0 引言

近年来, 关于生物数学领域的捕食食饵模型的研究已经成为热点, 尤其是对于种群扩散影响下的捕食模型, 国内外学者均已取得了一些符合实际的研究成果.文献[1]研究了一类捕食模型的正常数平衡态解的稳定性及分歧; 文献[2-3]利用极大值原理和分歧定理研究了一类捕食模型局部解的延拓; 文献[4-7]利用分歧定理研究了模型在交叉扩散影响下的正解的存在性问题.在文献[8]中, 作者提出了一类具有扩散项的捕食食饵模型, 通过给出正解的先验估计及局部分歧解存在条件, 进而得到该系统平衡态的全局分歧解及其走向; 文献[9]则在上述基础上研究了该类模型在交叉扩散项影响下的分歧.

在同时考虑交叉扩散和自扩散项时，本文将继续研究如下捕食-食饵模型在齐次Dirichlet边界条件下正解的存在性，即

$ \left\{ \begin{array}{l} {u_t} - \Delta \left[ {\left( {1 + {m_1}u + {m_2}\upsilon } \right)u} \right] = \left( {a - u - \frac{{b\upsilon }}{{\left( {1 + au} \right)\left( {1 + \beta u} \right)}}} \right)u,\;\;\;\;x \in \mathit{\Omega ,t > }0\mathit{,}\\ {\upsilon _t} - \Delta \left[ {\left( {1 + {m_3}\upsilon + {m_4}u} \right)\upsilon } \right] = \left( {c - \upsilon + \frac{{du}}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right)\upsilon ,\;\;\;x \in \mathit{\Omega ,t > }0\mathit{,}\\ u = \upsilon = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega ,t > }0\mathit{,}\\ u\left( {x,0} \right) = {u_0}\left( x \right) \ge 0,\upsilon \left( {x,0} \right) = {\upsilon _0}\left( x \right) \ge 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega }. \end{array} \right. $

(1)

其中:Ω为R^N中具有光滑边界∂Ω上的有界区域; u, v分别表示食饵和捕食者的种群密度; a, b, c, d, α, β都是正常数, m₁, m₃表示自扩散系数; m₂, m₄表示交叉扩散系数, 反应函数$\frac{{uv}}{{(1 + \alpha u)(1 + \beta v)}}$是Bazykin研究捕食者的饱和不稳定性与食饵的稳定性时建立的功能反应函数, 生物背景参见文献[10].

本文将针对模型(1) 的如下平衡态方程展开讨论.

$ \left\{ \begin{array}{l} - \Delta \left[ {\left( {1 + {m_1}u + {m_2}\upsilon } \right)u} \right] = \left( {a - u - \frac{{b\upsilon }}{{\left( {1 + au} \right)\left( {1 + \beta \upsilon } \right)}}} \right)u,\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,}\\ - \Delta \left[ {\left( {1 + {m_3}\upsilon + {m_4}u} \right)\upsilon } \right] = \left( {c - \upsilon + \frac{{du}}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right)\upsilon ,\;\;\;\;\;\;\;x \in \mathit{\Omega ,}\\ u = \upsilon = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }. \end{array} \right. $

(2)

注:对于问题(2) 的解(u, v), 如果在Ω中, (u, v) 中只有一个分量为0, 则称其为半平凡解.

1 预备知识

记C₀¹ (Ω)={u∈C¹(Ω):u|∂Ω=0}.定义C₀¹ (Ω) 中的范数为通常的Banach空间C¹ (Ω) 中的范数, 令X=C₀¹ (Ω)×C₀¹ (Ω), 则X是Banach空间.

首先, 考虑特征值问题

$ \left\{ \begin{array}{l} - p\Delta \varphi + q\left( x \right)\varphi = \lambda \varphi ,\;\;\;\;\;x \in \mathit{\Omega },\\ \varphi = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }\mathit{.} \end{array} \right. $

(3)

引理1^[11]  假设q(x)∈C(E), p为常数, 则问题(3) 的所有特征值满足λ₁(p, q) < λ₂(p, q)≤λ₃(p, q)≤…→∞, 相应的特征函数为φ₁, φ₂, ….由文献[11]知λ₁(p, q) 是简单的且关于q(x) 严格单调递增.为方便起见, 简记λ₁=λ₁(0), 相应的主特征函数φ₁ > 0.

