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数学物理学报, 2019, 39(3): 545-559 doi:

论文

一类带有交叉扩散的捕食-食饵模型的正解

袁海龙,1,2, 王玉萍1, 李艳玲3

Positive Solutions of a Predator-Prey Model with Cross Diffusion

Yuan Hailong,1,2, Wang Yuping1, Li Yanling3

通讯作者: 袁海龙, E-mail: yuanhailong@sust.edu.cn

收稿日期: 2018-04-20  

基金资助: 国家自然科学基金.  11271236
国家自然科学基金.  61672021
国家自然科学基金.  61872227
陕西科技大学博士科研启动基金.  2017BJ-44

Received: 2018-04-20  

Fund supported: the NSFC.  11271236
the NSFC.  61672021
the NSFC.  61872227
the Natural Science Foundation of Shaanxi University of Science and Technology.  2017BJ-44

摘要

该文研究了一类在齐次Dirichlet边界条件下的带有交叉扩散的捕食-食饵模型.首先,根据Leray-Schauder度理论,建立了系统的正解的存在性;其次,当参数m=β且充分大时,分别研究了正则扰动方程和奇异扰动方程的正解的存在性,和借助分歧理论说明奇异系统的正解在a*处爆破;最后,建立了系统正解的多解性.

关键词: 交叉扩散 ; 分歧 ; 正解

Abstract

A predator-prey model with cross diffusion under homogeneous Dirichlet boundary conditions is investigated. Firstly, the existence of positive solutions can be established by the Leray-Schauder degree theory. Secondly, we consider that the existence of positive solutions of the regular perturbation system and the singular perturbation system when m=β is sufficiently large, respectively, and moreover, we show that the positive solutions of the singular perturbation system will blow up along the continuum at a* by the bifurcation theory. Finally, the multiplicity results of positive solutions of system is also considered.

Keywords: Cross diffusion ; Bifurcation ; Positive solutions

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本文引用格式

袁海龙, 王玉萍, 李艳玲. 一类带有交叉扩散的捕食-食饵模型的正解. 数学物理学报[J], 2019, 39(3): 545-559 doi:

Yuan Hailong, Wang Yuping, Li Yanling. Positive Solutions of a Predator-Prey Model with Cross Diffusion. Acta Mathematica Scientia[J], 2019, 39(3): 545-559 doi:

1 引言

在本文中,我们将考虑下列带有交叉扩散的捕食-食饵模型

{Δ[(1+αv)u]=u(aucv1+mu),xΩ,Δ[(μ+11+βu)v]=v(bv+du1+mu),xΩ,u=v=0,xΩ,
(1.1)

其中Ω是在RN下带有光滑边界Ω的有界区域; u,v分别代表食饵和捕食者的密度.参数a,c,dμ是正常数;参数α,βm是非负常数;参数b可能变号.

当参数m=0时,则系统(1.1)退化为带有交叉扩散的经典的具有Lotka-Volterra的捕食-食饵模型,该模型已经被许多生物数学家所研究(见参考文献[13-16, 22]及其参考文献).特别地, Kadota和Kuto[13]利用分歧理论研究了系统正解的存在性,并证明该系统的共存区域随着交叉扩散系数β的变大而变大,随着交叉扩散系数α的变大而变小.接着, Kuto和Yamada[14]表明:该系统存在一个分歧解,在合适的条件下,分歧解曲线随着分歧参数其形状为S型或型.进一步, Kuto[15]讨论了当交叉扩散系数α=0,β时的阴影系统,从全局分歧的角度出发,他断言该系统的正解在参数a=λ1(cμθb/μ)时发生爆破.对于在齐次Neumann边界且带有交叉扩散的竞争模型的研究见文献[17-21]及其参考文献.

α=β=0时,则系统(1.1)变成了经典的捕食-食饵模型,其已经被许多生物数学家所研究.特别地, Du和Lou[4]讨论了当参数m不太小时,系统(1.1)正解的存在性、唯一性与多解性.接着,文献[5]证明:在合适的条件下,该系统存在Hopf分歧解的发生,且至少存在3个非退化的正解.对于其他的模型的研究,可参见文献[6-11, 25]及其参考文献.

最近, Wang和Li[24]以参数b为分歧参数,他们利用全局分歧理论说明了系统从半平凡解分歧出来的正解情况.同时,他们考虑了当交叉扩散系数β充分大时,极限系统的全局分歧解的结构及形状.

在本文中,我们考虑了系统(1.1)正解的一些性质.利用Leray-Schauder度理论,我们给出了系统正解存在的充分条件.特别地,我们考虑了当交叉扩散系数m=β充分大时,系统(1.1)可以被看成是一些极限系统的扰动系统.我们以参数a为分歧参数研究了奇异扰动系统的正解,从全局分歧的角度来看,全局分歧解关于(a,v)是一致有界的.然而,解将在 a = a^{*} 处爆破.最后,我们建立了系统的多解性.

我们在此介绍一些记号和基本事实.令 \lambda_{1} (p) < \lambda_{2} (p) \leq \lambda_{3} (p) \leq \cdot\cdot\cdot 是下列特征问题的特征值

- \Delta u + p (x) u = \lambda u, \; u|_{\partial \Omega} = 0,

其中, p \in C^{\sigma} (\bar{\Omega}) .我们知道 \lambda_{1} (p) 是简单的,实的,且 \lambda_{1} (p) 关于势函数 p 是严格单调递增的,即如果 p_{1} \leq \not\equiv p_{2} ,则我们有 \lambda_{1} (p_{1}) < \lambda_{1} (p_{2}) . p \equiv 0 ,我们记 \lambda_{1} (0) \lambda_{1} .进一步,我们记 \phi_{1} > 0 \Omega 是主特征值 \lambda_{1} 对应的主特征函数且满足 \| \phi_{1} \|_{\infty} = 1 .

我们知道如果 a > \mu \lambda_{1} ,则下列问题

\begin{equation} - \mu \Delta u = u (a - u), \; u|_{\partial \Omega} = 0 \end{equation}
(1.2)

存在唯一的正解,我们记为 \mu \theta_{a/ \mu } .显然, a \rightarrow \mu \theta_{a/ \mu } 关于参数 a \in (\mu \lambda_{1}, + \infty) 是连续的,且当 a_{1} < a_{2} 时,我们有 \mu \theta_{a_{1}/ \mu } < \mu \theta_{a_{2}/ \mu } .进一步,该唯一正解是非退化,线性稳定的.

全文安排如下:第二节是预备工作;第三节通过度理论研究了系统(1.1)正解的存在性;最后,第四节建立了当 m = \beta 且充分大时,模型(1.1)正解的多解性.

2 预备工作

\begin{equation} U = (1+\alpha v) u, \; V = \Big(\mu + \frac{1}{1 + \beta u}\Big)v, \end{equation}
(2.1)

则系统(1.1)可以改写为

\begin{equation} \left\{\begin{array}{ll} -\Delta U = u\Big(a-u-\frac{cv}{1+mu}\Big), & x \in \Omega, \\ -\Delta V = v\Big(b-v+\frac{du}{1+mu}\Big), \quad & x \in \Omega, \\ U = V = 0, & x\in\partial\Omega. \end{array}\right. \end{equation}
(2.2)

下面,我们通过最大值原理得到系统(2.2)非负解的先验估计,由于证明过程是标准的,我们在此省去其证明仅陈述其结论.

引理2.1  令 (U, V) 是系统(2.2)的非负解,则

0 \leq u(x) \leq U(x) \leq M (a, \alpha) = a \Big[1 + \alpha (\mu + 1)\Big(b + \frac{d}{m}\Big)\Big],

0 \leq v(x) \leq V(x) \leq (\mu + 1)\Big(b + \frac{d}{m}\Big).

引理2.2 如果 a \leq \lambda_{1} (b + \frac{d}{m}) \leq \mu \lambda_{1} ,则系统(2.2)无正解.

