Omega}\}, D = \{(z', u, v)\in W: z'\leq \max \limits_{\bar{\Omega}}z(x)+1, $

$ u\leq \max \limits_{\bar{\Omega}}\Theta(x)+1, v\leq (1-k)\mathop{\max }\limits_{\bar{\Omega}}z(x)+1\}. $

定义算子$ {\bf A}:D\rightarrow W $为

$ {\bf A}(z', u, v) = (-\Delta+M)^{-1}\left(\begin{array} {cc}auf(z-z', u)+Mz'\\ auf(z-z', u)-bvg(u, v)-\gamma\beta uv+Mu\\ b(1-k)vg(u, v)+Mv\end{array}\right), $

其中$ M $是充分大的正常数.由紧算子理论可知$ {\bf A} $是紧的.而且$ {\bf A} $是连续可微的.则由引理2.4可知,系统(2.5)有非负解当且仅当$ {\bf A} $在$ D $中有不动点.

接下来运用Dancer指数定理^[26]给出算子$ {\bf A} $在$ (0, 0, 0), D $和$ (\Theta, \Theta, 0) $的不动点指数.由于证明与文献[3,引理3.3-3.6]类似,故在此省略.

引理 2.5   设$ a>\lambda_{1}. $则

(i)  $ {\rm{inde}}{{\rm{x}}_W}({\bf{A}},(0,0,0)) = 0;$

(ii)  ${\rm{inde}}{{\rm{x}}_W}({\bf{A}},D) = 1; $

(iii) 若$ b>\frac{\lambda^{*}_{1}}{1-k} $,则$ {\rm{minde}}{{\rm{x}}_W}({\bf{A}},(\Theta ,\Theta ,0)) = 0;$若$ b<\frac{\lambda^{*}_{1}}{1-k} $,则${\rm{inde}}{{\rm{x}}_W}({\bf{A}},(\Theta ,\Theta ,0)) = 1. $

{\heiti\bf定理2.1}{\quad}若$ a>\lambda_{1}, b>\frac{\lambda^{*}_{1}}{1-k}, $则系统$ (1.3) $至少存在一个正解.

证   由前提条件、引理2.5和不动点指数的可加性有

$ 0 = {\rm index}_W({\bf A}, (0, 0, 0))+{\rm index}_{W}({\bf A}, (\Theta, \Theta, 0)) = {\rm index}_{W}({\bf A}, D), $

矛盾.于是当$ a>\lambda_{1}, b>\frac{\lambda^{*}_{1}}{1-k} $时系统(2.5)至少存在一个正解.因此若$ a>\lambda_{1}, b>\frac{\lambda^{*}_{1}}{1-k}, $则系统$ (1.3) $至少存在一个正解.

本小节通过详细分析毒素的影响来讨论系统$ (1.3) $共存解的稳定性和唯一性.首先将证明若$ (S, u, v) $是系统$ (1.3) $的正解,则当$ \gamma\rightarrow\infty $时, $ (S, u, \gamma v) $趋于如下极限问题的正解

(3.1) $ \begin{equation} \left\{\begin{array}{lll} \Delta S-auf(S, u) = 0, &x\in\Omega, \\ \Delta u+auf(S, u)-\beta uw = 0, &x\in\Omega, \\ \Delta w+b(1-k)wg(u, 0) = 0, &x\in\Omega, \\ { }\frac{\partial S}{\partial n}+rS = S^{0}(x), \frac{\partial u}{\partial n}+ru = 0, \frac{\partial w}{\partial n}+rw = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

引理 3.1   设$ a>\lambda_{1}, (S, u, w) $是系统(3.1)的非负解.则

(i) $ z-\Theta<S<z, 0<u<\Theta, b>\frac{\lambda^{*}_{1}}{1-k} $;

(ii) 存在正常数$ C $使得$ \|w\|_{\infty}<C. $

证    (i) 证明与引理2.4类似.