再考虑边值问题

$ \left\{ \begin{array}{l} \Delta \left[ {\left( {1 + {m_1}u} \right)u} \right] + u\left( {a - bu} \right) = 0,\;\;\;\;\;\;x \in \mathit{\Omega },\\ u\left( x \right) = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }. \end{array} \right. $

(4)

$ \left\{ \begin{array}{l} \Delta \left[ {\left( {1 + {m_3}\upsilon } \right)\upsilon } \right] + \upsilon \left( {a - d\upsilon } \right) = 0,\;\;\;\;\;\;x \in \mathit{\Omega },\\ \upsilon \left( x \right) = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }. \end{array} \right. $

(5)

引理2^[11]  (1) 如果a≤λ₁, 则u=0是问题(4) 的唯一非负解; 若a > λ₁, 则问题(4) 的唯一正解为θ_a.

(2) 如果c≤λ₁, 则v=0是问题(5) 的唯一非负解; 当c > λ₁时, 其存在唯一正解θ_c.因此, 当a > λ₁, 问题(2) 存在半平凡解(θ_a, 0);当c > λ₁, 问题(2) 存在半平凡解(0, θ_c).

定义Z=(U, V), 其中U=(1+m₁u+m₂v)u, V=(1+m₃v+m₄u)v, 则

$ \frac{{\partial \left( {U,V} \right)}}{{\partial \left( {u,\upsilon } \right)}} = \left( {1 + 2{m_1}u + {m_2}\upsilon } \right)\left( {1 + 2{m_3}u + {m_4}\upsilon } \right) - {m_2}u \cdot {m_4}\upsilon \ge 1 > 0. $

即(u, v)≥0与(U, V)≥0之间存在一一对应的关系.现在, 引入和问题(2) 等价的半线性椭圆系统

$ \left\{ \begin{array}{l} - \Delta U = \left( {a - u - \frac{{b\upsilon }}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right)u,\;\;\;\;\;x \in \mathit{\Omega ,}\\ - \Delta V = \left( {c - \upsilon - \frac{{du}}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right)\upsilon ,\;\;\;\;\;x \in \mathit{\Omega ,}\\ U = V = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }\mathit{.} \end{array} \right. $

(6)

易知, 当a, c > λ₁时, 问题(6) 的两个半平凡解分别为($\overline {{\theta _a}} $, 0), (0, $\overline {{\theta _c}} $), 其中

$ {{\bar \theta }_a} = \left( {1 + {m_1}{\theta _a}} \right){\theta _a},{{\bar \theta }_c} = \left( {1 + {m_3}{\theta _c}} \right){\theta _c}. $

引理3^[12]  假设a > λ₁, 令L(a)=-Δ $ - \frac{{a - 2{\theta _a}}}{{1 + 2{m_1}{\theta _a}}}$, 则L(a) 的特征值均大于0.

引理4  设c > λ₁, 则当cm₄ > d时, 存在唯一的a=a^*(c)∈(λ₁, ∞), 满足${\lambda _1}\left( { - \frac{{c + \left( {c\alpha + d} \right){\theta _a}}}{{(1 + {m_4}{\theta _a})(1 + \alpha {\theta _a})}}} \right) = 0$, 且a=a^*(c) 关于c严格单调递增.此外, ∃ψ^*≥0满足

$ \left\{ \begin{array}{l} - \Delta {\psi ^ * } - \frac{{c + \left( {ac + d} \right){\theta _a}}}{{\left( {1 + \alpha {\theta _a}} \right)\left( {1 + {m_4}{\theta _a}} \right)}}{\psi ^ * } = 0,{\psi ^ * } > 0,\;\;\;\;\;x \in \mathit{\Omega },\\ {\psi ^ * } = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }. \end{array} \right. $

