本文研究一个众所周知的反应扩散系统,即 Gierer-Meinhardt 模型. 该系统是由Gierer 和Meinhardt在研究图灵模式的时候提出来的$^{^[1]}$. 已经有很多作者研究了这个系统,获得了很多有意义的结果,见文献^[7,8,9,10]. 在文献^[7,8]中,陈,史和魏研究了带有基因表达时滞和活化剂 生产饱和的Gierer-Meinhardt 模型的平衡点全局吸引性,并完成了该情况下的Hopf分支分析. 在文献^[9]里,徐和魏研究了一维 Schnakenberg反应扩散模型的Hopf分支. 在文献^[10]里, 刘等研究了反应扩散Schnakenberg模型的分支问题. 本文主要研究扩散的Gierer Meinhardt模型的Hopf分支,我们主要研究如下系统 $$\begin{equation}λbel{0} \left\{\begin{array}{ll} U_s = D_1\Delta U + \rho_U\frac{U^2V}{1+K^2U^2}-\mu_UU,&\;\; x\inΩega,\; s>0,\\ [3mm] V_s = D_2\Delta V -\rho_V\frac{U^2V}{1+K^2U^2}+\sigma_V,& \;\; x\inΩega,\; s>0,\\[2mm] U(x,0)=U_0(x)\geq0,\;\; V(x,0)=V_0(x)\geq0,&\;\; x\in Ωega,\\ \partial_\nu U=\partial_\nu V=0,&\;\; x\in\partialΩega,\; s>0, \end{array}\right.(1.1) \end{equation}$$ 这里,$Ωega$ 在${\Bbb R}^n$ ($n\geq 1$) 中是有光滑边界$\partialΩega$的一个有界连通集 (反应器); 假定反应器是闭的,且符合自反的Neumann边界条件 (这里,$\partial_{\nu}U$是$u$的外法向导数). $U(x,s)$ 和 $V(x,s)$ 表示$s>0$,$x\inΩega$时的成形基因浓度. 参数 $\rho_U$,$\rho_V$,$\mu_U$,$\sigma_V$,和 $K$ 都是非负常数.也就是说,假设按着$U^2V$ 的形式,$U$ 和 $V$ 有有效的交流作用,当$U$比较大的时候,这个交流作用就会饱和, 且饱和常数$K$不等于零. 假定U以速率$\rho_U$增加促使V以速率 $\rho_V$被消耗. $U$以与$U$成正比的速度$\mu_U$移除. 对变量$V$, 存在一个常数生产期$\sigma_V$. 对于这个系统的更多更详细的论述可参见 文献 ^[2] 及它的参考文献.

我们引入如下新的变量 $$u=KU,~~v=\frac{\mu_UV}{\sigma_V},~~u_0=KU_0,~~v_0=\frac{\mu_UV_0}{\sigma_V},$$ $$t=\mu_Us,~~a=\frac{\rho_U\sigma_V}{K\mu_U^2},~~ b=\frac{\rho_V}{K^2\mu_U},~~d_i=\frac{D_i}{\mu_U},$$ 系统 (1.1) 写成如下系统 $$\begin{equation}λbel{1} \left\{\begin{array}{ll} u_t-d_1\Delta u=\frac{au^2v}{1+u^2}-u,\;\; & x\in Ωega,\; t>0,\\ [3mm] v_t-d_2\Delta v=-\frac{bu^2v}{1+u^2}+1,\;\; & x\in Ωega,\; t>0,\\ [2mm] \partial_\nu u=\partial_\nu v=0,\;\; & x\in ,\partialΩ; t>0,\\ u(x,0)=u_0(x)\ge 0,\; v(x,0)=v_0(x)\ge 0,~~& x\in Ωega, \end{array}\right.(1.2) \end{equation}$$

本文的目的是利用半线性偏微分方程组标准的Hpof分支定理,考虑空间齐次和非齐次周期轨道对唯一正平衡点的分支的存在性. 结果表明扩散Gierer-Meinhardt模型有丰富的振动模式. 过去几年的很多研究重点在于讨论不同类型Gierer-Meinhardt模型Hpof分支, 有兴趣的读者可以阅读文献^[3,4,5] 了解更多的细节.

本文由如下几个部分构成. 在第二部分,我们给出一般反应扩散 方程组的几个Hpof定理. 在第三部分,我们对反应扩散系统的Hpof分 支做了分析,并给出计算机模拟支持我们的结论.

本部分对反应扩散Gierer-Meinhardt模型 (1.1) 进行Hpof分支分析.

为了方便起见,我们复制系统 (1.2) 为如下方程 $$\begin{equation}λbel{2} \left\{\begin{array}{ll} u_t-d_1\Delta u=\frac{au^2v}{1+u^2}-u,\;\; & x\in ,\; t>0,\\ [3mm] v_t-d_2\Delta v=-\frac{bu^2v}{1+u^2}+1,\;\; & x\in ,\; t>0,\\ [2mm] \partial_\nu u=\partial_\nu v=0,\;\; & x\in ,\; t>0,\\ u(x,0)=u_0(x)\ge 0,\; v(x,0)=v_0(x)\ge 0,~~ & x\in ega, \end{array}\right.(2.1) \end{equation}$$

显然,系统 (2.1) 有唯一正平衡解 $$(u_,v_):=(,(^2+1)/a),$$ 给定 $a$,$:=a/b$ 相应的空间定义域为 $Ωega=(0,\ell\pi)$,$\ell\in{\Bbb R}^+$. 系统(2.1)在平衡状态$(u_,v_)$ 的线性算子为 $$\begin{equation} J():= \left(\begin{array}{cc} d_{1} \frac{\partial^{2}}{\partial x^{2}}+\frac{1-^2}{^2+1}, & ~~\frac{a^2}{^2+1} \\[3mm] -\frac{2}{(^2+1)} ,&~~ d_{2}\frac{\partial^{2}}{\partial x^{2}}-\frac{a}{^2+1} \end{array}\right),(2.2) \end{equation}$$ 其中 $b:=a/λ$. 在以下的讨论中,用 $λ$ 作为分支参数,并假设 $a>0$.