(ii) 显然系统(3.1)存在平凡解$ (z, 0, 0) $和半平凡解$ (z-\Theta, \Theta, 0)(a>\lambda_{1}) $.采用反证法.假设$ b_{i}\rightarrow b\in[\frac{\lambda^{*}_{1}}{1-k}, +\infty), (S_{i}, u_{i}, w_{i}) $是系统(3.1)带有$ b = b_{i} $和$ \|w_{i}\|_{\infty}\rightarrow\infty $的正解.令$ W_{i} = \frac{w_{i}}{\|w_{i}\|_{\infty}}, $则

(3.2) $ \begin{equation} \left\{\begin{array}{lll} \Delta S_{i}-au_{i}f(S_{i}, u_{i}) = 0, &x\in\Omega, \\ \Delta u_{i}+au_{i}f(S_{i}, u_{i})-\beta u_{i}W_{i}\|w_{i}\|_{\infty} = 0, &x\in\Omega, \\ \Delta W_{i}+b_{i}(1-k)W_{i}g(u_{i}, 0) = 0, &x\in\Omega, \\ { }\frac{\partial S_{i}}{\partial n}+rS_{i} = S^{0}(x), \frac{\partial u_{i}}{\partial n}+ru_{i} = 0, \frac{\partial W_{i}}{\partial n}+rW_{i} = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

由(i)和$ L^{p} $估计以及Soblove嵌入定理得$ W_{i}\rightarrow W\geq0, \not\equiv0, u_{i}\rightarrow u>0 $且

$ \Delta W+b(1-k)Wg(u, 0) = 0, \ x\in\Omega, \ \ \ \frac{\partial W}{\partial n}+rW = 0, \ x\in\partial\Omega. $

根据强最大值原理可知$ W>0. $因此, $ W_{i}>0 $且$ -af(S_{i}, u_{i})+\beta W_{i}\|w_{i}\|_{\infty}>0 $.这表明算子$ -\Delta-af(S_{i}, u_{i})+\beta W_{i}\|w_{i}\|_{\infty} $可逆,则$ u_{i}\equiv0, $矛盾.因此结论成立.

利用与文献[14,引理4.3]类似的证明过程可得如下引理,在此省略其证明.

引理 3.2   设$ a>\lambda_{1} $.则系统(3.1)存在正解当且仅当$ b>\frac{\lambda^{*}_{1}}{1-k} $.并且存在充分小的$ \epsilon>0 $使得若$ \frac{\lambda^{*}_{1}}{1-k}<b\leq\frac{\lambda^{*}_{1}}{1-k}+\epsilon $,则系统(3.1)存在唯一正解且稳定.

引理 3.3   设$ a>\lambda_{1}, b_{i}\rightarrow b\in (\frac{\lambda^{*}_{1}}{1-k}, +\infty), \gamma_{i}\rightarrow\infty, (S_{i}, u_{i}, v_{i}) $是系统$ (1.3) $的正解,则$ v_{i}\rightarrow0 $且$ (S_{i}, u_{i}, \gamma_{i}v_{i})\rightarrow (S, u, w) $,这里$ (S, u, w) $是系统(3.1)的正解.

证   考虑如下问题

(3.3) $ \begin{equation} \left\{\begin{array}{lll} \Delta S_{i}-au_{i}f(S_{i}, u_{i}) = 0, &x\in\Omega, \\ \Delta u_{i}+au_{i}f(S_{i}, u_{i})-b_{i}v_{i}g(u_{i}, v_{i})-\gamma_{i}\beta u_{i}v_{i} = 0, &x\in\Omega, \\ \Delta v_{i}+b_{i}(1-k)v_{i}g(u_{i}, v_{i}) = 0, &x\in\Omega, \\ { }\frac{\partial S_{i}}{\partial n}+rS_{i} = S^{0}, \frac{\partial u_{i}}{\partial n}+ru_{i} = 0, \frac{\partial v_{i}}{\partial n}+rv_{i} = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

由引理2.4可知$ \{\Delta S_{i}\}, \{\Delta v_{i}\} $在$ L^{\infty}(\Omega) $中有界.根据椭圆方程的正则性理论和Soblove嵌入定理可设在$ C^{1}(\overline{\Omega}) $中$ S_{i}\rightarrow S\geq0, v_{i}\rightarrow v\geq0. $又由强最大值原理得$ S>0. $而