证明  取A(a, c)=${\lambda _1}\left( { - \frac{{c + \left( {c\alpha + d} \right){\theta _a}}}{{(1 + {m_4}{\theta _a})(1 + \alpha {\theta _a})}}} \right)$.显然A(λ₁, c)=λ₁(-c)=λ₁-c < 0.由于当a→∞时θ_a→∞, 故有$\mathop {{\rm{lim}}}\limits_{a \to \infty } $A(a, c)=λ₁(0)=λ₁ > 0.经计算得

$ \frac{\partial }{{\partial a}}\left( { - \frac{{c + \left( {c\alpha + d} \right){\theta _a}}}{{\left( {1 + {m_4}{\theta _a}} \right)\left( {1 + \alpha {\theta _a}} \right)}}} \right) = \left( {\frac{{\left( {c\alpha + d} \right)\alpha {m_4}\theta _a^2 + 2c\alpha {m_4}{\theta _a} + c{m_4} - d}}{{{{\left( {1 + {m_4}{\theta _a}} \right)}^2}{{\left( {1 + \alpha {\theta _a}} \right)}^2}}}} \right)\frac{{\partial {\theta _a}}}{{\partial a}}. $

又因为q→λ₁(q):C(E)→R与a→θ_a:[λ₁, ∞)→C²(Ω)∩C₀(E) 均严格单调递增, 可知A(a, c) 关于a严格单调递增.从而存在唯一的a=a^*(c) > λ₁, 使得A(a^*(c), c)=0.

再对A(a^*(c), c)=0两边关于c求导, 得A_a(a^*(c), c)·a^*′(c)+A_c(a^*(c), c)=0.由于A_c(a, c) < 0, 结合A_a(a, c) > 0得知a^*′(c) > 0, 即a=a^*(c) 关于c严格单调递增.

类似可以证明以下引理.

引理5  假设c > λ₁, 则当aβ > b时, 就存在唯一的a=a_*(c)∈(λ₁, ∞), 满足${\lambda _1}\left( { - \frac{{a + \left( {\alpha \beta - b} \right){\theta _c}}}{{(1 + \beta {\theta _c})(1 + {m_2}{\theta _c})}}} \right) = 0$, 且a=a_*(c) 关于c严格单调递增.此外, ∃φ_*≥0满足

$ \left\{ \begin{array}{l} - \Delta {\phi _ * } - \frac{{a + \left( {a\beta - b} \right){\theta _c}}}{{\left( {1 + \beta {\theta _c}} \right)\left( {1 + {m_2}{\theta _c}} \right)}}{\phi _ * } = 0,{\phi _ * } > 0,\;\;\;\;\;\;x \in \mathit{\Omega },\\ {\phi _ * } = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }\mathit{.} \end{array} \right. $

2 正解的先验估计

现在, 结合文献[12-13]中的方法给出系统(6) 的正解存在的必要条件及先验估计.

定理1  当a≤λ₁, 或者c+$\frac{d}{\alpha }$≤λ₁, 则问题(6) 没有正解.

证明  若问题(6) 存在正解(U, V), 由问题(6) 中的第2个方程得

$ - \Delta V = \frac{V}{{1 + {m_3}\upsilon + {m_4}u}}\left( {c - \upsilon + \frac{{du}}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right) < \frac{V}{{1 + {m_3}\upsilon + {m_4}u}}\left( {c + \frac{d}{\alpha }} \right) < \left( {c + \frac{d}{\alpha }} \right)V, $

两边同乘以V, 分部积分得

$ \int_\mathit{\Omega } {{{\left| {\nabla V} \right|}^2}{\rm{d}}x = \left\| {\nabla V} \right\|_2^2 < \left( {c + \frac{d}{\alpha }} \right)} \left\| V \right\|_2^2, $

由Poincare不等式‖∇V‖₂²≥λ₁‖V‖₂²，可得c+ $ \frac{d}{\alpha }$ > λ₁, 同理可证a > λ₁.与已知条件矛盾, 则定理1得证.