根据文献 ^[6] 可知,$J(λ)$ 的特征值由如下算子 $J_{n}(λ)$给出 $$\begin{equation} J_{n}(λ)=: \left(\begin{array}{cc} -\frac{d_1n^2}{\ell^2}+\frac{1-λ^2}{λ^2+1}, & ~~\frac{aλ^2}{λ^2+1} \\[3mm] -\frac{2}{λ(λ^2+1)} ,&~~ -\frac{d_2n^2}{\ell^2}-\frac{aλ}{λ^2+1} \end{array}\right),(2.3) \end{equation}$$ 其特征方程为 $$\begin{equation} \beta^{2}- \beta T_{n}(λ)+D_{n}(λ)=0~(n=0,1,2\cdots),(2.4) \end{equation}$$ 其中, $$\begin{equation}λbel{3} \left\{\begin{array}{ll} T_{n}(λ):={\rm tr} J_{n} = \frac{1-aλ-λ^2}{λ^2+1}-\frac{(d_1+d_2)n^2}{\ell^2} ,\\ [3mm] D_{n}(λ):= {\rm det} J_{n} =\frac{d_1d_2n^4}{\ell^4}+\left(\frac{ad_1λ}{λ^2+1}-\frac{d_2(1-λ^2)}{λ^2+1}\right)\frac{n^2}{\ell^2}+\frac{aλ}{λ^2+1}. \end{array}\right.(2.5) \end{equation}$$

下面构建Hpof分支发生的潜在临界点元素.

存在 $n \in {\Bbb N}\cup\{0\}$,使得 $$T_n(λ) = 0,\;\; D_n(λ) > 0;\;\; T_j(λ) \neq 0,\;\; D_j(λ) \neq 0,~~ j \neq n;$$

设靠近虚轴的唯一一对复特征根为 $α(λ) \pm {\rm i} Ωega(λ)$,它穿过虚轴的横截条件为 $$\begin{equation} α'(λ) \neq 0.(2.63) \end{equation}$$

由式 (2.5),$T_{n}(λ)<0$ 和条件 $D_{n}(λ)>0$ ,$λ\geq1$, 其中后面这个条件意味着唯一正平衡解 $(λ,(λ^2+1)/aλ)$ 是局部渐近稳定的. 因此,任何一个潜在的分支点$λ_{0}$一定位于区间 $(0,1)$. 对于区间 $(0,1)$上的任何一个分支点 $λ_{0}$, $αpha(λ)\pm {\rm i}Ωega(λ)$ 一定是 $L_{n}(λ)$ 的特征值,于是 $$\begin{equation} αpha(λ)=\frac{A(λ)}{2}-\frac{(d_{1}+d_{2})n^2}{2\ell^2},~~~~~~ Ωega(λ)=\sqrt{D_{n}(λ)-αpha^2(λ)},(2.7) \end{equation}$$ 其中,$A(λ)=\frac{1-aλ-λ^2}{1+λ^2}$. 计算表明, 对于所有的$0<λ<1$,下列等式总是成立的 $$\begin{equation} αpha'(λ)=\frac{A'(λ)}{2}=\frac{aλ^2-4λ-a}{2(1+λ^2)^2} <0.(2.8) \end{equation}$$

因此,横截条件总是成立的.

对于任意的 $\ell>0$,$λ_0^H:=(-a+\sqrt{a^2+4})/2\in(0,1)$ 是Hpof分支点. 对应于空间齐次周期解的Hpof分支,这和已经研究过的常微分方程组的结论是一 样的. 显然, 对于任意的 $\ell>0$,空间齐次周期解的Hpof分支唯一$λ$值记为$λ_0^H$.

下面,当 $n\geq1$时,寻找空间非齐次Hpof分支. 注意到 $$A(0)=1, \quad A(λ_{0}^{H})=0, \quad A'(λ)=\frac{aλ^2-4λ-a}{(1+λ^2)^2}<0,~~~λ\in(0,λ_{0}^{H}).(2.9)$$ 定义 $$\begin{equation}λbel{4}\ell_{n}=n\sqrt{d_{1}+d_{2}},~~n\inΝ.\end{equation}$$ 则对$\ell_{n}<\ell<\ell_{n+1}$,及$1\leq j\leq n$,定义 $λ_{j}^{H}$ 为方程 $$A(λ)=\frac{(d_{1}+d_{2})j^2}{\ell^2}.$$ 这些$n$ 值满足 $$0<λ_{n}^{H}<\cdots<λ_{2}^{H}<λ_{1}^{H}<λ_{0}^{H}.$$

显然,对$i\neq j$,$T_{j}(λ_{j}^{H})=0$ ,且 $T_{i}(λ_{j}^{H})\neq0$. 现在,我们仅需要证明对所有的 $i\in {\Bbb N}_0$, $D_i(λ_{j,\pm}^H)\ne 0$,且 $D_j(λ_{j,\pm}^H)>0$.

下面,推导参数满足的使得 $D_i(λ_{j,\pm}^H)>0$的条件. 假设 $d_1\geq d_2$,有 $$\begin{eqnarray}λbel{5} D_i(λ_j^H) &=&\frac{d_1d_2i^4}{\ell^4}+\left(\frac{ad_1λ_j^H}{(λ_j^H)^2+1}-\frac{d_2(1-(λ_j^H)^2)}{(λ_j^H)^2+1}\right)\frac{i^2}{\ell^2}+\frac{aλ_j^H}{(λ_j^H)^2+1} \nonumber\\ &\geq&\frac{d_1d_2i^4}{\ell^4}+d_2\left(\frac{aλ_j^H}{(λ_j^H)^2+1}-\frac{(1-(λ_j^H)^2)}{(λ_j^H)^2+1}\right)\frac{i^2}{\ell^2}+\frac{aλ_j^H}{(λ_j^H)^2+1} \nonumber\\ &=&\frac{d_1d_2i^4}{\ell^4}+\frac{aλ_j^H}{(λ_j^H)^2+1} \nonumber\\ &>&0.(2.10) \end{eqnarray}$$

综上,得到本文的主要结果:

定理2.1假定常数 $d_{1},d_{2},a>0$ 满足 $d_{1}\geq d_{2}$,及$\ell_{n}$ 由系统 (2.9) 定义. 则对于任意的 $\ell_{n}<\ell<\ell_{n+1}$,存在 $n$ 个点$λ_{j}^{H}(\ell),~~1\leq j\leq n$,满足 $$0<λ_{n}^{H}<\cdots<λ_{2}^{H}<λ_{1}^{H}<λ_{0}^{H}.$$ 使得系统 (1.1) 在 $λ=λ_{j}^H$ 或 $λ=λ_{0}^H$ 具有Hpof分支. 而且,

(1)~ 在 $λ=λ_0^H$ 产生的Hpof周期解是空间齐次的, 这个结论和相应的常微分方程系统的周期解结论一致;

(2)~ 在 $λ=λ_{j}^H$ 产生的分支周期解是空间非齐次的.

