0 引言

目前，通过捕食食饵模型来探究物种的动力学行为已十分普遍，并有诸多成果^[1-4].文献[1]采用扰动理论及分歧理论研究了一类捕食模型正常数平衡态解的分歧与稳定性；文献[2]讨论了一类带Beddington-DeAngelis反应项的捕食-食饵模型在Neumann边界条件下解的性质; 文献[3-4]则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.此后有学者提出带交叉扩散项的捕食模型，其在实际上更能准确的反应捕食者和食饵的关系^[5-8].文献[9]研究了一类Variable-Territory捕食-食饵模型的动力学性质, 但其并未考虑扩散影响.更多关于该模型的生物学意义可参考文献[10-12].文献[13]加入了扩散项, 研究了模型在齐次Dirichlet边界条件下平衡正解的存在性及稳定性.本文考虑一类带交叉扩散项的Variable-Territory捕食-食饵模型，即

$ \left\{ \begin{array}{l} {u_t} - \Delta \left[ {\left( {1 + \alpha v} \right)u} \right] = u\left( {a - u - \frac{{bv}}{{1 + mu}}} \right),\;\;\;\;\;\;\;x \in \mathit{\Omega ,t > }0,\\ {v_t} - \Delta \left[ {\left( {1 + \beta u} \right)v} \right] = v\left( {d - \frac{v}{u} + \frac{{cu}}{{1 + mu}}} \right),\;\;\;\;\;\;\;x \in \mathit{\Omega ,t > }0,\\ u = v = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega ,t > }0\\ u\left( {x,0} \right) = {u_0}\left( x \right) \ge 0,v\left( {x,0} \right) = {v_0}\left( x \right) \ge 0,\;\;\;\;\;\;\;\;x \in \mathit{\Omega }. \end{array} \right. $

(1)

在齐次Dirichlet边界条件下的解的性质.这里, Ω为Rⁿ中具有光滑边界∂Ω上的有界开区域, u, v分别表示食饵和捕食者的种群密度, 这里a, b, c, d, m都是正常数, α, β表示交叉扩散系数.

本文主要讨论系统(1) 所对应的平衡态问题

$ \left\{ \begin{array}{l} - \Delta \left[ {\left( {1 + \alpha v} \right)u} \right] = u\left( {a - u - \frac{{bv}}{{1 + mu}}} \right),\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,t > }0,\\ - \Delta \left[ {\left( {1 + \beta u} \right)v} \right] = v\left( {d - \frac{v}{u} + \frac{{cu}}{{1 + mu}}} \right),\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,t > }0,\\ u = v = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega ,t > }0. \end{array} \right. $

(2)

注 对于问题(2) 的解(u, v), 若在Ω中, (u, v)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∈C₀¹(Ω):u|∂Ω=0}.定义C¹ (Ω)中的范数为通常的Banach空间C₀¹ (Ω)中的范数, 令X=C₀¹ (Ω)×C₀¹ (Ω), 则X是Banach空间.

引理1^[13] 设u, v∈C¹(E), u|_∂Ω=v|_∂Ω=0, u>0, x∈Ω, $\frac{{\partial u}}{{\partial n}}$＜0, x∈∂Ω, 则存在常数Δ>0, 使得v≤δu, x∈E.

由引理1知, 若u, v满足上述条件, 则存在常数Δ₁>0, Δ₂>0, 使得x∈Ω时有${\delta _1} \le \frac{v}{u} \le {\delta _2}$.

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 }. \end{array} \right. $

(3)

引理2^[14] 假设q(x)∈C(E), p为常数, 问题(3) 的所有特征值满足0＜λ₁(p, q)＜λ₂(p, q)≤λ₃(p, q)≤…→∞, 相应的特征函数为φ₁, φ₂, ….由文献[14]可知, λ₁(p, q)是简单的.记λ₁(1, q)为λ₁(q), 则λ₁(q)关于q(x)为单调递增的, 且λ₁(0) 为λ₁.

再考虑边值问题

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

(4)

引理3 若a>λ₁, 则边值问题(4) 存在惟一正解θ_a, 并且∀x∈Ω, 有0＜θ_a＜a, θ_a关于a严格递增.即当a>λ₁时, 则问题(2) 存在半平凡解(θ_a, 0).