定理2  设a > λ₁, c+ $\frac{d}{\alpha }$ > λ₁且b-βa(1+αa) > 0.若(U, V) 是问题(6) 的任意正解, 则∀x∈Ω, 有

$ \begin{array}{*{20}{c}} {0 < u\left( x \right) <U\left( x \right) < M\left( a \right) = \left[ {1 + {m_1}a + \frac{{{m_2}a\left( {1 + \alpha a} \right)}}{{b - \beta a\left( {1 + \alpha a} \right)}}} \right]a,}\\ {0 < \upsilon \left( x \right) < V\left( x \right)\left[ {1 + {m_3}\left( {c + \frac{{dM\left( a \right)}}{{1 + \alpha M\left( a \right)}}} \right) + {m_4}M\left( a \right)} \right]\left( {c + \frac{{dM\left( a \right)}}{{1 + \alpha M\left( a \right)}}} \right).} \end{array} $

证明  设∃x₀∈Ω, 使得U(x₀)=$\mathop {{\rm{max}}}\limits_{x \in \Omega } $U(x).由于

$ 0 \le - \Delta U\left( {{x_0}} \right) = u\left( {{x_0}} \right)\left( {a - u\left( {{x_0}} \right) - \frac{{b\upsilon \left( {{x_0}} \right)}}{{\left[ {1 + \alpha u\left( {{x_0}} \right)} \right]\left[ {1 + \beta \upsilon \left( {{x_0}} \right)} \right]}}} \right), $

故有u(x₀) < a, $\frac{{bv({x_0})}}{{\left[ {1 + \alpha u\left( {{x_0}} \right)} \right]\left[ {1 + \beta v\left( {{x_0}} \right)} \right]}} < a$, 则

$ U\left( x \right) \le U\left( {{x_0}} \right) = \left[ {1 + {m_1}u\left( {{x_0}} \right) + {m_2}\upsilon \left( {{x_0}} \right)} \right]u\left( {{x_0}} \right) < \left[ {1 + {m_1}a + \frac{{{m_2}a\left( {1 + \alpha a} \right)}}{{b - \beta a\left( {1 + \alpha a} \right)}}} \right]a. $

同理可得

$ V\left( x \right) < \left[ {1 + {m_3}\left( {c + \frac{{dM\left( a \right)}}{{1 + \alpha M\left( a \right)}}} \right) + {m_4}M\left( a \right)} \right]\left( {c + \frac{{dM\left( a \right)}}{{1 + \alpha M\left( a \right)}}} \right), $

由(u, v) 与(U, V) 之间的关系知定理2成立.

3 分歧正解的存在性

现在以a为分歧参数, 参考文献[14-19], 利用Crandall-Rabinowitz局部分歧定理, 给出问题(6) 发自半平凡解($\overline {{\theta _a}} $, 0) 与(0, $\overline {{\theta _c}} $) 的局部分歧正解的存在性.

定理3  设a > λ₁, c+ $\frac{d}{\alpha }$ > λ₁且cm₄ > d, 则(a^*; $\overline {{\theta _{a*}}} $, 0)∈X×R₊为问题(6) 的分歧点, 且(a^*; $\overline {{\theta _{a*}}} $, 0) 的领域内存在正解

$ \mathit {\Gamma} * = \{ (a(s);{\theta _{a*}} + s({\phi ^*} + {\mathit{\Phi} _1}(s)),s({\psi ^*} + {\mathit{\Psi} _1}(s))):0 < s < \delta \} . $

其中a^*由${\lambda _1}\left( { - \frac{{c + \left( {c\alpha + d} \right){\theta _a}}}{{(1 + {m_4}{\theta _a})(1 + \alpha {\theta _a})}}} \right) = 0$唯一确定, ψ^* > 0满足