$ -\Delta u_{i} = au_{i}f(S_{i}, u_{i})-b_{i}v_{i}g(u_{i}, v_{i})-\gamma_{i}\beta u_{i}v_{i}\leq au_{i}f(S_{i}, u_{i}). $

上式同乘以$ u_{i} $并在$ \Omega $上积分得

$ \int_{\Omega}|\nabla u_{i}|^{2}{\rm d}x+\int_{\partial\Omega}ru^{2}_{i}{\rm d}s\leq a\int_{\Omega}f(S_{i}, u_{i})u^{2}_{i}{\rm d}x. $

因此$ \{u_{i}\} $在$ W^{1, 2}(\Omega) $上有界,则设在$ W^{1, 2}(\Omega) $中$ u_{i}\rightarrow u\geq0 $ (弱).下面证明几乎处处在$ \Omega $上有$ uv = 0. $令$ P_{i} = S_{i}+u_{i}+\frac{v_{i}}{1-k}, $则在$ W^{1, 2}(\Omega) $中$ P_{i}\rightarrow P = S+u+\frac{v}{1-k} $ (弱)且$ P_{i} $满足

$ \Delta P_{i}-\beta\gamma_{i}u_{i}v_{i} = 0, \ x\in \Omega, \ \ \frac{\partial P_{i}}{\partial n}+rP_{i} = S^{0}, \ x\in\partial\Omega. $

上式两边同乘以$ \chi\in C^{\infty}_{0}(\Omega) $并在$ \Omega $上积分,得

$ \int_{\Omega}\nabla P_{i}\cdot\nabla\chi {\rm d}x+\beta\gamma_{i}\int_{\Omega}u_{i}v_{i}\chi {\rm d}x = 0. $

上式取极限得对任意的$ \chi\in C^{\infty}_{0}(\Omega) $有$ \int_{\Omega}uv\chi {\rm d}x = 0. $于是几乎处处在$ \Omega $上有$ uv = 0. $再对$ v_{i} $的方程取弱极限,得

$ \Delta v+b(1-k)vg(u, v) = 0, \ x\in \Omega, \ \ \frac{\partial v}{\partial n}+rv = 0, \ x\in\partial\Omega. $

若$ v\geq0, \not\equiv0, $则由强最大值原理知$ v>0. $因此, $ v $有两种可能: $ v>0 $或$ v\equiv0. $

假设$ v>0. $由几乎处处在$ \Omega $上$ uv = 0 $可得几乎处处在$ \Omega $上$ u = 0. $即在$ W^{1, 2}(\Omega) $上$ u_{i}\rightarrow0 $ (弱).因为在$ C^{1}(\overline{\Omega}) $中$ S_{i}\rightarrow S, v_{i}\rightarrow v, $故当$ \gamma_{i}\rightarrow\infty $时, $ \Delta+af(S_{i}, u_{i})-\frac{bv_{i}}{(1+h_{2}u_{i})(1+k_{2}v_{i})} -\beta\gamma_{i}v_{i} $可逆.因此, $ u_{i}\equiv0, $这与$ (S_{i}, u_{i}, v_{i}) $是系统$ (1.3) $的正解矛盾,故在$ C^{1}(\overline{\Omega}) $中$ v_{i}\rightarrow 0 $.接着证明$ u\not\equiv0. $反证,假设在$ W^{1, 2}(\Omega) $中$ u_{i}\rightarrow 0 $ (弱).给$ S_{i} $的方程乘以$ \chi\in C^{\infty}_{0}(\Omega) $并在$ \Omega $上积分,得

$ \int_{\Omega}\nabla S_{i}\cdot\nabla\chi {\rm d}x+a\int_{\Omega}f(S_{i}, u_{i})u_{i}\chi {\rm d}x = 0. $

