在本文中,我们将考虑下列带有交叉扩散的捕食-食饵模型

(1.1) $ \begin{equation} \left\{\begin{array}{ll} -\Delta[(1+\alpha v) u] = u \Big(a-u-\frac{cv}{1+mu}\Big), & x \in \Omega, \\ -\Delta \Big [\Big(\mu + \frac{1}{1 + \beta u}\Big)v\Big ] = 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} $

其中$ \Omega $是在$ {\Bbb R}^{N} $下带有光滑边界$ \partial \Omega $的有界区域; $ u, v $分别代表食饵和捕食者的密度.参数$ a, c, d $和$ \mu $是正常数;参数$ \alpha, \beta $和$ m $是非负常数;参数$ b $可能变号.

当参数$ m = 0 $时,则系统(1.1)退化为带有交叉扩散的经典的具有Lotka-Volterra的捕食-食饵模型,该模型已经被许多生物数学家所研究(见参考文献[13-16, 22]及其参考文献).特别地, Kadota和Kuto^[13]利用分歧理论研究了系统正解的存在性,并证明该系统的共存区域随着交叉扩散系数$ \beta $的变大而变大,随着交叉扩散系数$ \alpha $的变大而变小.接着, Kuto和Yamada^[14]表明:该系统存在一个分歧解,在合适的条件下,分歧解曲线随着分歧参数其形状为$ S $型或$ \supset $型.进一步, Kuto^[15]讨论了当交叉扩散系数$ \alpha = 0, \beta \rightarrow \infty $时的阴影系统,从全局分歧的角度出发,他断言该系统的正解在参数$ a = \lambda_{1} (c \mu \theta_{b/\mu}) $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.

当$ \alpha = \beta = 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 $为分歧参数,他们利用全局分歧理论说明了系统从半平凡解分歧出来的正解情况.同时,他们考虑了当交叉扩散系数$ \beta $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 = \beta $充分大时,系统(1.1)可以被看成是一些极限系统的扰动系统.我们以参数$ 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) $是一致有界的.然而,解$ \|w\|_{\infty} $将在$ 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}) $.当$ p \equiv 0 $,我们记$ \lambda_{1} (0) $为$ \lambda_{1} $.进一步,我们记$ \phi_{1} > 0 $在$ \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

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

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.

令

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

则系统(1.1)可以改写为

(2.2) $ \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} $

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} $或$ (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 $和$ \mu $,并且定义集合$ S_{0} $和$ S_{1} $:

(2.3a) $ 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\}, $

(2.3b) $ 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\}. $

下列引理表明函数$ a^{*}(b) $和$ 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^{*} $和$ \phi^{**} $分别满足下列方程

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

(2.4b) $ - \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. $

令$ p > N $,我们定义Banach空间$ X $和$ 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 $是$ 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} $.令$ L = A'(y_{0}) $是$ A $在$ y_{0} $处的Fréchet导数,则$ L: W_{y_{0}} \rightarrow W_{y_{0}} $.若$ I - L $在$ E $上可逆,并且

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

(ⅱ) $ L $在$ 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 $是$ L $大于1的特征值的代数重数之和.

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 $.

证  (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}, $

且$ r(A_{0}'(0, 0)) < 1 $.因此,算子$ I - A_{0}'(0, 0) $在$ \bar{P}_{(0, 0)} $是可逆的且算子$ A_{0}'(0, 0) $在$ \bar{P}_{(0, 0)} $无$ \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) $在$ \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 $.令$ t_{0} = r_{a}^{-1} $,则$ 0 < t_{0} < 1 $且$ (I - t_{0} A'(0, 0)) (U, 0) = (0, 0) \in S_{(0, 0)} $.从而, $ A'(0, 0) $有$ \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) $在$ \bar{P}_{(\theta_{a}, 0)} $是可逆的.

下面我们断言算子$ A'(\theta_{a}, 0) $在$ \bar{P}_{(\theta_{a}, 0)} $有$ \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). $

则$ r_{b} = r(L) > 1 $是算子$ L $的特征值,且其对应的特征函数$ V > 0 $.令$ 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) $有$ \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) $在$ \bar{P}_{(\theta_{a}, 0)} $可逆.下面我们仅说明算子$ A'(\theta_{a}, 0) $在$ \bar{P}_{(\theta_{a}, 0)} $无$ \alpha $性质.我们采用反证法.假设算子$ L $在$ \bar{P}_{(\theta_{a}, 0)} $有$ \alpha $性质,从而存在某个$ 0 < t < 1 $和$ (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 $在$ \bar{P}_{(\theta_{a}, 0)} $无$ \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

(3.1) $ \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} $

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

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

(4.1) $ \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} $

和奇异扰动问题

(4.2) $ \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} $

引理4.1 假设$ b > \mu \lambda_{1} $且$ 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.

证 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. $

令$ \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 $和$ 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 $.

证 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 $在$ L^{2} $空间弱收敛成立.令$ \tilde{w}_{i} = \frac{w_{i}}{\|w_{i}\|_{\infty}} $,则

(4.3) $ \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} $

令$ \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 $在$ 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 $在$ 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 $在$ \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.

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

(4.4) $ \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} $

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.

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

(4.5) $ \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} $

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)} $.令$ (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}) $是$ 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 $是$ 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 $且$ \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 $.

证 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*} $

令$ 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*} $

令$ 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.

令

$ \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, $

且$ \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

(4.6) $ \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} $

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 $.令$ 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*} $

证 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 $在$ {\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} $是$ \{(- \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 $在$ \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 $在$ \Omega $上成立.类似地,对于其他情形我们有$ \tilde{V} \equiv 0 $在$ \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} $在$ C^{1} $空间上成立且$ \|\tilde{W}^{*}\|_{\infty} = 1 $, $ \tilde{v}_{i} \rightharpoonup \tilde{v} $在$ 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}^{*} $在$ C^{1} $空间成立且满足$ \| \tilde{V}^{*} \|_{\infty} = 1 $, $ \tilde{w}_{i} \rightharpoonup \tilde{w} $在$ 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*} $

则$ \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 $在$ \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 $在$ 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} $.令$ 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^{*} $在$ C^{1} $空间成立且$ \frac{1}{1 + \alpha v_{i} + W_{i}} \rightharpoonup h $在$ 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 $在$ \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} $且$ 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} $且$ a \in [a^{*} + \epsilon, a^{**}) $时,有$ \|u - \tilde{\theta}_{a, \alpha}\|_{C^{1}} + \|v - \mu \theta_{b/\mu}\|_{C^{1}} < \delta $或$ \|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.

证 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 $在$ 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

(4.7) $ \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} $

令$ \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 $在$ C^{1} $空间成立且$ v_{i} \rightharpoonup v $, $ \frac{1}{1 + m_{i} u_{i}} \rightharpoonup h $在$ 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 $在$ \Omega $上成立.因此, $ u > 0 $在$ \Omega $上成立且$ h = 0 $在$ 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 $在$ 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 $在$ 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} $且$ \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} $在$ 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} $在$ 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 $在$ \Omega $上成立.因此, $ \tilde{v} > 0 $在$ \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} $当$ 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^{**} $.当$ 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.