例2.2 设 $Ωega=(0,\ell\pi)$,$d_1=4$,$d_2=1$,$a=2.25$,通过计算可知,$λ_0^H=0.3802$,$\ell_n:=n\sqrt{(d_1+d_2)}=\sqrt{5}n\approx2.2361n$, $D_i(λ_{j}^H)>0$.

(1)~ 设 $\ell=3$,则 $\ell\in(\ell_1,\ell_{2}]\approx (2.236,4.470]$. 解方程 $A(λ)=(d_1+d_2)n^2/{\ell^2}$,可知 $λ_{1}^H=0.1760$. 于是Hpof分支点集为 $$λmbda_1=\{λ_{1}^H,λ_0^H\}=\{0.1760,\,0.3802\}.$$

(2)~ 设 $\ell=5$,则 $\ell\in(\ell_2,\ell_{3}]\approx (4.470,6.7082]$. 同样, 可得 $λ_{1}^H\approx0.3057,\;λ_{2}^H=0.0833$. 则Hpof分支点集为 $$λmbda_1=\{λ_{1}^H,λ_{2}^H λ_0^H\}=\{0.0833,0.3057,\; 0.3802\}.$$

定理3.1 当$0<a<2/\sqrt{3}$时,系统 (1.2) 在点$λ=λ_0^H$的分支 是超临界的且空间齐次的分支周期解是不稳定的; 当 $a>2/\sqrt{3}$ 时, 在点$λ=λ_0^H$的Hpof分支是次临界的,空间齐次的分支周期解是稳定的.

证 根据文献[6,定理2.1],为了确定分支周期解的稳定性和分支方向, 需要计算 ${\rm Re}(c_1(λ_0^H))$, 其中$c_1(λ_0^H)$是按文献^[6]定义的. 当 $λ=λ_0^H$, 为方便起见,记 $λ_0^H=λ_0$. 于是 $$\begin{equation}q:=\left(\begin{array}{c} a_0\\ b_0\\ \end{array}\right)=\left(\begin{array}{c} 1\\[2mm] \frac{1}{aλ_0^2}[λ_0^2-1+{\rm i}Ωega_0(1+λ_0^2)] \end{array}\right), \end{equation}$$ (3.1) $$\begin{equation} q^*:=\left(\begin{array}{c} a^*_0\\ b^*_0\\ \end{array}\right)=\left(\begin{array}{c} \frac{1+{\rm i}Ωega_0}{2\ell\pi}\\[3mm] \frac{{\rm i}Ωega_0λ_0}{2\ell\pi} \end{array}\right),(3.2) \end{equation}$$ 其中,$Ωega_0=\sqrt{\frac{aλ_0}{(λ_0)^2+1}}$.

设 $\hat{u}=u-u_{λ}$,$\hat{v}=v-v_{λ}$,并仍记 $\hat{u}$ 和$\hat{v}$ 为 $u$ 和 $v$,系统 (1.2) 变为 $$\begin{equation}λbel{9} \left\{\begin{array}{ll} u_t-d_1u_{xx}=\frac{a(u+λ)^2(v+(λ^2+1)/aλ)}{1+(u+λ)^2}-(u+λ),\;\; & x\in Ω,\; t>0,\\ [3mm] v_t-d_2v_{xx}=-\frac{b(u+λ)^2(v+(λ^2+1)/aλ)}{1+(u+λ)^2}+1,\;\; & x\in Ω,\; t>0,\\ [2mm] \partial_\nu u=\partial_\nu v=0,\;\; & x\in Ω,\; t>0,\\ u(x,0)=u_0(x)\ge 0,\; v(x,0)=v_0(x)\ge 0,& x\in Ωega. \end{array}\right.(3.3) \end{equation}$$

设 $$\begin{equation} \begin{array}{l} f(λ,u,v)=\frac{a(u+λ)^2(v+(λ^2+1)/aλ)}{1+(u+λ)^2}-(u+λ),\\ [3mm] g(λ,u,v)=-\frac{b(u+λ)^2(v+(λ^2+1)/aλ)}{1+(u+λ)^2}+1. \end{array}(3.4) \end{equation}$$

在点$(λ_0,0,0)$,函数$f$和$g$的所有偏导数为 $$f''_{uu}(λ_0,0,0)=\frac{2-6λ_0^2}{λ_0(1+λ_0^2)^2},\;\;\;\;f''_{uv}(λ_0,0,0)=\frac{2aλ_0}{(1+λ_0^2)^2};$$ $$f'''_{uuu}(λ_0,0,0)=\frac{24(λ_0^2-1)}{(1+λ_0^2)^3},\;\;\;\;f'''_{uuv}(λ_0,0,0)=\frac{2a(1-3λ_0^2)}{(1+λ_0^2)^3};$$ $$g''_{uu}(λ_0,0,0)=\frac{-2+6λ_0^2}{λ_0^2(1+λ_0^2)^2},\;\;\;\;g''_{uv}(λ_0,0,0)=\frac{-2a}{(1+λ_0^2)^2};$$ $$g'''_{uuu}(λ_0,0,0)=\frac{24(1-λ_0^2)}{λ_0(1+λ_0^2)^3},\;\;\;\;g'''_{uuv}(λ_0,0,0)=\frac{-2a(1-3λ_0^2)}{λ_0(1+λ_0^2)^3};$$ $$f''_{vv}(λ_0,0,0)=f'''_{uvv}(λ_0,0,0)=f'''_{vvv}(λ_0,0,0)=0,$$ $$g''_{vv}(λ_0,0,0)=g'''_{uvv}(λ_0,0,0)=g'''_{vvv}(λ_0,0,0)=0.$$ 于是 $$c_0=\frac{-2+4{\rm i}Ωega_0}{λ_0(1+λ_0^2)}=-λ_0d_0,\;\;\;\; e_0=\frac{-2}{λ_0(1+λ_0^2)}=-λ_0f_0,$$ $$g_0=\frac{1}{(1+λ_0^2)^3}\bigg[24(λ_0^2-1) +\frac{2}{λ_0^2}(1-3λ_0^2)[{\rm i}Ωega_0(1+λ_0^2) -3(1-λ_0^2)]\bigg]=-λ_0h_0.$$ 且 $$Q_{qq}=\left(\begin{array}{c} c_0\\ d_0\\ \end{array}\right)=\frac{2-4{\rm i}Ωega_0}{λ_0^2(1+λ_0^2)}\left(\begin{array}{c} -λ_0\\ 1\\ \end{array}\right),\;\;\;\;Q_{q\bar{q}}=\left(\begin{array}{c} e_0\\ f_0\\ \end{array}\right)=\frac{2}{λ_0^2(1+λ_0^2)}\left(\begin{array}{c} -λ_0\\ 1 \end{array}\right),$$ $$C_{qq\bar{q}}=\left(\begin{array}{c} g_0\\ h_0\\ \end{array}\right)=\frac{-1}{λ_0(1+λ_0^2)^3} \bigg[24(λ_0^2-1)+\frac{2}{λ_0^2}(1-3λ_0^2) [{\rm i}Ωega_0(1+λ_0^2)-3(1-λ_0^2)]\bigg]\left(\begin{array}{c} -λ_0\\ 1 \end{array}\right),$$ $$\begin{equation} \begin{array}{l} λngle q^*,Q_{qq}\rangle=\frac{-1+2{\rm i}Ωega_0}{λ_0(1+λ_0^2)},\;\; λngle q^*,Q_{q\bar{q}}\rangle=\frac{-1}{λ_0(1+λ_0^2)},\;\; λngle \bar q^*,Q_{qq}\rangle=\frac{-1+2{\rm i}Ωega_0}{λ_0(1+λ_0^2)},\\[3mm] λngle q^*,C_{qq\bar{q}}\rangle=\frac{1}{(1+λ_0^2)^3} \bigg[12(λ_0^2-1)+\frac{1}{λ_0^2}(1-3λ_0^2) [{\rm i}Ωega_0(1+λ_0^2)-3(1-λ_0^2)]\bigg]. \end{array}(3.5) \end{equation}$$