上式取极限,则$ \int_{\Omega}\nabla S\cdot\nabla\chi {\rm d}x = 0 $,这表明$ S $是如下问题的弱解

$ -\Delta S = 0, \ x\in \Omega, \ \ \frac{\partial S}{\partial n}+r S = S^{0}, \ x\in\partial\Omega. $

因此$ S = z. $即在$ C^{1}(\overline{\Omega}) $中$ S_{i}\rightarrow z $.令$ V_{i} = \frac{v_{i}}{\|v_{i}\|_{\infty}}, $则

$ \Delta V_{i}+b_{i}(1-k)V_{i}g(u_{i}, v_{i}) = 0, \ x\in\Omega, \ \ \frac{\partial V_{i}}{\partial n}+rV_{i} = 0, \ x\in\partial\Omega. $

由$ L^{p} $估计和Soblove嵌入定理可设存在$ V\in C^{1}_{B}(\overline{\Omega}) $使得在$ C^{1}(\overline{\Omega}) $中$ V_{i}\rightarrow V $且$ V\geq0, \not\equiv0 $满足

$ \Delta V = 0, \ x\in\Omega, \ \ \frac{\partial V}{\partial n}+rV = 0, \ x\in\partial\Omega. $

由此可得$ V\equiv0, $矛盾.于是$ u\not\equiv0. $令$ w_{i} = \gamma_{i}v_{i}, $则$ (S_{i}, u_{i}, w_{i}) $满足

(3.4) $ \begin{equation} \left\{\begin{array}{lll} \Delta S_{i}-au_{i}f(S_{i}, u_{i}) = 0, &x\in\Omega, \\ \Delta u_{i}+au_{i}f(S_{i}, u_{i})-b_{i}v_{i}g(u_{i}, v_{i})-\beta u_{i}w_{i} = 0, &x\in\Omega, \\ \Delta w_{i}+b_{i}(1-k)w_{i}g(u_{i}, v_{i}) = 0, &x\in\Omega, \\ { }\frac{\partial S_{i}}{\partial n}+rS_{i} = S^{0}, \frac{\partial u_{i}}{\partial n}+ru_{i} = 0, \frac{\partial w_{i}}{\partial n}+rw_{i} = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

下面证明存在$ \tilde{M}>0 $使得$ \|w_{i}\|_{\infty}\leq\tilde{M}. $假设$ \|w_{i}\|_{\infty}\rightarrow\infty. $令$ W_{i} = \frac{w_{i}}{\|w_{i}\|_{\infty}}, $则

$ \Delta W_{i}+b_{i}(1-k)W_{i}g(u_{i}, v_{i}) = 0, \ x\in\Omega, \ \ \frac{\partial W_{i}}{\partial n}+rW_{i} = 0, \ x\in\partial\Omega. $

再由$ L^{p} $估计和Soblove嵌入定理可设存在$ W\in C^{1}_{B}(\overline{\Omega}) $使得在$ C^{1}(\overline{\Omega}) $中$ W_{i}\rightarrow W $且$ W\geq0, \not\equiv0 $满足

$ \Delta W+b(1-k)Wg(u, 0) = 0, \ x\in\Omega, \ \ \frac{\partial W}{\partial n}+rW = 0, \ x\in\partial\Omega. $

由强最大值原理可知$ W>0. $给系统(3.4)的第二个方程乘以$ u_{i} $并在$ \Omega $上积分,得

$ \begin{eqnarray*} &&\int_{\Omega}|\nabla u_{i}|^{2}{\rm d}x+\int_{\partial\Omega}ru^{2}_{i}{\rm d}s\\ & = &a\int_{\Omega}f(S_{i}, u_{i})u^{2}_{i}{\rm d}x-b_{i}\int_{\Omega}g(u_{i}, v_{i})u_{i}v_{i}{\rm d}x -\beta\|w_{i}\|_{\infty}\int_{\Omega}W_{i}u^{2}_{i}{\rm d}x\rightarrow-\infty. \end{eqnarray*} $