$ - \Delta {\psi ^ * } - \frac{{c + \left( {c\alpha + d} \right){\theta _a}}}{{\left( {1 + {m_4}{\theta _a}} \right)\left( {1 + \alpha {\theta _a}} \right)}}{\psi ^ * } = 0,x \in \mathit{\Omega ,} $

ψ^*=0, x∈∂Ω, ∫_Ωψ^*2dx=1, φ^*∈C₀¹ (Ω), δ > 0充分小.这里(a(s); Φ₁(s), Ψ₁(s)) 是C¹连续函数, 满足a(0)=a^*, Φ₁(0)=0, Ψ₁(0)=0, ∫_Ωψ₁φ^*dx=0, 且

$ {\phi ^*} = L_{{a^ * }}^{ - 1}\left[ { - \frac{{{m_2}{\theta _{{a^ * }}}\left( {{a^ * } - 2{\theta _{{a^ * }}}} \right)\left( {1 + \alpha {\theta _{{a^ * }}}} \right) + b{\theta _{{a^ * }}}\left( {1 + 2{m_1}{\theta _{{a^ * }}}} \right)}}{{\left( {1 + 2{m_1}{\theta _{{a^ * }}}} \right)\left( {1 + {m_4}{\theta _{{a^ * }}}} \right)\left( {1 + \alpha {\theta _{{a^ * }}}} \right)}}} \right]. $

证明令

$ f\left( {u,\upsilon } \right) = \left( {a - u - \frac{{b\upsilon }}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right)u,g\left( {u,\upsilon } \right) = \left( {c - \upsilon + \frac{{du}}{{\left( {1 + \alpha u} \right)\left( {1 + \beta \upsilon } \right)}}} \right)\upsilon , $

其中u, v均为(U, V) 的函数.将问题(6) 在(U, V)=($\overline {{\theta _a}} $, 0) 处Taylor展开为

$ \left( {\begin{array}{*{20}{c}} {\Delta U}\\ {\Delta V} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} {f\left( {\overline {{\theta _a}} ,0} \right)}\\ {g\left( {\overline {{\theta _a}} ,0} \right)} \end{array}} \right) + \left[ {\left( {\begin{array}{*{20}{c}} {{f_u}}&{{f_\upsilon }}\\ {{g_u}}&{{g_\upsilon }} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{u_U}}&{{u_V}}\\ {{\upsilon _U}}&{{\upsilon _V}} \end{array}} \right)} \right]\left| {_{\left( {{\theta _a},0} \right)}} \right. \cdot \left( {\begin{array}{*{20}{c}} {U - \overline {{\theta _a}} }\\ V \end{array}} \right) + \\ \left( {\begin{array}{*{20}{c}} {{F^1}\left( {a;U - \overline {{\theta _a}} ,V} \right)}\\ {{F^2}\left( {a;U - \overline {{\theta _a}} ,V} \right)} \end{array}} \right) = \left( \begin{array}{l} 0\\ 0 \end{array} \right). $

这里, 偏导数为($\overline {{\theta _a}} $, 0) 处的导数值, Fⁱ(U-$\overline {{\theta _a}} $, V) 满足Fⁱ(0, 0)=F_{(U, V)}ⁱ (0, 0)=0, i=1, 2.

同时对(U, V) 求导, 得

$ \left( {\begin{array}{*{20}{c}} {{u_U}}&{{u_V}}\\ {{\upsilon _U}}&{{\upsilon _V}} \end{array}} \right)\left| {_{\left( {{\theta _a},0} \right)}} \right. = \frac{1}{{\left( {1 + 2{m_1}{\theta _a}} \right)\left( {1 + {m_4}{\theta _a}} \right)}}\left( {\begin{array}{*{20}{c}} {1 + {m_4}{\theta _a}}&{ - {m_2}{\theta _a}}\\ 0&{1 + 2{m_1}{\theta _a}} \end{array}} \right). $