因此,可以直接计算得出如下结果 $$\begin{equation} \begin{array}{l} H_{20}=\left(\begin{array}{c} c_0\\ d_0\\ \end{array}\right)-λngle q^*,Q_{qq}\rangle\left(\begin{array}{c} a_0\\ b_0\\ \end{array}\right)-λngle \overline{q}^*,Q_{qq}\rangle\left(\begin{array}{c} \overline{a_0}\\ \overline{b_0}\\ \end{array}\right)=0,\\ [6mm] H_{11}=\left(\begin{array}{c} e_0\\ f_0\\ \end{array}\right)-λngle q^*,Q_{q\overline{q}}\rangle\left(\begin{array}{c} a_0\\ b_0\\ \end{array}\right)-λngle \overline{q}^*,Q_{q\overline{q}}\rangle\left(\begin{array}{c} \overline{a_0}\\ \overline{b_0}\\ \end{array}\right)=0, \end{array}(3.6) \end{equation}$$ 其中 $w_{20}=w_{11}=0$. 于是 $$\begin{equation} λngle q^*,Q_{w_{11},q}\rangle=λngle q^*,Q_{w_{20},\overline{q}}\rangle=0.(3.7) \end{equation}$$ 从而, $$\begin{equation} {\rm Re}(c_1(λ_0^H)) ={\rm Re}\left\{\frac{\rm i}{2Ωega_0}λngle q^*,Q_{qq}\rangle\cdotλngle q^*,Q_{q\overline{q}}\rangle+\frac{1}{2}λngle q^*,C_{q,q,\overline{q}}\rangle\right\} =\frac{2λ_0-a}{2λ_0(1+λ_0^2)^2}.(3.8) \end{equation}$$

注意到 $$λ_0:=λ_0^H=(-a+\sqrt{a^2+4})/2,$$ 容易得,若 $0<a<2/\sqrt{3}$,${\rm Re}(c_1(λ_0^H))>0$, 若 $a>2/\sqrt{3}$,${\rm Re}(c_1(λ_0^H))<0$. 由文献[6,定理2.1],即可知道结论成立, 证明完毕.

例3.2 令$d_1=4$,$d_2=1$,$a=2.25>\sqrt{3}/2$. 此时,$λ_0^H=0.3802$, 且 $${\rm Re}(c_1(λ_0^H))\approx-1.4954<0,$$ 则相应的分支周期解是稳定的. 由于 $$αpha'(λ_0^H)=A'(λ_0^H)/2=-1.3151<0,$$ 我们有 $$\frac{1}{αpha'(λ_0^H)}{\rm Re}(c_1(λ_0^H))>0.$$ 这样在点$λ=0.3802$处的分支方向是次临界的,该结果见图 1.

图 1

Fig. 1

图(a) $u(x,t)$; (b) $v(x,t)$. 系统 {\rm(1.1)}在点 $(u_λ,v_λ)\approx(0.3802,1.3379)$ 经历了一个空间齐次的Hpof分支周期轨道. 初始数据是 $(u_0,v_0)=(0.40,1.35)$}

定理3.3 对于系统 (1.2),如果条件$\frac{1}{αpha'(λ_j^H) }{\rm Re}c_1(λ_j^H)<0$ (或$>0$),成立,则位于点$λ=λ_j^H,$ $j\neq0$ 的Hpof分支是超临界的(或者次临界的),并且空间非齐次分支周期 解是不稳定的.

证 当 $λ=λ_j^H$ $(j\in {\Bbb N})$时,为方便起见,记 $λ_j^H=λ_j$. 由此我们得到 $$q:=\cos\frac{j}{\ell}x(a_{j},b_{j})^T =\cos\frac{j}{\ell}x\left(1, \frac{1+λ_j^2}{aλ_j^2}({\rm i}Ωega_0-\frac{d_2j^2}{\ell^2}-\frac{aλ_j}{1+λ_j^2})\right)^T,$$ $$\begin{eqnarray*} q^*&:=&\cos\frac{j}{\ell}x\left( a^*_{j},b^*_{j}\right)^T\\ &=&\cos\frac{j}{\ell}x \left( -\frac{\rm i}{\ell\piΩega_0}({\rm i}Ωega_0+\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}), \frac{λ_j(1+λ_j^2){\rm i}}{2\ell\piΩega_0} \bigg({\rm i}Ωega_0+\frac{d_2j^2}{\ell^2}+\frac{aλ_j} {1+λ_j^2}\bigg)^2\right)^T, \end{eqnarray*}$$ 这里 $$Ωega_0=\left(\frac{2aλ_j}{(1+λ_j^2)^2}- \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)^2\right)^{1/2}.$$