下面,我们固定参数 b, c, \alpha \mu ,并且定义集合 S_{0} S_{1} :

S_{0} (a, b, \alpha, \mu) = \bigg\{(a, b) \in {\Bbb R}^{2}: \lambda_{1} \Big(\frac{-a}{1 + \alpha \mu \theta_{b/\mu}}\Big) = 0, \; b \geq \mu\lambda_{1} \bigg\},
(2.3a)

S_{1} (a, b, c, \alpha, \mu) = \bigg\{(a, b) \in {\Bbb R}^{2}: \lambda_{1}\Big (\frac{c (\mu + 1)\theta_{b/(\mu + 1)} -a}{1 + \alpha (\mu + 1) \theta_{b/(\mu + 1)}}\Big) = 0, \; b \geq (\mu + 1)\lambda_{1}\bigg\}.
(2.3b)

下列引理表明函数 a^{*}(b) a^{**}(b) 的一些性质,由于证明过程是标准的,我们略去其证明仅陈述其结果.

引理2.3 假设 b > \mu \lambda_{1} ,则 S_{0} 可以改写为

\begin{eqnarray*} S_{0} = \{(a, b) \in {\Bbb R}^{2}: a = a^{*}(b) \; \mbox{当}\; b \geq \mu\lambda_{1}\}, \end{eqnarray*}

其中, a = a^{*}(b) 关于参数 b \in [\mu\lambda_{1}, +\infty) 是正的连续函数,且满足下列两条性质:

(ⅰ) a^{*}(b) 关于参数 b \in [\mu\lambda_{1}, +\infty) 是严格单调递增的;

(ⅱ) a^{*}(\mu\lambda_{1}) = \lambda_{1}; \lim\limits_{b \rightarrow +\infty} a^{*}(b) = \infty .

引理2.4 假设 b > (\mu + 1)\lambda_{1} ,则 S_{1} 可以改写为

\begin{eqnarray*} S_{1} = \{(a, b) \in {\Bbb R}^{2}: a = a^{**}(b) \; \mbox{当}\; b \geq (\mu+1) \lambda_{1}\}, \end{eqnarray*}

其中, a = a^{**}(b) 关于参数 b \in [(\mu+1)\lambda_{1}, +\infty) 是正的连续函数,且满足下列两条性质:

(ⅰ) a^{**}(b) 关于参数 b \in [(\mu+1)\lambda_{1}, +\infty) 是严格单调递增的;

(ⅱ) a^{**}((\mu+1)\lambda_{1}) = \lambda_{1}; \lim\limits_{b \rightarrow +\infty} a^{**}(b) = \infty .

为方便起见,我们在此定义两个正的函数 \phi^{*} \phi^{**} 分别满足下列方程

- \Delta \phi^{*} - \frac{ a^{*}}{1 + \alpha \mu \theta_{b/ \mu }} \phi^{*} = 0, \; \; \|\phi^{*}\|_{\infty} = 1, \; \phi^{*}|_{\partial \Omega} = 0,
(2.4a)

- \Delta \phi^{**} + \frac{c (\mu + 1)\theta_{b/(\mu + 1)} - a^{**}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \phi^{**} = 0, \; \; \|\phi^{**}\|_{\infty} = 1, \; \phi^{**}|_{\partial \Omega} = 0.
(2.4b)

p > N ,我们定义Banach空间 X Y

\begin{eqnarray*} \left\{\begin{array}{ll} X = [W^{2, p}(\Omega) \cap W^{1, p}_{0}(\Omega)] \times [W^{2, p}(\Omega) \cap W^{1, p}_{0}(\Omega)], \\ Y = L^{p}(\Omega) \times L^{p}(\Omega). \end{array}\right. \end{eqnarray*}

根据嵌入定理,我们知道 X \subseteq C^{1}(\bar{\Omega}) \times C^{1}(\bar{\Omega}) .

在此陈述一些关于不动点指标计算的理论.设 X 是实的Banach空间, W \subset X 为闭凸子集.若对任意的 \alpha \geq 0 均有 \alpha W \subset W ,则称 W 为楔.如果 W \cap \{-W\} = 0 ,那么楔 W 称之为锥.设 y \in W ,定义楔

W_{y}: = {\rm cl}\{x \in X:\ \mbox{存在}\; \nu > 0, \ \mbox{使得}\; y + \nu x \in W\},

其中, "cl"表示集合的闭包.

引理2.5[2, 9-10]  设 W E 中的一个楔, A:W \rightarrow W 是紧映射,且存在不动点 y_{0} \in W ,使得 Ay_{0} = y_{0} . L = A'(y_{0}) A y_{0} 处的Fréchet导数,则 L: W_{y_{0}} \rightarrow W_{y_{0}} . I - L E 上可逆,并且

(ⅰ) L W_{y_{0}} 上具有 \alpha 性质,则 {\rm index}_{W} (A, y_{0}) = 0 ;

(ⅱ) L W_{y_{0}} 不上具有 \alpha 性质,则 {\rm index}_{W} (A, y_{0}) = {\rm index}_{E}(L, 0) = (-1)^{\sigma} ,其中, \sigma L 大于1的特征值的代数重数之和.

3 正解的存在性

在本节中,我们将通过Leray-Schauder度理论来建立系统(2.2)的正解的存在性.首先,我们给出下列记号:

E = C_{0} (\bar{\Omega}) \times C_{0} (\bar{\Omega}) .

P = K \times K, \; \mbox{其中}, \; K = \{U \in C_{0} (\bar{\Omega}): U (x) \geq 0\; \mbox{当}\; x \in \bar{\Omega} \} .

D = \{(U, V) \in P : U < M (a, \alpha) + 1, \; V < (\mu + 1)(b + \frac{d}{m}) + 1\} .

我们定义正的紧算子 A_{t} : \bar{D} \rightarrow E

\begin{eqnarray*} A_{t} { U \choose V }: & = & (-\Delta + p I)^{-1}\bigg(U\bigg \{p + t \frac{1}{1 + \alpha v} \Big (a - u - \frac{c v}{1 + m u}\Big) \bigg\}, \\ &&V\bigg\{p + t \frac{1}{\mu + \frac{1}{1 + \beta u}} \Big(b - v + \frac{d u}{1 + m u}\Big)\bigg\}\bigg) \\ & = &(-\Delta + p I)^{-1}(p U + t \tilde{f} (u, v), p V + t \tilde{g} (u, v)), \end{eqnarray*}

其中, p 是充分大的数满足 A_{t} 是正的紧算子

\tilde{f} (u, v) = u \Big(a - u - \frac{c v}{1 + m u}\Big), \quad \tilde{g} (u, v) = v \Big(b - v + \frac{d u}{1 + m u}\Big).

为了计算算子 A_{t} 的不动点指标,我们在此给出算子 A_{t} 的Fréchet导数.令 Q = (1+\alpha v)(\mu + \frac{1}{1+\beta u}) + \frac{\alpha \beta u v}{(1 + \beta u)^{2}} ,则

A'_{t}{U \choose V} = (-\Delta+pI)^{-1} \bigg[ p + t {\tilde{f}_{u} \quad \tilde{f}_{v} \choose \tilde{g}_{u} \quad \tilde{g}_{v}}{u_{U} \quad u_{V} \choose v_{U} \quad v_{V}}\bigg ] {U \choose V}.

根据(2.1)式,我们有

{1 \quad 0 \choose 0 \quad 1} = \left(\begin{array}{cc} 1+\alpha v \quad & \alpha u \\ - \frac{\beta v}{(1 + \beta u)^{2}} \quad & \mu + \frac{1}{1+\beta u} \end{array}\right) {u_{U} \quad u_{V} \choose v_{U} \quad v_{V}}.