引理4^[15] 若a>λ₁, 令L_a=-Δ+2θ_a-a为问题(4) 在θ_a处的线性化算子, 则L_a的特征值均大于0, 即L_a可逆.

现令Z=(U, V), 其中

$ U = \left( {1 + \alpha v} \right)u,V = \left( {1 + \beta u} \right)v. $

(5)

因为在R₊²={u≥0, v≥0}中, 映射(u, v)→(U, V)是连续可逆的, 故(u, v)≥0与(U, V)≥0之间存在一一对应的关系.因此, 引入和问题(2) 等价的半线性椭圆系统

$ \left\{ \begin{array}{l} - \Delta U = u\left( {a - u - \frac{{bv}}{{1 + mu}}} \right),\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,t > }0,\\ - \Delta V = v\left( {d - \frac{v}{u} + \frac{{cu}}{{1 + mu}}} \right),\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,t > }0,\\ U = V = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega ,t > }0. \end{array} \right. $

(6)

则易知, 当a>λ₁时, 问题(6) 就存在半平凡解(θ_a, 0).

2 正解的先验估计

下面给出正解的先验估计和正解存在的必要条件.

定理1 设a>λ₁, 若(U, V)是问题(6) 的任意正解, 那么∀x∈Ω, 有

$ \begin{array}{l} 0 < u\left( x \right) < U\left( x \right) < M = a + \frac{{{a^2}\alpha \left( {1 + ma} \right)}}{b},\\ 0 < v\left( x \right) < V\left( x \right) < M\left( {1 + \beta M} \right)\left( {d + \frac{{cM}}{{1 + mM}}} \right). \end{array} $

证明 当a>λ₁时, 由引理3知θ_a是边值问题(4) 的唯一正解, 又

$ - \Delta u = u\left( {a - u - \frac{{bv}}{{1 + mu}}} \right) < u\left( {a - u} \right), $

可知u为问题(4) 的下解, 那么根据θ_a的唯一性及上下解法知0＜u＜θ_a.

现设∃x₀∈Ω, 使得$U\left( {{x_0}} \right) = \mathop {\max }\limits_{x \in \mathit{\Omega }} U\left( x \right)$.因为

$ 0 \le - \Delta U\left( {{x_0}} \right) = u\left( {{x_0}} \right)\left( {a - u\left( {{x_0}} \right) - \frac{{bv\left( {{x_0}} \right)}}{{1 + mu\left( {{x_0}} \right)}}} \right), $

故有u(x₀)＜a, 且$v\left( {{x_0}} \right) ＜ \frac{{a\left( {1 + mu\left( {{x_0}} \right)} \right)}}{b}$, 那么

$ U\left( x \right) \le U\left( {{x_0}} \right) = \left[ {1 + \alpha v\left( {{x_0}} \right)} \right]u\left( {{x_0}} \right) < a + \frac{{{a^2}\alpha \left( {1 + ma} \right)}}{b} = M. $

同理, 可得$V\left( x \right) ＜ M\left( {1 + \beta M} \right)\left( {d + \frac{{cM}}{{1 + mM}}} \right)$, 那么由(u, v)与(U, V)之间的关系可知定理成立.

定理2 如果问题(6) 存在正解, 那么a>λ₁且$d > {\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right)$.

证明 假设问题(6) 存在正解(U, V), 那么由问题(6) 的第一个方程得

$ - \Delta U = \frac{U}{{1 + \alpha v}}\left( {a - u - \frac{{bv}}{{1 + mu}}} \right) < aU, $

两边同乘以U, 分部积分得

$ \int_\mathit{\Omega } {{{\left| {\nabla U} \right|}^2}{\rm{d}}x} = \left\| {\nabla U} \right\|_2^2 < a\left\| U \right\|_2^2. $

那么由Poincare不等式‖$\nabla $U‖₂²≥λ₁‖U‖₂²得a>λ₁.