经过计算,得 $$\left[2{\rm i}Ωega_0I-L_{2j}(λ_{j})\right]^{-1}=(αpha_1+αpha_2{\rm i})^{-1} \left(\begin{array}{cc} 2{\rm i}Ωega_0+\frac{4d_2{j}^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}&\frac{aλ_j^2}{1+λ_j^2}\\ [3mm] -\frac{2}{λ_j(1+λ_j^2)} & 2{\rm i}Ωega_0+\frac{4d_1{j}^2}{\ell^2}-\frac{1-λ_j^2}{1+λ_j^2} \end{array}\right),$$ 其中 $$αpha_1:=-4Ωega_0^2+ \bigg(\frac{4d_1j^2}{\ell^2}-\frac{1-λ_j^2}{1+λ_j^2}\bigg) \bigg(\frac{4d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)+\frac{2aλ_j}{(1+λ_j^2)^2},αpha_2=\frac{6Ωega_0(d_1+d_2){j}^2}{\ell^2};$$ 且 $$\left[2{\rm i}Ωega_0I-L_{0}(λ_{j})\right]^{-1}=(αpha_3+αpha_4{\rm i})^{-1} \left(\begin{array}{cc} 2{\rm i}Ωega_0+\frac{aλ_j}{1+λ_j^2} ~~&\frac{aλ_j^2}{1+λ_j^2}\\ [3mm] -\frac{2}{λ_j(1+λ_j^2)} ~~& 2{\rm i}Ωega_0-\frac{1-λ_j^2}{1+λ_j^2} \end{array}\right),$$ 其中 $$αpha_3:=-4Ωega_0^2+\frac{aλ_j}{1+λ_j^2},\;\;αpha_4:= -\frac{2Ωega_0(d_1+d_2){j}^2}{\ell^2}.$$ $$\left[L_{2j}(λ_{j})\right]^{-1}=(αpha_5)^{-1} \left(\begin{array}{cc} -\frac{4d_2j^2}{\ell^2}-\frac{aλ_j}{1+λ_j^2} ~~&-\frac{aλ_j^2}{1+λ_j^2}\\ [3mm] \frac{2}{λ_j(1+λ_j^2)} ~~& -\frac{4d_1j^2}{\ell^2}+\frac{1-λ_j^2}{1+λ_j^2} \end{array}\right),$$ 且 $$\left[L_{0}(λ_{j})\right]^{-1}=(αpha_6)^{-1} \left(\begin{array}{cc} -\frac{aλ_j}{1+λ_j^2} ~~&-\frac{aλ_j^2}{1+λ_j^2}\\[3mm] \frac{2}{λ_j(1+λ_j^2)} ~~& \frac{1-λ_j^2}{1+λ_j^2} \end{array}\right),$$ 其中 $$αpha_5:=\bigg(\frac{4d_1j^2}{\ell^2}-\frac{1-λ_j^2} {1+λ_j^2}\bigg)\bigg (\frac{4d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)+\frac{2aλ_j}{(1+λ_j^2)^2},αpha_6:=\frac{aλ_j}{(1+λ_j^2)}.$$