这与$ \{u_{i}\} $有界矛盾,故存在$ \tilde{M}>0 $使得$ \|w_{i}\|_{\infty}\leq\tilde{M}. $再由$ L^{p} $估计和Soblove嵌入定理可设存在$ w\in C^{1}_{B}(\overline{\Omega}) $使得在$ C^{1}(\overline{\Omega}) $中$ w_{i}\rightarrow w $,则根据$ \|w_{i}\|_{\infty}\leq\tilde{M} $可知

$ -\Delta u_{i} = au_{i}f(S_{i}, u_{i})-b_{i}v_{i}g(u_{i}, v_{i})-\beta u_{i}w_{i}\in L^{\infty}(\Omega). $

则存在$ u\in C^{1}(\overline{\Omega}) $使得在$ C^{1}(\overline{\Omega}) $中$ u_{i}\rightarrow u $.系统(3.4)取弱极限,得

(3.5) $ \begin{equation} \left\{\begin{array}{lll} \Delta S-auf(S, u) = 0, &x\in\Omega, \\ \Delta u+auf(S, u)-\beta uw = 0, &x\in\Omega, \\ \Delta w+b(1-k)wg(u, 0) = 0, &x\in\Omega, \\ { }\frac{\partial S}{\partial n}+rS = S^{0}, \frac{\partial u}{\partial n}+ru = 0, \frac{\partial w}{\partial n}+rw = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

因此, $ (S, u, w) $是系统(3.1)的非负解.最后证明$ (S, u, w) $是系统(3.1)的正解.由强最大值原理易得$ u>0. $同理得$ w\geq0, \not\equiv0. $否则若$ w\equiv0, $则$ S = z-\Theta, u = \Theta, $于是

$ b_{i}(1-k) = \lambda_{1}(g(u_{i}, 0))\rightarrow\lambda_{1}(g(\Theta, 0)) = \lambda^{*}_{1}, $

这表明$ b_{i}\rightarrow b = \frac{\lambda^{*}_{1}}{1-k}, $与$ b\in (\frac{\lambda^{*}_{1}}{1-k}, +\infty) $矛盾.则$ w\geq0, \not\equiv0. $又由强最大值原理可得$ w>0. $因此$ (S, u, w) $是系统(3.1)的正解.

最后给出系统$ (1.3) $正解的稳定性和唯一性结果.

定理 3.1   设$ a>\lambda_{1} $.则存在充分小的$ \varepsilon_{0}>0 $和充分大的$ \gamma_{0}>0 $使得若$ \gamma>\gamma_{0} $且$ b\in(\frac{\lambda^{*}_{1}}{1-k}, \frac{\lambda^{*}_{1}}{1-k}+\varepsilon_{0}] $,则系统$ (1.3) $存在唯一正解且稳定.

证   首先由定理2.1得系统$ (1.3) $存在正解.设$ (S, u, v) $是系统$ (1.3) $的正解.由引理3.3可知$ (S, u, \gamma v) $接近于$ (\tilde{S}, \tilde{u}, \tilde{w}) $,这里$ (\tilde{S}, \tilde{u}, \tilde{w}) $是系统(3.1)的正解.接下来证明存在充分小的$ 0<\varepsilon_{0}\leq \varepsilon $和充分大的$ \gamma_{0}>0 $使得对于$ b\in(\frac{\lambda^{*}_{1}}{1-k}, \frac{\lambda^{*}_{1}}{1-k}+\varepsilon_{0}] $且$ \gamma\geq\gamma_{0} $存在唯一解$ (S, u, \gamma v) $接近于$ (\tilde{S}, \tilde{u}, \tilde{w}). $

令$ z' = z-S, \bar{v} = \gamma v, \varepsilon = \frac{1}{\gamma}, $则

(3.6) $ \begin{equation} \left\{\begin{array}{lll} \Delta z'+auf(z-z', u) = 0, &x\in\Omega, \\ \Delta u+auf(z-z', u)-b\varepsilon\bar{v}g(u, \varepsilon\bar{v})-\beta u\bar{v} = 0, &x\in\Omega, \\ \Delta \bar{v}+b(1-k)\bar{v}g(u, \varepsilon\bar{v}) = 0, &x\in\Omega, \\ [2mm] { } \frac{\partial z'}{\partial n}+rz' = 0, \frac{\partial u}{\partial n}+ru = 0, \frac{\partial \bar{v}}{\partial n}+r\bar{v} = 0, &x\in\partial\Omega. \end{array}\right. \end{equation} $