进一步,我们有

{u_{U} \quad u_{V} \choose v_{U} \quad v_{V}} = \frac{1}{Q} \left(\begin{array}{cc} \mu + \frac{1}{1+\beta u} \quad & - \alpha u \\ \frac{\beta v}{(1 + \beta u)^{2}} \quad & 1+\alpha v\end{array}\right).

因此

\begin{eqnarray*} A'_{t}{U \choose V} & = & (-\Delta + p I)^{-1}\left[p + t \frac{1}{Q} \left(\begin{array}{cc} a - 2 u - \frac{c v}{(1 + m u)^{2}} \quad & - \frac{c u}{1 + m u} \\ \frac{d v}{(1 + m u)^{2}} \quad & b - 2 v + \frac{d u}{1 + m u}\end{array} \right)\right.\\ &&\times \left. \left(\begin{array}{cc} \mu + \frac{1}{1+\beta u} \quad& - \alpha u \\ \frac{\beta v}{(1 + \beta u)^{2}} \quad & 1 + \alpha v\end{array} \right)\right ]{U \choose V}. \end{eqnarray*}

现在,我们建立算子 A 的不动点指标.

引理3.1 假设 a > \lambda_{1} ,则

(ⅰ) deg_{P} (I - A, D) = 1 ;

(ⅱ) ind_{P} (A, (0, 0)) = 0 如果 b \not = (\mu+1) \lambda_{1} ;

(ⅲ) ind_{P} (A, (\theta_{a}, 0)) = 0 如果 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) < 0 ;

(ⅳ) ind_{P} (A, (\theta_{a}, 0)) = 1 如果 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) > 0 .进一步,如果 b > (\mu + 1) \lambda_{1} ,则

(ⅴ) ind_{P} (A, (0, (\mu + 1)^{2} \theta_{b/(\mu + 1)})) = 0 如果 \lambda_{1} (\frac{c (\mu + 1) \theta_{b/(\mu + 1)} -a}{1 + \alpha (\mu + 1) \theta_{b/(\mu + 1)}}) < 0 ;

(ⅵ) ind_{P} (A, (0, (\mu + 1)^{2}\theta_{b/(\mu + 1)})) = 1 如果 \lambda_{1} (\frac{c (\mu + 1)\theta_{b/(\mu + 1)} -a}{1 + \alpha (\mu + 1) \theta_{b/(\mu + 1)}}) > 0 .

  (ⅰ)显然,通过引理2.1我们知道算子 A_{t} 在边界 \partial D 无不动点.因此,我们知道Leray-Schauder度 deg_{P}(I - A_{t}, D) 有定义,且我们有

deg_{P}(I - A, D) = deg_{P}(I - A_{1}, D) = deg_{P}(I - A_{0}, D).

特别地

deg_{P} (I - A_{0}, D) = ind_{P} (A_{0}, (0, 0)) .

经过简单计算我们有

A_{0}'(0, 0) = (-\Delta+pI)^{-1}{p \quad 0 \choose 0 \quad p},

r(A_{0}'(0, 0)) < 1 .因此,算子 I - A_{0}'(0, 0) \bar{P}_{(0, 0)} 是可逆的且算子 A_{0}'(0, 0) \bar{P}_{(0, 0)} \alpha 性质.因此, ind_{P} (A_{0}, (0, 0)) = 1 .也就是说, deg_{P} (I - A, D) = 1 .

(ⅱ)显然,我们有 A'(0, 0) (U, V) = (-\Delta + p I)^{-1} ((p + a) U, (p + \frac{b}{\mu + 1}) V) ,假设存在一些 (U, V) \in \bar{P}_{(0, 0)} 使得 A'(0, 0) (U, V) = (U, V) 成立,则 - \Delta U = a U, \; U|_{\partial \Omega} = 0 .如果 U > 0 ,则 a = \lambda_{1} ,矛盾.因此, U \equiv 0 .类似地, V \equiv 0 .从而, I - A'(0, 0) \bar{P}_{(0, 0)} 是可逆的.

由于 a > \lambda_{1} ,从而 r_{a} = r [(- \Delta + p)^{-1}(a + p)] > 1 ,且 r_{a} 是算子 (-\Delta + p)^{-1}(a + p) 的主特征值,其对应的主特征函数 U > 0 . t_{0} = r_{a}^{-1} ,则 0 < t_{0} < 1 (I - t_{0} A'(0, 0)) (U, 0) = (0, 0) \in S_{(0, 0)} .从而, A'(0, 0) \alpha 性质.因此, ind_{P} (A, (0, 0)) = 0 .

(ⅲ)经过简单的计算我们有

\begin{eqnarray*} A'(\theta_{a}, 0) (U, V) & = &(-\Delta + pI)^{-1}\bigg(\{p+(a-2\theta_{a})\} U \\ &&- \frac{\alpha\theta_{a}(1 + \beta \theta_{a})(a - 2 \theta_{a})(1 + m \theta_{a}) + c \theta_{a} (1 + \beta \theta_{a})}{(\mu (1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V, \\ &&\bigg\{p + \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu (1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\bigg\} V\bigg). \end{eqnarray*}

假设存在一些 (U, V) \in \bar{P}_{(\theta_{a}, 0)} 使得 A'(\theta_{a}, 0) (U, V) = (U, V) 成立,则

\begin{eqnarray*} \left\{\begin{array}{ll} - \Delta U + (2\theta_{a} - a) U = - \frac{\alpha \theta_{a}(1 + \beta \theta_{a})(a - 2 \theta_{a})(1 + m \theta_{a}) + c \theta_{a} (1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V , & x \in \Omega, \\ -\Delta V = \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu (1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V, & x \in \Omega, \\ U = V = 0, & x\in\partial \Omega. \end{array}\right. \end{eqnarray*}

如果 V \geq\not\equiv 0 ,则 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) = 0 .由于 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) < 0 ,从而 V \equiv 0 ,进而 U \equiv 0 .因此, (U, V) = (0, 0) .也就是说, I - A'(\theta_{a}, 0) \bar{P}_{(\theta_{a}, 0)} 是可逆的.

下面我们断言算子 A'(\theta_{a}, 0) \bar{P}_{(\theta_{a}, 0)} \alpha 性质.令

L = (- \Delta + p)^{-1} \Big(p + \frac{(b (1 + m\theta_{a}) + d\theta_{a})(1 + \beta\theta_{a})}{(\mu (1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\Big).

r_{b} = r(L) > 1 是算子 L 的特征值,且其对应的特征函数 V > 0 . t_{0} = r_{b}^{-1} ,则 t_{0} \in (0, 1) .由于 (0, V) \in \bar{P}_{(\theta_{a}, 0)} \setminus S_{(\theta_{a}, 0)} ,我们有

\begin{eqnarray*} (I - t_{0}L){0 \choose V} & = & \left(\begin{array}{cc} - (- \Delta + p)^{-1}\frac{\alpha \theta_{a}(1+\beta\theta_{a})(a - 2 \theta_{a})(1 + m \theta_{a}) + c \theta_{a} (1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V \\ V - t_{0} (- \Delta + p)^{-1} \Big(\frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} + p\Big) V \end{array} \right) \\ & = & \left(\begin{array}{cc} - (- \Delta + p)^{-1}\frac{\alpha \theta_{a}(1 + \beta \theta_{a})(a - 2 \theta_{a})(1 + m \theta_{a}) + c \theta_{a} (1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V \\ 0 \end{array} \right)\\ & \in& S_{(\theta_{a}, 0)}. \end{eqnarray*}

进而,算子 A'(\theta_{a}, 0) \alpha 性质.因此, ind_{P} (A, (\theta_{a}, 0)) = 0 .

(ⅳ)类似地,我们可以证明算子 I - A'(\theta_{a}, 0) \bar{P}_{(\theta_{a}, 0)} 可逆.下面我们仅说明算子 A'(\theta_{a}, 0) \bar{P}_{(\theta_{a}, 0)} \alpha 性质.我们采用反证法.假设算子 L \bar{P}_{(\theta_{a}, 0)} \alpha 性质,从而存在某个 0 < t < 1 (U, V) \in \bar{P}_{(\theta_{a}, 0)}\setminus S_{(\theta_{a}, 0)} 使得 (I - t L) (U, V) \in S_{(\theta_{a}, 0)} .因此

V - t (- \Delta + p)^{-1} \Big(p + \frac{(b(1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m\theta_{a})}\Big) V = 0.