通过计算,可得函数$f$ 和 $g$ 在点$(λ_j,0,0)$的所有偏导数 $$f''_{uu}=\frac{2-6λ_j^2}{λ_j(1+λ_j^2)^2},\;\;\;\;f''_{uv}=\frac{2aλ_j}{(1+λ_j^2)^2};$$ $$f'''_{uuu}=\frac{24(λ_j^2-1)}{(1+λ_j^2)^3},\;\;\;\;f'''_{uuv}=\frac{2a(1-3λ_j^2)}{(1+λ_j^2)^3};$$ $$g''_{uu}=\frac{-2+6λ_j^2}{λ_j^2(1+λ_j^2)^2},\;\;\;\;g''_{uv}=\frac{-2a}{(1+λ_j^2)^2};$$ $$g'''_{uuu}=\frac{24(1-λ_j^2)}{λ_j(1+λ_j^2)^3},\;\;\;\;g'''_{uuv}=\frac{-2a(1-3λ_j^2)}{λ_j(1+λ_j^2)^3};$$ $$f''_{vv}=f'''_{uvv}=f'''_{vvv}=0,~~g''_{vv}=g'''_{uvv}=g'''_{vvv}=0.$$ 故 $$c_j=f_{uu}+2f_{uv} \bigg({\rm i}Ωega_0-\frac{d_2j^2}{\ell^2}-\frac{aλ_j}{1+λ_j^2} \bigg)=-λ_jd_j,$$ $$e_j=f_{uu}-2f_{uv}\frac{1+λ_j^2}{aλ_j^2} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)=-λ_jf_j,$$ $$g_j=f_{uuu}+2f_{uuv}\frac{1+λ_j^2}{aλ_j^2} \bigg({\rm i}Ωega_0-\frac{3d_2j^2}{\ell^2}-\frac{3aλ_j} {1+λ_j^2}\bigg)=-λ_jh_j.$$ 由文献 ^[6] $$Q_{qq}=\cos^2\frac{j}{\ell}x\left(\begin{array}{c} c_j\\ d_j \end{array}\right) ,\;\;\;\; Q_{q\bar{q}}=\cos^2\frac{j}{\ell}x\left(\begin{array}{c} e_j\\ f_j \end{array}\right) ,\;\;\;\; C_{qq\bar{q}}=\cos^3\frac{j}{\ell}x\left(\begin{array}{c} g_j\\ h_j \end{array}\right),$$ 有 $$λngle q^*,Q_{qq}\rangle=λngle q^*,Q_{q\overline{q}}\rangle=λngle \overline{q}^*,Q_{qq}\rangle=λngle \overline{q}^*,\;Q_{\overline{q}\overline{q}}\rangle=0.$$ 这样, $$H_{20}=Q_{qq}=\cos^2\frac{j}{\ell}x\left(\begin{array}{c} c_{j}\\ d_{j}\\ \end{array}\right)=\left(\frac{1}{2}\cos\frac{2j}{\ell}x+\frac{1}{2}\right)\left(\begin{array}{c} c_{j}\\ d_{j}\\ \end{array}\right),$$ $$H_{11}=Q_{q\bar{q}}=\cos^2\frac{j}{\ell}x\left(\begin{array}{c} e_{j}\\ f_{j}\\ \end{array}\right)=\left(\frac{1}{2}\cos\frac{2j}{\ell}x+\frac{1}{2}\right)\left(\begin{array}{c} e_{j}\\ f_{j}\\ \end{array}\right),$$ $$\begin{eqnarray*} w_{20}&=&[2{\rm i}Ωega_0I-L(λ_j)]^{-1}H_{20}\\ &=&\left[\frac{[2{\rm i}Ωega_0I-L_{2j}(λ_j)]^{-1}}{2}\cos\frac{2j}{\ell}x+\frac{[2{\rm i}Ωega_0I-L_{0}(λ_j)]^{-1}}{2}\right]\cdot\left(\begin{array}{c} c_j\\ d_j\\ \end{array}\right)\\ &=&\left(\begin{array}{c} \xi\\ \eta\\ \end{array}\right)\cos\frac{2jx}{\ell} +\left(\begin{array}{c} \tau\\ \chi \end{array}\right), \end{eqnarray*}$$ $$\begin{eqnarray*} w_{11}&=&-[L(λ_j)]^{-1}H_{11}\\ &=&-\left[\frac{[L_{2j}(λ_j)]^{-1}}{2}\cos\frac{2j}{\ell}x+\frac{[L_{0}(λ_j)^{-1}]}{2}\right]\cdot\left(\begin{array}{c} e_j\\ f_j\\ \end{array}\right)\\ &=&\left(\begin{array}{c} \tilde{\xi}\\ \tilde{\eta}\\ \end{array}\right)\cos\frac{2jx}{\ell} +\left(\begin{array}{c} \tilde{\tau}\\ \tilde{\chi} \end{array}\right), \end{eqnarray*}$$ 及 $$\xi=\frac{1}{2}(αpha_1+{\rm i}αpha_2)^{-1}\left[\left(2{\rm i}Ωega_0+\frac{4d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\right)c_j+\frac{aλ_j^2}{1+λ_j^2}d_j\right],$$ $$\eta=\frac{1}{2}(αpha_1+{\rm i}αpha_2)^{-1}\left[-\frac{2}{λ_j(1+λ_j^2)}c_j+\left(2{\rm i}Ωega_0+\frac{4d_2j^2}{\ell^2}-\frac{1-λ_j^2}{1+λ_j^2}\right)d_j\right],$$ $$\tau=\frac{1}{2}(αpha_3+{\rm i}αpha_4)^{-1}\left[\left(2{\rm i}Ωega_0+\frac{aλ_j}{1+λ_j^2}\right)c_j+\frac{aλ_j^2}{1+λ_j^2}d_j\right],$$ $$\chi=\frac{1}{2}(αpha_3+{\rm i}αpha_4)^{-1}\left[-\frac{2}{λ_j(1+λ_j^2)}c_j+\left(2{\rm i}Ωega_0-\frac{1-λ_j^2}{1+λ_j^2}\right)d_j\right],$$ $$\widetilde{\xi}=\frac{1}{2αpha_5}\left[\left(\frac{4d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\right)e_j+\frac{aλ_j^2}{1+λ_j^2}f_j\right],$$ $$\widetilde{\eta}=\frac{1}{2αpha_5}\left[-\frac{2}{λ_j(1+λ_j^2)}e_j+\left(\frac{4d_1j^2}{\ell^2}-\frac{1-λ_j^2}{1+λ_j^2}\right)f_j\right],$$ $$\widetilde{\tau}=\frac{1}{2αpha_6}\left(\frac{aλ_j}{1+λ_j^2}e_j+\frac{aλ_j^2}{1+λ_j^2}f_j\right),$$ $$\widetilde{\chi}=\frac{1}{2αpha_6}\left(-\frac{2}{λ_j(1+λ_j^2)}e_j-\frac{1-λ_j^2}{1+λ_j^2}f_j\right).