设与${\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right)$对应的特征函数为Z(x), 用Z(x)乘以系统(6) 的第2个方程并积分可得

$ - \int_\mathit{\Omega } {Z\Delta V{\rm{d}}x} = \int_\mathit{\Omega } {Z\frac{V}{{1 + \beta u}}\left( {d - \frac{v}{u} + \frac{{cu}}{{1 + mu}}} \right){\rm{d}}x} , $

右端应用格林公式得

$ - \int_\mathit{\Omega } {Z\Delta V{\rm{d}}x} = - \int_\mathit{\Omega } {V\Delta Z{\rm{d}}x} = \int_\mathit{\Omega } {ZV\left[ {\frac{{c{\theta _a}}}{{1 + m{\theta _a}}} + {\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right)} \right]{\rm{d}}x} . $

根据定理1的证明知, 0＜u＜θ_a, 又$\frac{{cu}}{{1 + mu}}$是关于u的递增函数, 即$\frac{{cu}}{{1 + mu}} - \frac{{c{\theta _a}}}{{1 + m{\theta _a}}} ＜ 0$, 所以有

$ \begin{array}{l} \int_\mathit{\Omega } {\left[ {{\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right) - d} \right]ZV{\rm{d}}x} = \\ \int_\mathit{\Omega } {\left[ {\frac{{ - d\beta u}}{{1 + \beta u}} - \frac{v}{{\left( {1 + \beta u} \right)u}} + \frac{{cu}}{{\left( {1 + \beta u} \right)\left( {1 + mu} \right)}} - \frac{{c{\theta _a}}}{{1 + m{\theta _a}}}} \right]ZV{\rm{d}}x} \le \\ \int_\mathit{\Omega } {\left[ { - \frac{{d\beta u}}{{1 + \beta u}} - \frac{v}{{\left( {1 + \beta u} \right)u}} + \frac{{cu}}{{1 + mu}} - \frac{{c{\theta _a}}}{{1 + m{\theta _a}}}} \right]ZV{\rm{d}}x < 0,} \end{array} $

即$d > {\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right)$.可知系统(6) 没有平凡非负解, 当a>λ₁时, 仅有惟一的半平凡解(θ_a, 0).

3 局部分歧正解的存在性

以d为分歧参数, 结合文献[16-17], 利用Crandall-Rabinowitz局部分歧定理, 给出问题(6) 发自半平凡解(θ_a, 0) 的局部分歧正解的存在性.

定理3 设a>λ₁, $d > {\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right)$, 则(d^*; θ_a, 0)∈X×R₊为问题(6) 的一个局部分歧点, 即在(d^*; θ_a, 0) 的领域内系统(6) 至少存在一个正解

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

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

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

当ψ^*=0, x∈∂Ω, ∫_Ωψ^*2dx=1, $\phi $^*∈C₀¹ (Ω), Δ>0充分小.这里(d(s); Φ₁(s), Ψ₁(s))是C¹连续函数, 满足d(0)=d^*, Φ₁(0)=0, Ψ₁(0)=0, ∫_Ωψ₁$\phi $^*dx=0, 且

$ {\phi ^ * } = L_a^{ - 1}\left( { - \frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}{\psi ^ * }} \right). $

证明 令$f\left( {u,v} \right) = u\left( {a - u - \frac{{bv}}{{1 + mu}}} \right),g\left( {u,v} \right) = v\left( {d - \frac{v}{u} + \frac{{cu}}{{1 + mu}}} \right),$

其中u, v均为(U, V)的函数, 将问题(6) 在(U, V)=(θ_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( {{\theta _a},0} \right)}\\ {g\left( {{\theta _a},0} \right)} \end{array}} \right) + \left[ {\left( {\begin{array}{*{20}{c}} {{f_u}}&{{f_v}}\\ {{g_u}}&{{g_v}} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{u_U}}&{{u_V}}\\ {{v_V}}&{{v_U}} \end{array}} \right)} \right]\left| {_{\left( {{\theta _a},0} \right)}} \right. \cdot \left( {\begin{array}{*{20}{c}} {U - {\theta _a}}\\ V \end{array}} \right) +\\ \left( {\begin{array}{*{20}{c}} {{F^1}\left( {d;U - {\theta _a},V} \right)}\\ {{F^2}\left( {d;U - {\theta _a},V} \right)} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right). $

这里, 偏导数为(θ_a, 0) 处的导数值, Fⁱ(U-θ_a, V)满足Fⁱ(0, 0)=F_{(U, V)}ⁱ (0, 0)=0, i=1, 2.对式(5) 两边同时求导并求逆矩阵, 得