由于 V \in K \setminus \{0\} ,我们知道 \frac{1}{t} > 1 是算子 L 的特征值.由于 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) > 0 ,我们有 r(L) < 1 ,矛盾.因此, L \bar{P}_{(\theta_{a}, 0)} \alpha 性质.根据引理, ind_{P} (A, (\theta_{a}, 0)) = (-1)^{\sigma} ,其中, \sigma 是算子 L 所有大于1的特征值的代数重数之和.

假设 \frac{1}{\mu} > 1 是算子 L 的特征值且其对应的特征函数为 (U, V) ,则

\begin{eqnarray*} &&(-\Delta + p I)^{-1} \left(\begin{array}{cc} (p + a - 2 \theta_{a}) U - \frac{\alpha \theta_{a}(1 + \beta \theta_{a})(a - 2 \theta_{a})(1 + m \theta_{a}) + c \theta_{a} (1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V\\ \Big(p + \frac{(b(1 + m \theta_{a})+d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\Big) V \end{array}\right)\\ && = \frac{1}{\mu} {U \choose V}. \end{eqnarray*}

也就是说

\begin{eqnarray*} \left\{\begin{array}{ll} - \Delta U + p U = \mu \Big((p + a - 2 \theta_{a}) U - \frac{\alpha \theta_{a}(1 + \beta \theta_{a})(a - 2 \theta_{a})(1 + m \theta_{a}) + c \theta_{a} (1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})} V\Big), \\ \hskip 8.8cm x \in \Omega, \\ -\Delta V + p V = \mu \Big(p + \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\Big) V, \qquad x \in \Omega, \\ U = V = 0, \hskip 7cm x\in\partial\Omega. \end{array}\right. \end{eqnarray*}

如果 V \not\equiv 0 ,则

\begin{eqnarray*} 0 & = & \lambda_{1} \Big(p (1 - \mu) - \mu \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\Big)\\ & >& \lambda_{1}\Big (- \mu \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\Big) \\ & >& \lambda_{1} \Big(- \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}\Big). \end{eqnarray*}

因此, \lambda_{1} (- \frac{(b (1 + m \theta_{a}) + d \theta_{a})(1 + \beta \theta_{a})}{(\mu(1 + \beta \theta_{a}) + 1)(1 + m \theta_{a})}) < 0 ,矛盾.从而, V \equiv 0, U \equiv 0 .也就是说,算子 L 没有大于1的特征值.进而, ind_{P} (A, (\theta_{a}, 0)) = 1 .

由于情形(ⅴ)和(ⅳ)证明方法是类似地,我们在此省略其证明.

根据引理3.1我们可以建立系统(2.2)正解的存在性,由于证明过程是标准的,我们在此省略其证明仅陈述其结果.

定理3.1 假设 a > \lambda_{1} ,则系统(2.2)有正解的充分条件为

\begin{equation} \lambda_{1} \Big(- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}\Big) < 0 \; \mbox{且} \; \lambda_{1} \Big(\frac{c (\mu + 1)\theta_{b/(\mu + 1)} -a}{1 + \alpha (\mu + 1) \theta_{b/(\mu + 1)}}\Big) < 0 \end{equation}
(3.1)

成立.其中, \theta_{b/(\mu + 1)} \equiv 0 如果 b \leq (\mu + 1) \lambda_{1} .

4 正解的多解性

在本节中,我们考虑参数 \beta = m 且充分大时的情形,则当参数 \beta = m 且充分大时,系统(1.1)可以被看成是下列的正则扰动问题

\begin{equation} \left\{\begin{array}{ll} -\Delta [(1 + \alpha v) u] = u (a-u), \quad & x \in \Omega, \\ -\mu \Delta v = v (b - v), \quad & x \in \Omega, \\ u = v = 0, & x\in\partial\Omega, \end{array}\right. \end{equation}
(4.1)

和奇异扰动问题

\begin{equation} \left\{\begin{array}{ll} -\Delta [(1 + \alpha v) w] = w \Big(a - \frac{cv}{1 + w}\Big), & x \in \Omega, \\ -\Delta\Big [\Big(\mu + \frac{1}{1 + w}\Big)v\Big ] = v (b - v ), \quad & x \in \Omega, \\ w = v = 0, & x\in\partial\Omega. \end{array}\right. \end{equation}
(4.2)

引理4.1 假设 b > \mu \lambda_{1} a > a^{*} ,则系统(4.1)存在正解,记为 (\tilde{\theta}_{a, \alpha}, \mu \theta_{b/\mu}) .

引理4.2[15]  令 (w, v) 是系统(4.2)的任意正解,则我们有

\frac{\mu^{2}}{\mu + 1} \theta_{b/(\mu + 1)} < v < \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu}, \quad \mu^{2} \theta_{b/(\mu + 1)} < V < (\mu + 1)^{2} \theta_{b/\mu},

其中, V = (\mu + \frac{1}{1 + w}) v .

引理4.3 如果 a > \lambda_{1} (c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu}) ,则系统(4.2)无正解.

 我们采用反证法.假设当 a > \lambda_{1} (c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu}) 时,系统(4.2)有正解 (w, v) ,则

-\Delta [(1 + \alpha v) w] = w \Big(a - \frac{cv}{1 + w}\Big), \quad w|_{\partial \Omega} = 0.

\phi = (1 + \alpha v) w ,则 w = \frac{\phi}{1 + \alpha v} .因此

- \Delta \phi = \phi \Big(\frac{a}{1 + \alpha v} - \frac{c v}{1 + \alpha v + \phi}\Big), \quad \phi|_{\partial \Omega} = 0.

从而,我们有

0 = \lambda_{1}\Big (- \frac{a}{1 + \alpha v} + \frac{c v}{1 + \alpha v + \phi}\Big) < \lambda_{1} \Big(\frac{c v - a}{1 + \alpha v}\Big) < \lambda_{1} \Big(c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu} - a\Big).

也就是说, a < \lambda_{1} (c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu}) ,矛盾.证毕.

引理4.4  令 (w, v) 是系统(4.2)的任意正解,则对任意充分小的 \epsilon > 0 b > (\mu + 1) \lambda_{1} , a^{*} + \epsilon \leq a < \lambda_{1} (c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu}) ,存在常数 C = C(\epsilon) > 0 满足 \| w \|_{C^{1}} \leq C .

 我们采用反证法.假设存在 \epsilon_{0} > 0, a_{i} \rightarrow a \in [a^{*} + \epsilon, \lambda_{1} (c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu})] 且当 a = a_{i} 时,设 w_{i} 是系统(4.2)的解,满足 \| w_{i} \|_{\infty} \rightarrow \infty ,且 v_{i} \rightharpoonup v L^{2} 空间弱收敛成立.令 \tilde{w}_{i} = \frac{w_{i}}{\|w_{i}\|_{\infty}} ,则

\begin{equation} -\Delta [(1 + \alpha v_{i}) \tilde{w}_{i}] = \tilde{w}_{i}\Big (a_{i} - \frac{cv_{i}}{1 + w_{i}}\Big), \; \tilde{w}_{i}|_{\partial \Omega} = 0. \end{equation}
(4.3)

\phi_{i} = (1 + \alpha v_{i}) \tilde{w}_{i} ,则 \tilde{w}_{i} = \frac{\phi_{i}}{1 + \alpha v_{i}} .因此,系统(4.3)可以改写为

- \Delta \phi_{i} = \phi_{i} \Big(\frac{a_{i}}{1 + \alpha v_{i}} - \frac{c v_{i}}{1 + \alpha v_{i} + \phi_{i} \| w_{i} \|_{\infty} }\Big), \; \phi_{i}|_{\partial \Omega} = 0.