令U=U-$\overline {{\theta _a}} $, 则有

$ T\left( {a;\bar U,V} \right) = \left( \begin{array}{l} \Delta \bar U + \frac{{\left( {a - 2{\theta _a}} \right)}}{{1 + 2{m_1}{\theta _a}}}\bar U - \frac{{\left( {a{m_2} + b + a\alpha {m_2}{\theta _a} - 2{m_2}{\theta _a} + 2{m_1}b{\theta _a} - 2\alpha {m_2}\theta _a^2} \right){\theta _a}}}{{\left( {1 + 2{m_1}{\theta _a}} \right)\left( {1 + \alpha {\theta _a}} \right)\left( {1 + {m_4}{\theta _a}} \right)}}V\\ + {F^1}\left( {a;\bar U,V} \right)\\ \Delta V + \frac{{c + \left( {ac + d} \right){\theta _a}}}{{\left( {1 + a{\theta _a}} \right)\left( {1 + {m_4}{\theta _a}} \right)}}V + {F^2}\left( {a;\bar U,V} \right) \end{array} \right) = 0, $

显然T(a; 0, 0)=0.记T(a; U, V) 关于(U, V) 在(a^*; 0, 0) 处的Frechlet导数是L(a^*; 0, 0).经计算, L(a^*; 0, 0)·(φ, ψ)=0等价于

$ \left\{ \begin{array}{l} - \Delta \phi - \left( {\frac{{{a^ * } - 2{\theta _{{a^ * }}}}}{{1 + 2{m_1}{\theta _{{a^ * }}}}}} \right)\phi = \\ \;\;\;\;\;\;\; - \frac{{\left( {{a^ * }{m_2} + b + {a^ * }\alpha {m_2}{\theta _{{a^ * }}} - 2{m_2}{\theta _{{a^ * }}} + 2{m_1}b{\theta _{{a^ * }}} - 2\alpha {m_2}{\theta _{{a^ * }}}^2} \right){\theta _{{a^ * }}}}}{{\left( {1 + 2{m_1}{\theta _{{a^ * }}}} \right)\left( {1 + \alpha {\theta _{{a^ * }}}} \right)\left( {1 + {m_4}{\theta _{{a^ * }}}} \right)}}\psi ,\;\;\;x \in \mathit{\Omega ,}\\ - \Delta \psi - \frac{{c + \left( {\alpha c + d} \right){\theta _{{a^ * }}}}}{{\left( {1 + \alpha {\theta _{{a^ * }}}} \right)\left( {1 + {m_4}{\theta _{{a^ * }}}} \right)}}\psi = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,}\\ \phi = \psi = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega ,} \end{array} \right. $

如果ψ≡0, 那么由算子L_a^*可逆知φ≡0, 矛盾, 所以ψ不恒为零.又${\lambda _1}\left( { - \frac{{c + \left( {c\alpha + d} \right){\theta _a}}}{{(1 + {m_4}{\theta _a})(1 + \alpha {\theta _a})}}} \right) = 0$, 故有

$ \psi = {\psi ^ * },\phi = {\phi ^ * } = {L_{{a^ * }}}^{ - 1}\left( { - \frac{{\left( {{a^ * }{m_2} + b + {a^ * }\alpha {m_2}{\theta _{{a^ * }}} - 2{m_2}{\theta _{{a^ * }}} + 2{m_1}b{\theta _{{a^ * }}} - 2\alpha {m_2}{\theta _{{a^ * }}}^2} \right){\theta _{{a^ * }}}}}{{\left( {1 + 2{m_1}{\theta _{{a^ * }}}} \right)\left( {1 + \alpha {\theta _{{a^ * }}}} \right)\left( {1 + {m_4}{\theta _{{a^ * }}}} \right)}}{\psi ^ * }} \right). $