显然, $ (S, u, v) $是系统$ (1.3) $的正解当且仅当$ (z', u, \gamma v) $是系统(3.6)带有$ \varepsilon = \frac{1}{\gamma} $的正解.因此只需证明系统(3.6)的正解$ (z', u, \gamma v) $唯一即可.对于固定的$ \varepsilon\geq0, $取$ b $为分歧参数由局部分歧理论^[27]可得$ (\frac{\lambda^{*}_{1}}{1-k}, \Theta, \Theta, 0) $是系统(3.6)的分歧点,因此存在$ \delta_{0}>0 $和$ C^{1} $曲线

$ \Gamma_{\varepsilon} = \{(b(\varepsilon, s), z'(\varepsilon, s), u(\varepsilon, s), \bar{v}(\varepsilon, s)):0<s<\delta_{0}, 0\leq\varepsilon\leq\delta_{0}\}, $

这里$ b(0, 0) = \frac{\lambda^{*}_{1}}{1-k}, z'(\varepsilon, s) = \Theta-s(\varphi_{2}+\kappa_{1}(\varepsilon, s)), $$ u(\varepsilon, s) = \Theta-s(\phi_{2}+\kappa_{2}(\varepsilon, s)), $$ \bar{v}(\varepsilon, s) = s(\psi_{1}+\kappa_{3}(\varepsilon, s)), $其中

$ \begin{eqnarray*} \varphi_{2}& = &L^{-1}(-\frac{\varepsilon a}{1-k}[f(z-\Theta, \Theta)+\Theta f'_{2}(z-\Theta, \Theta)]\psi_{1}\\ &&-a[f(z-\Theta, \Theta)+\Theta f'_{2}(z-\Theta, \Theta)]K(\beta\Theta\psi_{1})) <0, \end{eqnarray*} $

$ \phi_{2} = L^{-1}((b\varepsilon g(\Theta, 0)+\beta\Theta)\psi_{1}+a\Theta f'_{1}(z-\Theta, \Theta)(\frac{\varepsilon\psi_{1}}{1-k}+K(\beta\Theta\psi_{1})))>0. $

显然系统(3.6)在邻域$ (\frac{\lambda^{*}_{1}}{1-k}, \Theta, \Theta, 0) $内的所有正解都在曲线$ \Gamma_{\varepsilon} $上.因此只需证明对固定的$ \varepsilon, \Gamma_{\varepsilon} $一致覆盖$ b $ -区间:$ b\in(\frac{\lambda^{*}_{1}}{1-k}, \frac{\lambda^{*}_{1}}{1-k}+\varepsilon_{0}] $仅一次.通过计算可得

$ \frac{\partial b}{\partial s}(0, 0) = \frac{\lambda^{*}_{1}\int_{\Omega}g'_{1}(\Theta, 0)\phi_{2}\psi^{2}_{1}{\rm d}x}{(1-k)\int_{\Omega}g(\Theta, 0)\psi^{2}_{1}{\rm d}x}>0. $

取$ \delta_{1} $充分小,对于$ 0\leq\varepsilon, s\leq\delta_{1} $可假设$ \frac{\partial b}{\partial s}(\varepsilon, s)>0. $于是

$ b(0, \delta_{1})-\frac{\lambda^{*}_{1}}{1-k} = b(0, \delta_{1})-b(0, 0)>0. $

因为$ b(\varepsilon, s) $连续,所以存在$ \delta\in(0, \delta_{1}] $使得

$ \varepsilon_{0} = \min\limits_{0\leq\varepsilon\leq\delta}(b(\varepsilon, \delta_{1})-\frac{\lambda^{*}_{1}}{1-k})>0. $