$ \left( {\begin{array}{*{20}{c}} {{u_U}}&{{u_V}}\\ {{v_V}}&{{v_U}} \end{array}} \right)\left| {_{\left( {{\theta _a},0} \right)}} \right. = \frac{1}{{1 + \beta {\theta _a}}}\left( {\begin{array}{*{20}{c}} {1 + \beta {\theta _a}}&{ - \alpha {\theta _a}}\\ 0&1 \end{array}} \right). $

即有

$ \begin{array}{l} \left( {\begin{array}{*{20}{c}} {\Delta U + {\theta _a}\left( {a - {\theta _a}} \right)}\\ {\Delta V} \end{array}} \right) + \\ \left( {\begin{array}{*{20}{c}} {a - 2{\theta _a}}&{\frac{{ - \alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}}\\ 0&{\frac{{d\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {U - {\theta _a}}\\ V \end{array}} \right) +\\ \left( {\begin{array}{*{20}{c}} {{F^1}\left( {d;U - {\theta _a},V} \right)}\\ {{F^2}\left( {d;U - {\theta _a},V} \right)} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right). \end{array} $

又令U=U-θ_a, 由引理2知ΔU=ΔU+θ_a(a-θ_a), 则有

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

(7)

显然T(d; 0, 0)=0.记T(d; U, V)关于(U, V)在(d^*; 0, 0) 处的Fréchlet导数是L(d^*; 0, 0) 经计算, L(d^*; 0, 0)·($\phi $, ψ)^T=0等价于

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

由算子L_a是可逆的, 可以推算出ψ不恒为零.又${\lambda _1}\left( { - \frac{{{d^*}(1 + m{\theta _a}) + c{\theta _a}}}{{(1 + m{\theta _a})(1 + \beta {\theta _a})}}} \right) = 0$, 故有

$ \psi = {\psi ^ * },\phi = {\phi ^ * } = L_a^{ - 1}\left( { - \frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}{\psi ^ * }} \right). $

因此, 算子L(d^*; 0, 0) 的核空间W(L(d^*; 0, 0))=span{U₀}, U₀=($\phi $^*, ψ^*)^T, 其中

$ {\phi ^ * } = L_a^{ - 1}\left( { - \frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}{\psi ^ * }} \right). $

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

$ W\left( {{L^ * }\left( {{d^ * };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( {d;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dim W(L(d^*; 0, 0))=1, codim R(L(d^*; 0, 0))=1.

再令L′(d^*; 0, 0)=D_{d(U, V)}² T(d^*; 0, 0), 下面采用反证法证明

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

假设存在(h, k)∈X, 使得L′(d^*; 0, 0)·($\phi $^*, ψ^*)=L(d^*; 0, 0)·(h, k).经计算得

$ L'\left( {{d^ * };0,0} \right) \cdot \left( {{\phi ^ * },{\psi ^ * }} \right) = \left( {\begin{array}{*{20}{c}} 0\\ {\frac{{\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}{\psi ^ * }} \end{array}} \right). $

那么有

$ - \Delta k - \frac{{{d^ * }\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}k = \frac{{\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}{\psi ^ * }. $

两边同时乘以ψ^*, 分部积分得

$ \int_\mathit{\Omega } {\frac{{\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}{\psi ^{ * 2}}{\rm{d}}x} = 0, $

由于上式左端大于0, 故矛盾.

最后, L(d^*; 0, 0) 和L′(d^*; 0, 0) 均是连续的.

因此由文献[16]中的Crandall-Rabinowitz局部分歧定理知, 存在充分小的δ>0及C¹连续曲线(d(s):Φ₁(s), Ψ₁(s)):(-δ, δ)→R×X满足d(0)=d^*, Φ₁(0)=0, Ψ₁(0)=0, 其中X=Z$ \oplus $W(L(d^*; 0, 0)), Φ₁(s), Ψ₁(s)∈Z使得(d(s):U(s), V(s))=(d(s); s($\phi $^*+Φ₁(s)), s(ψ^*+Ψ₁(s)))是T(d(s):U(s), V(s))=0的零点, 由于U=U-θ_a, 因此可得到发自(d^*; θ_a, 0) 的局部分歧正解

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

4 局部分歧正解的延拓

下面利用文献[18]中的方法, 将定理3中的局部分歧延拓为全局分歧.