根据二阶椭圆型正则化理论,我们可以假设 \phi_{i} \rightarrow \phi C^{1} 空间上成立且 \frac{1}{1 + \alpha v_{i} + \phi_{i} \| w_{i} \|_{\infty} } \rightharpoonup h , v_{i} \rightharpoonup v L^{2} 空间上弱收敛成立.在上述方程两边同时取极限,我们有

- \Delta \phi = \phi\Big (\frac{a}{1 + \alpha v} - c v h\Big), \; \phi|_{\partial \Omega} = 0.

根据最大值原理,我们知道 \phi > 0 \Omega 上成立.因此, h = 0 , - \Delta \phi = \frac{a}{1 + \alpha v} \phi ,且 \lambda_{1} (- \frac{a}{1 + \alpha v}) = 0 .经过简单计算我们知道, v = \mu \theta_{b/\mu} .因此, \lambda_{1} (- \frac{a}{1 + \alpha \mu \theta_{b/\mu}}) = 0 .也就是说, a = a^{*} ,矛盾.证毕.

W = (1 + \alpha v) w, V = (\mu + \frac{1}{1 + w})v ,则系统(4.2)可以改写为

\begin{equation} \left\{\begin{array}{ll} -\Delta W = w\Big (a - \frac{cv}{1 + w}\Big), \quad & x \in \Omega, \\ -\Delta V = v (b - v ), & x \in \Omega, \\ W = V = 0, & x \in \partial \Omega. \end{array}\right. \end{equation}
(4.4)

显然,当 b > (\mu + 1)\lambda_{1} 时,系统(4.4)存在半平凡解 (W, V) = (0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) .特别地,我们令参数 a 为分歧参数且固定其他参数.

f (w, v) = w (a - \frac{cv}{1 + w}), g(u, v) = v (b - v) ,其中, w, v W, V 的函数.根据在 (W, V) (W^{*}, V^{*}) 的Taylor展式,我们可以将系统(4.4)改写为

\begin{eqnarray} { \Delta W \choose \Delta V } & +& { f(w(W^{*}, V^{*}), v(W^{*}, V^{*}) \choose g(w(W^{*}, V^{*}), v(W^{*}, V^{*}) }+ { f^{*}_{w} \quad f^{*}_{v} \choose g^{*}_{w} \quad g^{*}_{v} } { w^{*}_{W} \quad w^{*}_{V} \choose v^{*}_{W} \quad v^{*}_{V} } { W-W^{*} \choose V-V^{*} } \\ & + &{ \rho^{1}(W-W^{*}, V-V^{*}) \choose \rho^{2}(W-W^{*}, V-V^{*}) } = { 0 \choose 0 }, \end{eqnarray}
(4.5)

其中, f^{*}_{w} = f_{w}(w(W^{*}, V^{*}), v(W^{*}, V^{*})), w^{*}_{W} = w_{W}(W^{*}, V^{*}) ,其他记号定义类似.因此, \rho^{i}(W-W^{*}, V-V^{*})(i = 1, 2) 是光滑函数且满足 \rho^{i}(0, 0) = \rho^{i}_{(W, V)}(0, 0) = 0\; (i = 1, 2) ,且

\begin{eqnarray*} { 1 \quad 0 \choose 0 \quad 1 } = \left(\begin{array}{cc} 1+ \alpha v \quad & \alpha w \\ - \frac{v}{(1 + w)^{2}} \quad& \mu + \frac{1}{1 + w} \end{array}\right) { w_{W} \quad w_{V} \choose v_{W} \quad v_{V} }. \end{eqnarray*}

如果 (w, v) = (0, (\mu + 1) \theta_{b/(\mu + 1)}) ,则

\begin{eqnarray*} { w_{W} \quad w_{V} \choose v_{W} \quad v_{V} } = \left(\begin{array}{cc} \frac{1}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)} } \quad& 0 \\ \frac{ \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \quad & \frac{1}{\mu + 1} \end{array}\right). \end{eqnarray*}

显然, f(0, (\mu + 1)\theta_{b/(\mu + 1)}) = 0, \; g(0, (\mu + 1)\theta_{b/(\mu + 1)}) = - (\mu + 1)^{2} \Delta \theta_{b/(\mu + 1)} . (W^{*}, V^{*}) = (0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}), \overline{V} = V - (\mu + 1)^{2} \theta_{b/(\mu + 1)} ,则

\begin{eqnarray*} { \Delta W \choose \Delta \overline{V} }& +& \left(\begin{array}{cc} \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \quad &0 \\ \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \quad & \frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)} \end{array}\right) { W \choose \overline{V} } \\ &+& { \rho^{1}(a;W, \overline{V}) \choose \rho^{2}(a;W, \overline{V}) } = { 0 \choose 0 }, \end{eqnarray*}

其中, \rho^{i}(a; W, \overline{V})(i = 1, 2) 是光滑函数且满足 \rho^{1}_{(W, \overline{V})}(a; 0, 0) = \rho^{2}_{(W, \overline{V})}(a; 0, 0) = 0 .

令算子 K 是算子 - \Delta 在齐次Dirichlet边界条件下的逆算子,则

W = K \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W + K \rho^{1}(a; W, \overline{V}),

\overline{V} = K\Big (\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)}\Big) \overline{V} + K \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W + K \rho^{2}(a; W, \overline{V}).

我们定义算子 T : {\Bbb R} \times X \rightarrow X

\begin{eqnarray*} T(a; W, \overline{V}) & = & \bigg( K \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W + K \rho^{1}(a; W, \overline{V}), K \Big(\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)}\Big ) \overline{V}\\ && + K \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W + K \rho^{2}(a; W, \overline{V}) \bigg). \end{eqnarray*}

则算子 T(a; W, \overline{V}) X 空间的紧算子.令 G (a; W, \overline{V}) = (W, \overline{V}) - T (a; W, \overline{V}) ,显然,算子 G C^{1} 函数且满足 G (a; 0, 0) = 0 并将其Fréchet导数记为 D G_{(W, \overline{V})}(a; 0, 0) .

根据局部分歧理论表明系统(4.4)在 a = a^{**} 附近存在局部分歧解.

定理4.1  系统(4.4)从半平凡解 (0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) 产生分歧的充分必要条件是 a = a^{**} .进一步,系统(4.4)在 (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \in {\Bbb R} \times X 附近的正解可以写为

\begin{eqnarray*} \Gamma : = \{ (a;W, V):( a(s); s(\phi^{**} + \phi(s)), (\mu + 1)^{2} \theta_{b/(\mu + 1)} + s(\psi^{**} + \psi(s)) ): 0 < s \leq \delta \}, \end{eqnarray*}

其中, \delta > 0 \phi^{**}, \psi^{**} \in X . (a(s); \phi(s), \psi(s)) 关于参数 s 是光滑函数且满足 a(0) = a^{**}, \phi(0) = \psi(0) = 0 .

 令 L(a; 0, 0) = D G_{(W, \overline{V})}(a; 0, 0) ,则

\begin{eqnarray*} L(a;0, 0) (W, \overline{V}) & = &\bigg (W - K \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W, \\ & & \overline{V} - K\Big (\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)} \Big) \overline{V} - K \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W\bigg). \end{eqnarray*}

L(a; 0, 0)(W, \overline{V}) = 0 ,则

\begin{eqnarray*} \left\{\begin{array}{ll} \Delta W + \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W = 0, & x \in \Omega, \\ \Delta \overline{V} +\Big (\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)}\Big ) \overline{V} + \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} W = 0, \ & x \in \Omega, \\ W = \overline{V} = 0, & x\in\partial\Omega. \end{array}\right. \end{eqnarray*}

如果 W = 0 ,则 \overline{V} = 0 ,矛盾.因此 a = a^{**} .进而, {\rm Ker}\; L(a^{**}; 0, 0) = span \{\phi^{**}, \psi^{**}\} ,其中, \psi^{**} = (-\Delta - \frac{b}{\mu + 1} + 2 \theta_{b/(\mu + 1)})^{-1}\frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \phi^{**} .