$$ 则有 $$Q_{w_{20},\overline{q}} =\left(\begin{array}{c} f_{uu}\xi+f_{uv}\eta+f_{uv}\overline{b_j}\xi\\ g_{uu}\xi+g_{uv}\eta+g_{uv}\overline{b_j}\xi\\ \end{array}\right)\cos\frac{jx}{\ell}\cos\frac{2jx}{\ell} +\left(\begin{array}{c} f_{uu}\tau+f_{uv}\chi+f_{uv}\overline{b_j}\tau\\ g_{uu}\tau+g_{uv}\chi+g_{uv}\overline{b_j}\tau\\ \end{array}\right)\cos\frac{jx}{\ell},$$ $$Q_{w_{11},q} =\left(\begin{array}{c} f_{uu}\widetilde{\xi}+f_{uv}\widetilde{\eta}+f_{uv}b_j\widetilde{\xi}\\ g_{uu}\widetilde{\xi}+g_{uv}\widetilde{\eta}+g_{uv}b_j\widetilde{\xi}\\ \end{array}\right)\cos\frac{2jx}{\ell}\cos\frac{jx}{\ell} +\left(\begin{array}{c} f_{uu}\widetilde{\tau}+f_{uv}\widetilde{\chi}+f_{uv}b_j\widetilde{\tau}\\ g_{uu}\widetilde{\tau}+g_{uv}\widetilde{\chi}+f_{uv}b_j\widetilde{\tau}\\ \end{array}\right)\cos\frac{jx}{\ell},$$ 注意对任意的 $j\in {\Bbb N}$, $$\int_0^{\ell\pi}\cos^2\frac{jx}{\ell}{\rm d}x=\frac{1}{2}\ell\pi,\int_0^{\ell\pi}\cos\frac{2jx}{\ell}\cos^2\frac{jx}{\ell}{\rm d}x=\frac{1}{4}\ell\pi, \int_0^{\ell\pi}\cos^4\frac{{j}x}{\ell}{\rm d}x=\frac{3}{8}\ell\pi,$$ 有 $$\begin{eqnarray*} λngle q^*,Q_{w_{20},\overline{q}}\rangle &=&\frac{\ell\pi}{4}\left\{\overline{a_j^*}(f_{uu}\xi+f_{uv}\eta+f_{uv}\xi\overline{b_j})+\overline{b_j^*}(g_{uu}\xi+g_{uv}\eta+g_{uv}\xi\overline{b_j})\right\}\\ &&+\frac{\ell\pi}{2}\left\{\overline{a_j^*}(f_{uu}\tau+f_{uv}\chi+f_{uv}\tau\overline{b_j})+\overline{b_j^*}(g_{uu}\tau+g_{uv}\chi+g_{uv}\tau\overline{b_j})\right\}, \end{eqnarray*}$$ $$\begin{eqnarray*} λngle q^*,Q_{w_{11},q}\rangle &=&\frac{\ell\pi}{4}\left\{\overline{a_j^*}(f_{uu}\widetilde{\xi}+f_{uv}\widetilde{\eta}+f_{uv}\widetilde{\xi}b_j) +\overline{b_j^*}(g_{uu}\widetilde{\xi}+g_{uv}\widetilde{\eta}+g_{uv}\widetilde{\xi}b_j)\right\}\\ &&+\frac{\ell\pi}{2}\left\{\overline{a_j^*}(f_{uu}\widetilde{\tau}+f_{uv}\widetilde{\chi}+f_{uv}\widetilde{\tau}b_j) +\overline{b_j^*}(g_{uu}\widetilde{\tau}+g_{uv}\widetilde{\chi}+g_{uv}\widetilde{\tau}b_j)\right\}, \end{eqnarray*}$$ $$λngle q^*,C_{q,q,\overline{q}}\rangle =\frac{3\ell\pi}{4}(\overline{a_j^*}g_j +\overline{b_j^*}h_j).$$ $$\begin{eqnarray*} &&{\rm Re}λngle q^*,Q_{w_{20},\bar{q}}\rangle \\ &=&\frac{1}{4}\bigg\{f _{uu}(\xi_R+2\tau_R) +f_{uv}(\eta_R+2\chi_R) -f_{uv}\frac{1+λ_j^2}{aλ_j^2} \bigg[(\xi_R+2\tau_R) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \\ && -(\xi_I+2\tau_I)Ωega_0\bigg]\bigg\} -\frac{1}{4Ωega_0} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\\ &&\cdot\bigg\{f_{uu}(\xi_I+2\tau_I)+f_{uv}(\eta_I+2\chi_I) -f_{uv}\frac{1+λ_j^2}{aλ_j^2}\bigg[(\xi_I+2\tau_I) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \\ && +(\xi_R+2\tau_R)Ωega_0\bigg]\bigg\} -\frac{λ_j(1+λ_j^2)}{4} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\\ &&\cdot\bigg\{g_{uu}(\xi_R+2\tau_R)+g_{uv}(\eta_R+2\chi_R) -g_{uv}\frac{1+λ_j^2}{aλ_j^2}\bigg[(\xi_R+2\tau_R) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\\ &&- (\xi_I+2\tau_I)Ωega_0\bigg]\bigg\} +\frac{λ_j(1+λ_j^2)}{8Ωega_0} \bigg[\bigg(\frac{d_2j^2}{\ell^2}+ \frac{aλ_j}{1+λ_j^2}\bigg)^2-Ωega_0^2\bigg]\\ &&\cdot\bigg\{g_{uu}(\xi_I+2\tau_I) +g_{uv}(\eta_I+2\chi_I)\\ &&-g_{uv}\frac{1+λ_j^2}{aλ_j^2} \bigg[(\xi_I+2\tau_I)\bigg(\frac{d_2j^2}{\ell^2}+ \frac{aλ_j}{1+λ_j^2}\bigg) +(\xi_R+2\tau_R)Ωega_0\bigg]\bigg\}, \end{eqnarray*}$$ $$\begin{eqnarray*} &&{\rm Re}λngle q^*,Q_{w_{11},q}\rangle \\ &=&\frac{1}{4}\bigg[f _{uu}(\tilde{\xi}+2\tilde{\tau})+f_{uv}(\tilde{\eta}+2\tilde{\chi}) -f_{uv}(\tilde{\xi}+2\tilde{\tau}) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg(\frac{1+λ_j^2}{aλ_j^2}+1\bigg) \bigg]\\ &&-\frac{λ_j(1+λ_j^2)}{4}\bigg[g _{uu}(\tilde{\xi}+2\tilde{\tau})+g_{uv} (\tilde{\eta}+2\tilde{\chi})\bigg] \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\\ &&+\frac{λ_j(1+λ_j^2)}{4}g_{uv}(\tilde{\xi}+2\tilde{\tau}) \bigg[\bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)^2 \bigg(\frac{1+λ_j^2}{aλ_j^2}+1 \bigg)-\frac{aλ_j} {(1+λ_j^2)^2} \bigg], \end{eqnarray*}$$ $$\begin{eqnarray*} &&{\rm Re}λngle q^*,C_{q,q,\overline{q}}\rangle\\ &=&\frac{3}{8}\bigg[f _{uuu}-6f_{uuv}\frac{1+λ_j^2}{aλ_j^2} \bigg( \frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\bigg] \bigg[1+(1+λ_j^2)\bigg( \frac{d_2j^2}{\ell^2}+\frac{2aλ_j}{1+λ_j^2}\bigg)\bigg]\\ &&-f_{uuv}\frac{3(1+λ_j^2)}{4aλ_j^2} \bigg[\frac{d_2j^2}{\ell^2}+\frac{2aλ_j}{1+λ_j^2} -Ωega_0^2(1+λ_j^2)\bigg]. \end{eqnarray*}$$