定理4 在定理3的条件下, 系统(6) 发自(d^*; θ_a, 0) 的局部分歧正解Γ^*可以沿参数d延拓为全局分歧解.

证明 在式(7) 中, 令K为(-Δ)^-1, 则其等价于

$ \left\{ \begin{array}{l} \bar U + K\left[ {\left( {a - 2{\theta _a}} \right)\bar U - \frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}V} \right] + K{F^1}\left( {d;\bar U,V} \right) = 0,\\ V + K\frac{{d\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}V + K{F^2}\left( {d;\bar U,V} \right) = 0. \end{array} \right. $

定义算子T:R₊×X→X为

$ T\left( {d;\bar U,V} \right) = \\ \left[ \begin{array}{l} K\left[ {\left( {a - 2{\theta _a}} \right)\bar U - \frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}V} \right] + K{F^1}\left( {d;\bar U,V} \right)\\ K\frac{{d\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}V + K{F^2}\left( {d;\bar U,V} \right) \end{array} \right], $

则T(d; U, V)为X上的紧可微算子.令G(d; U, V)=(U, V)^T-T(d; U, V), 则G是C¹函数, 且G(d; 0, 0)=0, 易知G(d; U, V)满足≥0, V≥0的零点恰好是问题(6) 的非负解.要证(d; θ_a, 0) 是一个全局分歧点, 首先令T′(d)=D_{(U, V)}T(d; 0, 0).设ξ≥1是T′(d)的一个特征值, 相应的特征函数设为(μ, η), 经计算得(μ, η)满足

$ \left\{ \begin{array}{l} - \xi \Delta \mu - \left( {a - 2{\theta _a}} \right)\mu = - \frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}\eta ,\;\;\;\;\;x \in \mathit{\Omega ,}\\ - \xi \Delta \eta - \frac{{d\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}\eta = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,}\\ \mu = \eta = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }\mathit{.} \end{array} \right. $

如果η≡0, 那么由算子(-ξΔ+2θ_a-a)可逆知μ≡0, 与已知矛盾, 则η不恒为零.令${q_d}\left( x \right) = - \frac{{d(1 + m{\theta _a}) + c{\theta _a}}}{{(1 + m{\theta _a})(1 + \beta {\theta _a})}}$, 则存在i(i=1, 2, …), 使得λ_i(ξ, q_d)=0.

对任意i, λ_i(ξ, q_d)关于ξ≥1, $d > {\lambda _1}\left( {\frac{{ - c{\theta _a}}}{{1 + m{\theta _a}}}} \right)$均严格单调递增, 且λ_i(ξ, q_d)可列为λ₁(ξ, q_d)＜λ₂(ξ, q_d)≤λ₃(ξ, q_d)≤…→∞.特别地, λ₁(1, q_d)=0.此外, 若存在i有λ_i(ξ, q_d)=0, 则ξ≥1为T′(d)的特征值.因此ξ≥1为T′(d)的特征值的充分必要条件是存在某个i(i=1, 2, …), 使得λ_i(ξ, q_d)=0.

令d>d^*, ∀ξ≥1, i≥2, 有λ_i(ξ, q_d)>λ₁(ξ, q_d^*)>λ₁(1, q_d^*)=0.因此, T′(d)没有大于或等于1的特征值.此时i((T(d); ·), 0)=1.

设存在充分小的ε>0, 使得d^*-ε＜d＜d^*, λ₂(ξ, q_d^*-ε)≥λ₁(ξ, q_d^*), 则∀ε≥1, i≥2，有λ_i(ξ, q_d)≥λ₂(ξ, q_d)>λ₂(ξ, q_d^*-ε)≥λ₁(ξ, q_d^*)≥λ₁(1, q_d^*)=0.又由于λ₁(1, q_d)＜λ₁(1, q_d^*)=0, $\mathop {\lim }\limits_{\xi \to \infty } {\lambda _1}\left( {\xi ,{q_d}} \right) = + \infty $, 且λ₁(ξ, q_d)关于ξ单调递增, 故存在唯一的ξ₁>1, 有λ₁(ξ₁, q_d)=0, 从而得到