记算子 L (a^{**}; 0, 0) 的伴随算子为 L^{*}(a^{**}; 0, 0) ,则

\begin{eqnarray*} L^{*}(a^{**};0, 0) (W, \overline{V}) & = &\bigg ( W - \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} K W - \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} K \overline{V}, \\ && \overline{V} - \Big (\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)} \Big) K \overline{V}\bigg). \end{eqnarray*}

L^{*}(a^{**}; 0, 0)(W, \overline{V}) = 0 ,则 {\rm Ker}\; L^{*}(a^{**}; 0, 0) = span \{ - \Delta \phi^{**}, 0 \} .根据Fredholm选择公理,我们知道算子 L (a^{**}; 0, 0) 的值域为 R (L (a^{**}; 0, 0)) = \{(h, k) \in X : \int_{\Omega} h \Delta \phi^{**} = 0 \} .因此,算子 L (a^{**}; 0, 0) 的值域 R (L (a^{**}; 0, 0)) 的余维数为1.令 L_{1} (a^{**}, 0, 0) (\phi^{**}, \psi^{**}) = D^{2}_{a, (W, \overline{V})} G(a^{**}; 0, 0) (\phi^{**}, \psi^{**}) = (- K \frac{1}{1 + \alpha (\mu + 1)\theta_{b /(\mu + 1)}} \phi^{**}, 0) .由于 L_{1} (a^{**}; 0, 0) (\phi^{**}, \psi^{**}) \not\in R (L (a^{**}; 0, 0)) .根据局部分歧理论[1]我们知道,结论成立.证毕.

下面,我们根据全局分歧理论将上述的局部分歧解延拓为全局分歧解.特别地,我们断言:该全局分歧解将在 a = a^{*} 处爆破.

\begin{eqnarray*} T'(a)(\xi, \eta)& = & D T_{(W, \overline{V})}(a;0, 0) \\ & = & \bigg(K \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \xi, K\Big (\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)}\Big) \eta \\ &&+ K \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \xi\bigg). \end{eqnarray*}

假设 \sigma \geq 1 是算子 T'(a) 的特征值且其对应的特征函数为 (\xi, \eta) ,则

\begin{eqnarray*} \left\{\begin{array}{ll} - \sigma \Delta \xi = \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \xi, & x\in\Omega, \\ - \sigma \Delta \eta = \Big (\frac{b}{\mu + 1} - 2 \theta_{b/(\mu + 1)} \Big) \eta + \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \xi, \ & x\in\Omega, \\ \xi = \eta = 0, & x\in\partial\Omega. \end{array}\right. \end{eqnarray*}

如果 \xi = 0 ,则 \eta = 0 ,矛盾.也就是说, a = a_{i}(\sigma) (i = 1, 2, \cdot \cdot \cdot) ,其中, a_{i}(\sigma) 关于参数 \sigma \geq 1 是单调递增的且可排成 0 < a_{1}(\sigma) < a_{2}(\sigma) \leq \cdot \cdot \cdot, \; a_{1}(1.1) = a^{**} .进一步,如果 \sigma \geq 1 ,我们知道算子 - \sigma \Delta - \frac{b}{\mu + 1} + 2 \theta_{b/(\mu + 1)} 的所有特征值是正的.因此,我们有 \eta = (- \sigma \Delta - \frac{b}{\mu + 1} + 2 \theta_{b/(\mu + 1)})^{-1} \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \xi .从而,存在一些 i = 1, 2, \cdot \cdot \cdot 使得 \sigma \geq 1 是算子 T'(a) 特征值的充要条件是 a = a_{i}(\sigma) .

假设 a < a^{**} ,则对任意的 \sigma \geq 1 , i \geq 1 我们有 a < a_{1}(1.1) \leq a_{i}(\sigma) .因此, T'(a) 没有大于1的特征值,从而, i (T(a, \cdot), 0) = 1 .

假设 a^{**} < a < a_{2}(1.1) ,则对任意的 \sigma \geq 1, i \geq 2 我们有 a < a_{i}(\sigma) .由于 \lim\limits_{\sigma \rightarrow \infty} a_{1} (\sigma) = + \infty ,存在唯一的 \sigma_{1} 使得 a = a_{1}(\sigma_{1}) .因此, {\rm Ker} (\sigma_{1} I - T'(a)) = {\rm span} \{(\bar{\xi}, \bar{\eta}) \}, {\rm dim}\; {\rm Ker} (\sigma_{1} I - T'(a)) = 1 ,其中, \bar{\xi} > 0 是下列问题的主特征函数

- \sigma_{1} \Delta \bar{\xi} = \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \bar{\xi}, \; \bar{\xi}|_{\partial \Omega} = 0,

\bar{\eta} = (- \sigma_{1} \Delta - \frac{b}{\mu + 1} + 2 \theta_{b/(\mu + 1)})^{-1} \frac{(b - 2(\mu + 1)\theta_{b/(\mu + 1)}) \theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \bar{\xi} .

下面,我们断言: R (\sigma_{1} I - T'(a)) \cap {\rm Ker} (\sigma_{1} I - T'(a)) = 0 .我们采用反证法.假设 (\bar{\xi}, \bar{\eta}) \in R (\sigma_{1} I - T'(a)) ,存在 (\xi, \eta) \in X 使得 (\sigma_{1} I - T'(a))(\xi, \eta) = (\bar{\xi}, \bar{\eta}) .也就是说

\begin{equation} \sigma_{1} \Delta \xi + \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \xi = \Delta \bar{\xi}, \; \xi|_{\partial \Omega} = 0. \end{equation}
(4.6)

在(4.6)式两边同时乘以 \bar{\xi} 且分部积分,有 \int_{\Omega} \bar{\xi} \Delta \bar{\xi} = 0 .也就是说, \int_{\Omega} \frac{a - c (\mu + 1)\theta_{b/(\mu + 1)}}{1 + \alpha (\mu + 1)\theta_{b/(\mu + 1)}} \bar{\xi}^{2} = 0 ,矛盾.因此,当 a^{**} < a < a_{2}(1.1) 时,我们有 i (T(a, \cdot), 0) = - 1 .

根据全局分歧理论[23],在 {\Bbb R}^{+} \times X 中,存在从半平凡解 (a^{**}; 0, 0) 出发的连通分支 C_{0} 满足 G(a; \xi, \eta) = 0 ,且在 (a^{**}; 0, 0) 的附近, G(a; \xi, \eta) 的所有零点都在定理4.1中得到的那条分歧曲线 \{(a(s); s(\phi^{**}+ \phi(s)), s(\psi^{**}+ \psi(s))): |s| < \delta\} ,其中, a(0) = a^{**}, \phi(0) = \psi(0) = 0 . C_{1} = C_{0} - \{(a(s); s(\phi^{**}+ \phi(s)), s(\psi^{**}+ \psi(s))) : - \delta < s < 0 \} , C = \{(a; W, V) : W = s(\phi^{**} + \phi(s)), V = (\mu + 1)^{2} \theta_{b/(\mu + 1)} + s(\psi^{**} + \psi(s)), (a; \xi, \eta) \in C_{1} \} ,则 C 是系统(4.4)从 \{(a^{**}; 0, (\mu + 1)^{2} \theta_{b/ (\mu + 1)}) \} 分歧出来的解曲线,且在 \{(a^{**}; 0, (\mu + 1)^{2} \theta_{b/ (\mu + 1)}) \} 的小邻域内,解曲线 C \subset P_{0} ,其中, P_{0} = \{W, V: W > 0, V > 0, x \in \Omega, \frac{\partial W}{\partial \upsilon} < 0, \frac{\partial V}{\partial \upsilon} < 0, x \in \partial \Omega\} .