因此, 算子L(a^*; 0, 0) 的核空间N(L(a^*; 0, 0))=span{U₀}, U₀=(φ^*, ψ^*)^T, 其中

$ {\phi ^ * } = {L_{{a^ * }}}^{ - 1}\left( { - \frac{{\left( {{a^ * }{m_2} + b + {a^ * }\alpha {m_2}{\theta _{{a^ * }}} - 2{m_2}{\theta _{{a^ * }}} + 2{m_1}b{\theta _{{a^ * }}} - 2\alpha {m_2}{\theta _{{a^ * }}}^2} \right){\theta _{{a^ * }}}}}{{\left( {1 + 2{m_1}{\theta _{{a^ * }}}} \right)\left( {1 + \alpha {\theta _{{a^ * }}}} \right)\left( {1 + {m_4}{\theta _{{a^ * }}}} \right)}}{\psi ^ * }} \right). $

又令L^*(a^*; 0, 0) 为L(a^*; 0, 0) 的自伴算子, 类似可得

$ N\left( {{L^ * }\left( {{a^ * };0,0} \right)} \right) = {\rm{span}}\left\{ {{U^ * }} \right\},{U^ * } = {\left( {0,{\psi ^ * }} \right)^{\rm{T}}}. $

由Fredholm选择公理知

$ {\rm{Range}}\left( {L\left( {{a^ * };0,0} \right)} \right) = \left\{ {\left( {\phi ,\psi } \right) \in X:\int_\mathit{\Omega } {\psi {\psi ^ * }{\rm{d}}x = 0} } \right\}, $

因此可得dimN(L(a^*; 0, 0))=1, codimR(L(a^*; 0, 0))=1.

令L₁(a^*; 0, 0)=$D_{a\left( {\bar U,V} \right)}^2$ T(a^*; 0, 0), 下面采用反证法证明

$ {L_1}\left( {{a^ * };0,0} \right) \cdot \left( {{\phi ^ * },{\psi ^ * }} \right) \notin R\left( {L\left( {{a^ * };0,0} \right)} \right). $

假设∃(h, k)∈X, 使得L₁(a^*; 0, 0)·(φ^*, ψ^*)=L(a^*; 0, 0)·(h, k).经计算得

$ {L_1}\left( {{a^ * };0,0} \right) \cdot \left( {{\phi ^ * },{\psi ^ * }} \right) = \left( \begin{array}{l} - \left[ {{{\left( {\frac{{\left( {a{m_2} + b + a\alpha {m_2}{\theta _a} - 2{m_2}{\theta _a} + 2{m_1}b{\theta _a} - 2\alpha {m_2}\theta _a^2} \right){\theta _a}}}{{\left( {1 + 2{m_1}{\theta _a}} \right)\left( {1 + \alpha {\theta _a}} \right)\left( {1 + {m_4}{\theta _a}} \right)}}} \right)}_{{\theta _a}}}\frac{{\partial {\theta _a}}}{{\partial a}}{\psi ^ * }} \right]\left| {_{a = {a^ * }}} \right. + \\ \;\;\;\;\;\left[ {{{\left( {\frac{{a - 2{\theta _a}}}{{1 + 2{m_1}{\theta _a}}}} \right)}_{{\theta _a}}}\frac{{\partial {\theta _a}}}{{\partial a}}{\phi ^ * }} \right]\left| {_{a = {a^ * }}} \right.\\ \left[ {\frac{{c{m_4} - d + 2c{m_4}\alpha {\theta _a} + \left( {c\alpha + d} \right)\alpha {m_4}\theta _a^2}}{{{{\left( {1 + {m_4}{\theta _a}} \right)}^2}{{\left( {1 + \alpha {\theta _a}} \right)}^2}}}\frac{{\partial {\theta _a}}}{{\partial a}}} \right]\left| {_{a = {a^ * }}} \right.{\psi ^ * } \end{array} \right). $

那么有

$ - \Delta k - \frac{{c + \left( {\alpha c + d} \right){\theta _{{a^ * }}}}}{{\left( {1 + \alpha {\theta _{{a^ * }}}} \right)\left( {1 + {m_4}{\theta _{{a^ * }}}} \right)}}k = \left[ {\frac{{c{m_4} - d + 2c{m_4}\alpha {\theta _a} + \left( {c\alpha + d} \right)\alpha {m_4}\theta _a^2}}{{{{\left( {1 + {m_4}{\theta _a}} \right)}^2}{{\left( {1 + \alpha {\theta _a}} \right)}^2}}}\frac{{\partial {\theta _a}}}{{\partial a}}} \right]\left| {_{a = {a^ * }}} \right.{\psi ^ * }. $