$ N\left( {{\xi _1}I - T'\left( d \right)} \right) = {\rm{span}}\left\{ {{{\left( {\bar \mu ,\bar \eta } \right)}^{\rm{T}}}} \right\},\dim N\left( {{\xi _1}I - T'\left( d \right)} \right) = 1. $

其中

$ \bar \mu = {\left( {{\xi _1}\Delta + \left( {a - 2{\theta _a}} \right)} \right)^{ - 1}}\left[ {\frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}\bar \eta } \right], $

且η>0满足

$ \left\{ \begin{array}{l} - {\xi _1}\Delta \bar \eta + {q_d}\left( x \right)\bar \eta = 0,\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,}\\ \bar \eta = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }\mathit{.} \end{array} \right. $

接下来证明ξ₁的代数重数是1, 即只要证明R(ξ₁I-T′(ρ))∩M(ξ₁I-T′(ρ))=0成立即可.假设∃(μ, η)∈X, 使得(ξ₁I-T′(ρ))·(μ, η)=(μ, η)^T, 即

$ \left( {\begin{array}{*{20}{c}} {{\xi _1} - {\Delta ^{ - 1}}\left( {a - 2{\theta _a}} \right)}&{{\Delta ^{ - 1}}\frac{{\alpha {\theta _a}\left( {a - 2{\theta _a}} \right)\left( {1 + m{\theta _a}} \right) + b{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}}\\ 0&{{\xi _1} - {\Delta ^{ - 1}}\frac{{d\left( {1 + m{\theta _a}} \right) + c{\theta _a}}}{{\left( {1 + m{\theta _a}} \right)\left( {1 + \beta {\theta _a}} \right)}}} \end{array}} \right) \cdot \left( {\mu ,\eta } \right) = \left( {\begin{array}{*{20}{c}} {\bar \mu }\\ {\bar \eta } \end{array}} \right), $

那么

$ \left\{ \begin{array}{l} \Delta \bar \eta = {\xi _1}\Delta \eta - {q_d}\left( x \right)\eta \;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,}\\ \eta = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \mathit{\Omega }\mathit{.} \end{array} \right. $

两边同乘以η, 积分得

$ \int_\mathit{\Omega } {\bar \eta \Delta \bar \eta {\rm{d}}x} = \int_\mathit{\Omega } {\left( {{\xi _1}\Delta \eta - {q_d}\left( x \right)\eta } \right)\bar \eta {\rm{d}}x} = \int_\mathit{\Omega } {\left( {{\xi _1}\Delta \bar \eta - {q_d}\left( x \right)\bar \eta } \right)\eta {\rm{d}}x} = 0. $

又因为$\Delta \bar \eta = \frac{1}{{{\xi _1}}}{q_d}\left( x \right)\bar \eta $, 所以$\frac{1}{{{\xi _1}}}$∫_Ωq_d(x)η²dx=0, 与q_d(x)＜0矛盾, 所以ξ₁的代数重数是1.因此当d^*-ε＜d＜d^*时, i((T(d); ·), 0)=-1.

由全局分歧定理知, 定理3中得到的局部分歧正解Γ^*可以延拓为全局分歧, 令E为Γ^*沿d方向的连通分支, 则E为问题(6) 由(d^*; θ_a, 0) 出发的解曲线.令P=P₁×P₂, 其中

$ {P_1} = \left\{ {u \in C_0^1\left( \mathit{\Omega } \right);u > 0,u \in \mathit{\Omega };\frac{{\partial u}}{{\partial n}} < 0,x \in \partial \mathit{\Omega }} \right\}. $

易得在(d^*; θ_a, 0) 的小邻域内, E⊂P且E也包含定理3中给出的局部分歧解, 则E-(d^*; θ_a, 0) 包含系统(6) 发自(d^*; θ_a, 0) 的正解分支, 且必满足下列条件之一:

(A₁)E从(d^*; θ_a, 0) 连接另一个分歧点(${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over d} }$; θ_a, 0), 其中${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over d} }$≠d^*;

(A₂)E在R₊×P内由(d^*; θ_a, 0) 沿参数d延伸到∞.

注 由于$\frac{v}{u}$在R上的光滑性未知, 因此无法使用文献[16-17]中的判断分歧解分支走向的方法来证明(A₁)和(A₂)中的哪一个成立.故本定理仅给出定理3中的分歧解可以延拓成全局分歧的结论.