定理4.2   C - \{(a^{**}; 0, (\mu + 1)^{2} \theta_{b/ (\mu + 1)}) \} 连接半平凡解 \{(a^{**}; 0, (\mu + 1)^{2} \theta_{b/ (\mu + 1)}) \} 在正椎 P_{0} 内延伸至 \infty .进一步,对于序列 \{(a_{i}, W_{i}, V_{i})\} \subset C 满足

\begin{eqnarray*} \lim\limits_{i \rightarrow \infty} \|W_{i}\|_{\infty} = \infty, \; \lim\limits_{i \rightarrow \infty} V_{i} = \mu^{2} \theta_{b/\mu}, \; \lim\limits_{i \rightarrow \infty} a_{i} = a^{*}. \end{eqnarray*}

 根据Rabinowitz的全局分歧理论[23]和更加一般的全局分歧理论(López-Gómez[12]和Dancer[3]),我们知道全局分歧解必居以下三条之一:

(ⅰ) C 连接另一个半平凡解 \{a; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)} \} ,其中, a \neq a^{**}, I - T'(a) 是不可逆的;

(ⅱ) C {\Bbb R} \times X 内延伸至无穷;

(ⅲ)存在 a = \hat{a}, Z \in \overline{Y} \setminus \{0\} 满足 (\hat{a}; Z) \in C ,其中, \overline{Y} \{(- \Delta \phi^{**}, 0)\} 的补空间,其中, \{(- \Delta \phi^{**}, 0)\} 由定理4.1给出.

如果 C - \{ (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \} \not\subset P_{0} ,则存在 (\tilde{a}; \tilde{W}, \tilde{V}) \in \{C - (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \} \cap \partial P_{0} 是下列序列的极限 \{ (\tilde{a}_{i}; \tilde{W}_{i}, \tilde{V}_{i}) \} \subset C \cap P_{0}, \; \tilde{W}_{i} > 0, \; \tilde{V}_{i} > 0 \Omega .由于 (\tilde{a}; \tilde{W}, \tilde{V}) \in \partial P_{0} ,我们知道存在点 x_{0} \in \Omega 使得 \tilde{W}(x_{0}) = 0 或存在点 x_{0} \in \partial \Omega 使得 \frac{\partial \tilde{W}}{\partial \upsilon} (x_{0}) = 0 .根据强最大值原理,我们知道 \tilde{W} \equiv 0 \Omega 上成立.类似地,对于其他情形我们有 \tilde{V} \equiv 0 \Omega 上成立.

假设 \tilde{W} \equiv 0, \tilde{V} = (\mu + 1)^{2} \theta_{b/(\mu + 1)} ,令 \tilde{W}_{i}^{*} = \frac{\tilde{W}_{i}}{\|\tilde{W}_{i}\|_{\infty}} ,则

\begin{eqnarray*} \left\{\begin{array}{ll} -\Delta \tilde{W}_{i}^{*} = \frac{\tilde{W}_{i}^{*}}{1 + \alpha \tilde{v}_{i}} \Big(\tilde{a}_{i} - \frac{c\tilde{v}_{i}}{1 + w_{i}}\Big), \quad & x \in \Omega, \\ -\Delta \tilde{V}_{i} = \tilde{v}_{i} (b - \tilde{v}_{i} ), & x \in \Omega, \\ \tilde{W}_{i}^{*} = \tilde{V}_{i} = 0, & x\in\partial\Omega. \end{array}\right. \end{eqnarray*}

根据二阶椭圆型正则化理论我们可以假设 \tilde{W}_{i}^{*} \rightarrow \tilde{W}^{*} , \tilde{V}_{i} \rightarrow \tilde{V} C^{1} 空间上成立且 \|\tilde{W}^{*}\|_{\infty} = 1 , \tilde{v}_{i} \rightharpoonup \tilde{v} L^{p} 空间弱收敛成立.在上述方程两边同时取极限,我们有

\begin{eqnarray*} \left\{\begin{array}{ll} -\Delta \tilde{W}^{*} = \frac{\tilde{W}^{*}}{1 + \alpha\tilde{ v}} (\tilde{a} - c \tilde{v}), \quad & x \in \Omega, \\ - \Delta \tilde{V} = \tilde{v} (b - \tilde{v} ), & x \in \Omega, \\ \tilde{W}^{*} = \tilde{V} = 0, & x\in\partial\Omega. \end{array}\right. \end{eqnarray*}

因此, a = a^{**} ,矛盾.

假设 \tilde{W} > 0, \tilde{V} \equiv 0 ,令 \tilde{V}_{i}^{*} = \frac{\tilde{V}_{i}}{\|\tilde{V}_{i}\|_{\infty}} ,则

\begin{eqnarray*} \left\{\begin{array}{ll} -\Delta \tilde{W}_{i} = \frac{\tilde{W}_{i}}{1 + \alpha \tilde{v}_{i}}\Big (\tilde{a}_{i} - \frac{c \tilde{v}_{i}}{1 + \tilde{w}_{i}}\Big), \quad & x \in \Omega, \\ -\Delta \tilde{V}_{i}^{*} = \frac{\tilde{V}_{i}^{*}}{\mu + \frac{1}{1 + \tilde{w}_{i}}} (b - \tilde{v}_{i} ), & x \in \Omega, \\ \tilde{W}_{i} = \tilde{V}_{i}^{*} = 0, & x\in\partial\Omega. \end{array}\right. \end{eqnarray*}

根据二阶椭圆型正则化理论我们可以假设 \tilde{W}_{i} \rightarrow \tilde{W} , \tilde{V}_{i}^{*} \rightarrow \tilde{V}^{*} C^{1} 空间成立且满足 \| \tilde{V}^{*} \|_{\infty} = 1 , \tilde{w}_{i} \rightharpoonup \tilde{w} L^{p} 空间弱收敛成立.在上述方程两边同时取极限,我们有

\begin{eqnarray*} \left\{\begin{array}{ll} - \Delta \tilde{W} = \tilde{a} \tilde{W}, & x \in \Omega, \\ - \Delta \tilde{V}^{*} = b \frac{\tilde{V}^{*}}{\mu + \frac{1}{1 + \beta \tilde{w}}}, \quad & x \in \Omega, \\ \tilde{W} = \tilde{V}^{*} = 0, & x \in \partial \Omega. \end{array}\right. \end{eqnarray*}

\tilde{W} = s \phi_{1} ,其中, s \geq 0 .因此, \lambda_{1} (- b \frac{1 + \beta s \phi_{1}}{\mu (1 + \beta s \phi_{1}) + 1}) = 0 ,矛盾.

假设 (\tilde{W}, \tilde{V}) \equiv (0, 0) ,类似地,我们可以得到矛盾.

因此,我们必有 C - \{ (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \} \subset P_{0} .从而,情形(ⅰ)不会发生.由于 \phi^{**} > 0 \Omega 上成立,补空间 \overline{Y} 不可能包含不变号的元素,从而,情形(ⅲ)也不会发生.因此,情形(ⅱ)必然发生.也就是说,解曲线 C R \times X 空间延伸至无穷.根据引理4.3,我们知道参数 a 是有界的,从而我们有 \lim\limits_{i \rightarrow \infty} \|W_{i}\|_{\infty} = \infty .现在,我们证明: \lim\limits_{i \rightarrow \infty} a_{i} = a^{*} .显然, \lim\limits_{i \rightarrow \infty} v_{i} = \mu \theta_{b/\mu} ,因此, \lim\limits_{i \rightarrow \infty} V_{i} = \mu^{2} \theta_{b/\mu} . W_{i}^{*} = \frac{W_{i}}{\|W_{i}\|_{\infty}} ,则

-\Delta W_{i}^{*} = W_{i}^{*} \Big(\frac{a_{i}}{1 + \alpha v_{i}} - \frac{c v_{i}}{1 + \alpha v_{i} + W_{i}}\Big), \; \|W_{i}^{*}\|_{\infty} = 1, \; W_{i}^{*}|_{\partial \Omega} = 0.