两边同时乘以ψ^*, 分部积分得

$ \int_\mathit{\Omega } {\left[ {\frac{{c{m_4} - d + 2c{m_4}\alpha {\theta _a} + \left( {c\alpha + d} \right)\alpha {m_4}\theta _a^2}}{{{{\left( {1 + {m_4}{\theta _a}} \right)}^2}{{\left( {1 + \alpha {\theta _a}} \right)}^2}}}\frac{{\partial {\theta _a}}}{{\partial a}}} \right]} \left| {_{a = {a^ * }}} \right.{\psi ^{ * 2}}{\rm{d}}x = 0, $

由于cm₄-d > 0, 且θ_a关于a严格单调递增, 则上式左端大于0, 矛盾.

由Crandall-Rabinowitz局部分歧定理知, 存在充分小的δ > 0及C¹连续曲线(a(s):Φ₁(s), Ψ₁(s)):(-δ, δ)→R×X满足a(0)=a^*, Φ₁(0)=0, Ψ₁(0)=0, Φ₁(s), Ψ₁(s)∈Z使得(a(s):U(s), V(s))=(a(s); s(φ^*+Φ₁(s)), s(ψ^*+Ψ₁(s))) 是T(a(s):U(s), V(s)) 的零点, 其中X=Z⊕N(L(a^*; 0, 0)), 由于U=U-$\overline {{\theta _a}} $, 因此可得到发自(a^*; $\overline {{\theta _{a*}}} $, 0) 的局部分歧正解Γ^*.

同理可得到发自半平凡分支(a_*; 0, $\overline {{\theta _c}} $) 的局部分歧正解.

定理4  设a > λ₁, c+ $\frac{d}{\alpha }$ > λ₁且aβ > b, 则(a_*; 0, $\overline {{\theta _c}} $)∈X×R₊为问题(5) 的分歧点, 且(a_*; 0, $\overline {{\theta _c}} $) 的领域内存在正解

$ {\mathit{\Gamma }_ * } = \left\{ {\left( {a\left( s \right);s\left( {{\phi _ * } + {\mathit{\Phi }_2}\left( s \right)} \right),\overline {{\theta _c}} + s\left( {{\psi ^ * } + {\mathit{\Psi }_2}\left( s \right)} \right)} \right):0 < s < \delta } \right\}. $

a_*由${\lambda _1}\left( { - \frac{{a + \left( {\alpha \beta - b} \right){\theta _c}}}{{(1 + \beta {\theta _c})(1 + {m_2}{\theta _c})}}} \right) = 0$唯一确定, φ_* > 0满足-Δφ_*-${ - \frac{{a + \left( {\alpha \beta - b} \right){\theta _c}}}{{(1 + \beta {\theta _c})(1 + {m_2}{\theta _c})}}}$φ_*=0, x∈Ω, φ_*=0, x∈∂Ω, ∫_Ωφ_*²dx=1, ψ_*∈C₀¹ (Ω), δ > 0充分小.这里(Φ₂(s), Ψ₂(s); a(s)) 是连续函数, 满足a(0)=a_*, Φ₂(0)=0, Ψ₂(0)=0, ∫_Ωψ₂φ_*dx=0, 且

$ {\psi ^ * } = L_c^{ - 1}\left[ {\frac{{\left( {d - c{m_4}} \right){\theta _c} + \left( {2{m_3}d - c{m_4}\beta + 2{m_4}} \right)\theta _c^2 + 2{m_4}\beta \theta _c^3}}{{\left( {1 + \beta {\theta _c}} \right)\left( {1 + {m_2}{\theta _c}} \right)\left( {1 + 2{m_3}{\theta _c}} \right)}}{\phi _ * }} \right]. $