故若$ b\leq\frac{\lambda^{*}_{1}}{1-k}+\varepsilon_{0} $,则对任意$ \varepsilon\in[0, \delta] $都有$ b(\varepsilon, \delta_{1})\geq b. $因此对固定的$ \varepsilon, \Gamma_{\varepsilon} $一致覆盖$ b $ -区间:$ b\in(\frac{\lambda^{*}_{1}}{1-k}, \frac{\lambda^{*}_{1}}{1-k}+\varepsilon_{0}] $.又因为对于$ 0\leq\varepsilon, s\leq\delta_{1}, \frac{\partial b}{\partial s}(\varepsilon, s)>0, $所以每条曲线覆盖$ b $ -区间且仅一次.取$ \gamma_{0} = \frac{1}{\delta} $,可知若$ \gamma>\gamma_{0} $且$ b\in(\frac{\lambda^{*}_{1}}{1-k}, \frac{\lambda^{*}_{1}}{1-k}+\varepsilon_{0}], $则系统$ (1.3) $存在唯一正解.

最后证明正解的稳定性.令$ L_{1} $是系统$ (1.3) $在$ (S, u, v) $处带有特征值$ \mu_{1} $和特征函数$ (\chi_{1}, \xi_{1}, \varsigma_{1}) $的线性化算子,即

$ L_{1}(\chi_{1}, \xi_{1}, \varsigma_{1}) = \mu_{1}(\chi_{1}, \xi_{1}, \varsigma_{1}). $

令$ L_{2} $是系统(3.6)在$ (z-S, u, \gamma v) $处带有$ \mu_{2} $和特征函数$ (\chi_{2}, \xi_{2}, \varsigma_{2}) $的线性化算子,即

$ L_{2}(\chi_{2}, \xi_{2}, \varsigma_{2}) = \mu_{2}(\chi_{2}, \xi_{2}, \varsigma_{2}). $

容易验证$ \chi_{1} = \chi_{2}, \xi_{1} = \xi_{2}, \varsigma_{1} = \gamma\varsigma_{2}, \mu_{1} = \mu_{2}. $于是只需验证系统(3.6)的线性化在局部分歧曲线$ \Gamma_{\varepsilon} $的任意点处所有特征值$ \mu $均为正即可.通过类似文献[1,引理5.3]的讨论可得存在$ \iota>0 $和$ C^{1} $函数$ \rho:(-\iota, \iota)\times(\frac{\lambda^{*}_{1}}{1-k}-\iota, \frac{\lambda^{*}_{1}}{1-k}+\iota)\rightarrow {{\Bbb R}} ^{1} $以及$ \varrho:(-\iota, \iota)\times(-\iota, \iota)\rightarrow {{\Bbb R}} ^{1} $使得$ \rho(\varepsilon, b) $是系统(3.6)在$ (b, \Theta, \Theta, 0) $处线性化的主特征值, $ \varrho(\varepsilon, s) $是系统(3.6)在$ (b(\varepsilon, s), z'(\varepsilon, s), u(\varepsilon, s), \bar{v}(\varepsilon, s)) $处线性化的主特征值$ (0<\varepsilon, s\leq\iota) $.而且

$ \rho(\varepsilon, \frac{\lambda^{*}_{1}}{1-k}) = \varrho(\varepsilon, 0) = 0. $

另外

$ \begin{eqnarray*} \label{5.9} \mathop{\lim}\limits_{s\rightarrow0}(s\frac{\partial b}{\partial s}(0, s) \frac{\partial\rho}{\partial b}(0, \frac{\lambda^{*}_{1}}{1-k}))/\varrho(0, s) = -1. \end{eqnarray*} $

通过简单计算可知$ \rho(\varepsilon, b) $是如下问题的简单特征值

$ \begin{eqnarray*} \label{6} -\Delta\chi-b(1-k)g(\Theta, 0)\chi = \rho(\varepsilon, b)\chi. \end{eqnarray*} $

故, $ \frac{\partial\rho}{\partial b}(\varepsilon, \frac{\lambda^{*}_{1}}{1-k})<0. $则$ \varrho(0, s)>0. $这说明系统$ (1.3) $的唯一正解稳定.