根据二阶椭圆型正则化理论我们可以假设 W_{i}^{*} \rightarrow W^{*} C^{1} 空间成立且 \frac{1}{1 + \alpha v_{i} + W_{i}} \rightharpoonup h L^{2} 空间弱收敛成立.在上述方程两边同时取极限,我们有

\begin{eqnarray*} -\Delta W^{*} = W^{*} \Big(\frac{a}{1 + \alpha \mu \theta_{b/\mu}} - c \mu \theta_{b/\mu} h\Big), \; \|W^{*}\|_{\infty} = 1, \; W^{*}|_{\partial \Omega} = 0. \end{eqnarray*}

根据强最大值原理,我们有 W^{*} > 0 \Omega 上成立.因此, h = 0, -\Delta W^{*} = \frac{a}{1 + \alpha \mu \theta_{b/\mu}} W^{*} .从而, \lambda_{1} (- \frac{a}{1 + \alpha \mu \theta_{b/\mu}}) = 0 .也就是说, a = a^{*} .证毕.

下面的定理表明:当 a \in [a^{*} + \epsilon, a^{**}), b > (\mu + 1) \lambda_{1} m = \beta 充分大时,则系统(1.1)有两种类型的正解.

定理4.3 假设 b > (\mu + 1) \lambda_{1} ,则存在充分小的 \epsilon, \delta 和充分大的 M_{1} = M_{1} (\epsilon, \delta) ,使得当 m \geq M_{1} a \in [a^{*} + \epsilon, a^{**}) 时,有 \|u - \tilde{\theta}_{a, \alpha}\|_{C^{1}} + \|v - \mu \theta_{b/\mu}\|_{C^{1}} < \delta \|m u - \tilde{w}\|_{C^{1}} + \|v - \tilde{v}\|_{C^{1}} < \delta ,其中, (u, v) 是系统(1.1)的正解, (\tilde{\theta}_{a, \alpha}, \mu \theta_{b/\mu}) 是系统(4.1)的正解, (\tilde{w}, \tilde{v}) 是系统(4.2)的正解.

 我们采用反证法.假设存在 m_{i} \rightarrow \infty , a_{i} \rightarrow a \in [a^{*} + \epsilon, a^{**}] 且当 (a, m) = (a_{i}, m_{i}) 时,设 (u_{i}, v_{i}) 是系统(1.1)的解使得 (u_{i}, v_{i}) 满足 \|u_{i} - \tilde{\theta}_{a, \alpha} \|_{C^{1}} + \|v_{i} - \mu \theta_{b/\mu} \|_{C^{1}} \geq \delta ,且 \|m_{i} u_{i} - \tilde{w}\|_{C^{1}} + \|v_{i} - \tilde{v}\|_{C^{1}} \geq \delta .这里有两种情形我们需要考虑:

情形(ⅰ)如果 m_{i} \|u_{i}\|_{\infty} \rightarrow \infty ,则我们断言: u_{i} \rightharpoonup u > 0 L^{2} 空间弱收敛成立.反证法.假设 u_{i} \rightharpoonup 0 ,令 \tilde{u}_{i} = \frac{u_{i}}{\|u_{i}\|_{\infty}} ,则

\begin{equation} -\Delta [(1 + \alpha v_{i}) \tilde{u}_{i} ] = \tilde{u}_{i} \Big(a_{i} - u_{i}- \frac{c v_{i}}{1 + m_{i} u_{i}}\Big), \; \tilde{u}_{i}|_{\partial \Omega} = 0. \end{equation}
(4.7)

\phi_{i} = (1 + \alpha v_{i}) \tilde{u}_{i} ,则(4.7)式可以改写为

-\Delta \phi_{i} = \frac{\phi_{i}}{1 + \alpha v_{i}}\Big (a_{i} - u_{i}- \frac{c v_{i}}{1 + m_{i} u_{i}}\Big), \; \phi_{i}|_{\partial \Omega} = 0.

根据二阶椭圆型正则化理论我们知道, \phi_{i} \rightarrow \phi C^{1} 空间成立且 v_{i} \rightharpoonup v , \frac{1}{1 + m_{i} u_{i}} \rightharpoonup h L^{2} 空间弱收敛成立.在上述方程两边同时取极限,我们有

\begin{eqnarray*} -\Delta \phi = \frac{\phi}{1 + \alpha v} (a - c v h), \; \phi|_{\partial \Omega} = 0. \end{eqnarray*}

根据强最大值原理我们知道, \phi > 0 \Omega 上成立.因此, u > 0 \Omega 上成立且 h = 0 L^{2} 上成立.因此, -\Delta \phi = \frac{a}{1 + \alpha v} \phi .也就是说, \lambda_{1} (- \frac{a}{1 + \alpha v}) = 0 .显然, \lim\limits_{i \rightarrow \infty} (\mu + \frac{1}{1 + m_{i} u_{i}}) = \mu v C^{1} 空间上成立且 v = \mu \theta_{b/\mu} .因此, \lambda_{1} (- \frac{a}{1 + \alpha \mu \theta_{b/\mu}}) = 0 .从而, a = a^{*} ,矛盾.类似地,我们有 u = \tilde{\theta}_{a, \alpha} ,矛盾.

情形(ⅱ)如果 m_{i} \|u_{i}\|_{\infty} 是一致有界的,则 u_{i} \rightharpoonup 0 L^{2} 空间弱收敛成立.令 \tilde{w}_{i} = m_{i} u_{i} , \tilde{W}_{i} = (1 + \alpha v_{i}) w_{i} \tilde{V}_{i} = (\mu + \frac{1}{1 + w_{i}}) v_{i} .根据二阶椭圆型正则化理论我们知道, \tilde{w}_{i} \rightharpoonup \tilde{w} , \tilde{v}_{i} \rightharpoonup \tilde{v} L^{2}(\Omega) 空间上弱收敛成立且 \tilde{W}_{i} \rightarrow \tilde{W} = (1 + \alpha \tilde{v}) \tilde{w} , \tilde{V}_{i} \rightarrow \tilde{V} = (\mu + \frac{1}{1 + \tilde{w}}) \tilde{v} C^{1} 上成立.因此,我们说明 (\tilde{w}, \tilde{v}) 是系统(4.2)的解.根据最大值原理我们知道, \tilde{V} > 0 \Omega 上成立.因此, \tilde{v} > 0 \Omega 上成立.

进一步,我们说明当 a = a^{**} ,我们能够证明 m_{i} u_{i} \rightarrow \tilde{w}_{i} ,其中, \tilde{w}_{i} 是系统(4.2)的解且 a = a_{i} a_{i} \rightarrow a^{**} .如果 a < a^{**} ,则我们能够说明 \tilde{w} 是系统(4.2)的解.证毕.

事实上,根据局部分歧理论我们知道, a = a^{**} 是系统(2.2)的简单分歧点.从全局分歧的角度出发,我们说明当 a \in (a^{**}, \infty) 时,系统(2.2)至少有一个正解.进一步,我们断言当 m = \beta 且充分大时,局部分歧解将向左延拓,从而存在一个转向点 \tilde{a}^{*} 满足 \tilde{a}^{*} < a^{**} . a \in (\tilde{a}^{*}, a^{**}) 时,系统(2.2)至少有两个正解.由于证明过程是标准的,我们忽略其证明仅陈述其结果.

定理4.4 当 a \in (a^{**}, \infty) 时,系统(2.2)至少有一个正解.进一步,当 m = \beta 且充分大时,存在一个 \tilde{a}^{*} 满足 \tilde{a}^{*} < a^{**} ,当 a \in (\tilde{a}^{*}, a^{**}) 时,系统(2.2)至少有两个正解.

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