对 $\Gamma=\xi,\eta,\tau,\chi$,定义 $\Gamma_R:={\rm Re} \Gamma$ 和 $\Gamma_I:={\rm Im} \Gamma$. 进一步, $$\begin{eqnarray*} \xi_R&=&\frac{1}{2(αpha_1^2+αpha_2^2)} \bigg\{2\bigg[f_{uu}+2f_{uv} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg]\bigg(αpha_2Ωega_0+\frac{2αpha_1d_2j^2}{\ell^2}\bigg) \\ && +4f_{uv}\bigg(\frac{2αpha_2d_2j^2}{\ell^2}-αpha_1Ωega_0^2 \bigg)\bigg\}, \end{eqnarray*}$$ $$\begin{eqnarray*} \xi_I&=&\frac{1}{2(αpha_1^2+αpha_2^2)} \bigg\{2\bigg[f_{uu}+2f_{uv}\bigg(\frac{d_2j^2}{\ell^2} +\frac{aλ_j}{1+λ_j^2}\bigg)\bigg] \bigg(αpha_1Ωega_0-\frac{2αpha_2d_2j^2}{\ell^2}\bigg) \\ &&+4Ωega_0f_{uv}\bigg(\frac{2αpha_1d_2j^2} {\ell^2}-αpha_2Ωega_0\bigg)\bigg\}, \end{eqnarray*}$$ $$\begin{eqnarray*} \eta_R&=&-\frac{1}{2λ_j(αpha_1^2+αpha_2^2)} \bigg\{\bigg[f_{uu}+2f_{uv} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg] \bigg[2αpha_2Ωega_0+αpha_1 \bigg(1+\frac{4d_2j^2}{\ell^2}\bigg)\bigg]\bigg\}\\ &&-\frac{Ωega_0f_{uv}}{λ_j(αpha_1^2+αpha_2^2)} \bigg[αpha_2\bigg(1+\frac{4d_2j^2}{\ell^2}\bigg) -2αpha_1Ωega_0\bigg], \end{eqnarray*}$$ $$\begin{eqnarray*} \eta_I&=&-\frac{1}{2λ_j(αpha_1^2+αpha_2^2)} \bigg\{\bigg[f_{uu}+2f_{uv}\bigg(\frac{d_2j^2} {\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\bigg] \bigg[2αpha_1Ωega_0-αpha_2 \bigg(1+\frac{4d_2j^2}{\ell^2}\bigg)\bigg]\bigg\}\\ &&-\frac{Ωega_0f_{uv}}{λ_j(αpha_1^2+αpha_2^2)} \bigg[αpha_1\bigg(1+\frac{4d_2j^2}{\ell^2}\bigg) -2αpha_2Ωega_0\bigg], \end{eqnarray*}$$ $$\begin{eqnarray*} \tau_R=-\frac{Ωega_0}{(αpha_3^2+αpha_4^2)} \bigg\{2Ωega_0αpha_3f_{uv}-αpha_4\bigg[f_{uu}+2f_{uv} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg]\bigg\}, \end{eqnarray*}$$ $$\begin{eqnarray*} \tau_I=-\frac{Ωega_0}{(αpha_3^2+αpha_4^2)} \bigg\{2Ωega_0αpha_4f_{uv}-αpha_3\bigg[f_{uu}+2f_{uv} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg]\bigg\}, \end{eqnarray*}$$ $$\begin{eqnarray*} \chi_R&=&-\frac{1}{2λ_j(αpha_3^2+αpha_4^2)} \bigg\{(2Ωega_0αpha_4+αpha_3)\bigg[f_{uu}+2f_{uv} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg] \\ &&+2Ωega_0f_{uv}(αpha_4-2Ωega_0αpha_3)\bigg\}, \end{eqnarray*}$$ $$\begin{eqnarray*} \chi_I&=&-\frac{1}{2λ_j(αpha_3^2+αpha_4^2)} \bigg\{(2Ωega_0αpha_3-αpha_4) \bigg[f_{uu}+2f_{uv} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2} \bigg)\bigg] \\ &&+2Ωega_0f_{uv}(αpha_3+2Ωega_0αpha_4)\bigg\}, \end{eqnarray*}$$ 由此,得到 $$\begin{eqnarray*} &&{\rm Re}(c_1(λ_{j}))\\ &=&{\rm Re}λngle q^*,Q_{w_{11},q}\rangle +\frac{1}{2}{\rm Re}λngle q^*,Q_{w_{20},\overline{q}}\rangle+\frac{1}{2}{\rm Re}λngle q^*,C_{q,q,\overline{q}}\rangle\\ &=&\frac{1}{4}\bigg[f _{uu}(\tilde{\xi}+2\tilde{\tau})+f_{uv}(\tilde{\eta}+2\tilde{\chi}) -f_{uv}(\tilde{\xi}+2\tilde{\tau}) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg(\frac{1+λ_j^2}{aλ_j^2}+1\bigg) \bigg]\\ &&-\frac{λ_j(1+λ_j^2)}{4} \bigg[g _{uu}(\tilde{\xi}+2\tilde{\tau}) +g_{uv}(\tilde{\eta}+2\tilde{\chi})\bigg] \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)\\ &&+\frac{λ_j(1+λ_j^2)}{4}g_{uv}(\tilde{\xi}+2\tilde{\tau}) \bigg[\bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)^2 \bigg(\frac{1+λ_j^2}{aλ_j^2}+1\bigg)-\frac{aλ_j} {(1+λ_j^2)^2}\bigg]\\ &&+\frac{1}{8}\bigg\{f _{uu}(\xi_R+2\tau_R)+f_{uv}(\eta_R+2\chi_R) \\ && -f_{uv}\frac{1+λ_j^2}{aλ_j^2}\ \bigg[(\xi_R+2\tau_R) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) -(\xi_I+2\tau_I)Ωega_0\bigg]\bigg\}\\ &&-\frac{1}{8Ωega_0} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg\{f_{uu}(\xi_I+2\tau_I)+f_{uv}(\eta_I+2\chi_I) \\ &&-f_{uv}\frac{1+λ_j^2}{aλ_j^2} \bigg[(\xi_I+2\tau_I) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) +(\xi_R+2\tau_R)Ωega_0\bigg]\bigg\}\\ &&-\frac{λ_j(1+λ_j^2)}{8} \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg\{g_{uu}(\xi_R+2\tau_R)+g_{uv}(\eta_R+2\chi_R) \\ &&-g_{uv}\frac{1+λ_j^2}{aλ_j^2} \bigg[(\xi_R+2\tau_R) \bigg(\frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg)- (\xi_I+2\tau_I)Ωega_0\bigg]\bigg\} \\ &&+\frac{λ_j(1+λ_j^2)}{16Ωega_0} \bigg[\bigg(\frac{d_2j^2}{\ell^2}+ \frac{aλ_j}{1+λ_j^2}\bigg)^2-Ωega_0^2\bigg] \bigg\{g_{uu}(\xi_I+2\tau_I) +g_{uv}(\eta_I+2\chi_I)\\ &&-g_{uv}\frac{1+λ_j^2}{aλ_j^2} \bigg[(\xi_I+2\tau_I)\bigg(\frac{d_2j^2}{\ell^2} +\frac{aλ_j}{1+λ_j^2}\bigg) +(\xi_R+2\tau_R)Ωega_0\bigg]\bigg\}\\ &&+\frac{3}{16}\bigg[f _{uuu}-6f_{uuv}\frac{1+λ_j^2}{aλ_j^2}\bigg( \frac{d_2j^2}{\ell^2}+\frac{aλ_j}{1+λ_j^2}\bigg) \bigg]\bigg[1+(1+λ_j^2)\bigg( \frac{d_2j^2}{\ell^2}+\frac{2aλ_j}{1+λ_j^2}\bigg) \bigg]\\ &&-f_{uuv}\frac{3(1+λ_j^2)}{8aλ_j^2} \bigg[\frac{d_2j^2}{\ell^2}+\frac{2aλ_j}{1+λ_j^2} -Ωega_0^2(1+λ_j^2)\bigg], \end{eqnarray*}$$ 于是,如果 $\frac{1}{αpha'(λ_{j,\pm}^H)}{\rm Re}(c_1(λ_{j,\pm}^H))<0$ ($>0)$,分支的周期解是超临界的 (或次临界的). 证明完毕.

例3.4 令 $Ωega=\left(0,5\pi\right)$,$d_1=4$,$d_2=1$,$a=2.25$.则 根据文献 ^[7]中的例子 $$λmbda_1=\{ λ_2^H,λ_1^H, λ_0^H\}=\{0.0833,\;0.3057,\;0.3802\}.$$ $${\rm Re}(c_1(λ_1^H))\approx-36.9473<0,~~ {\rm Re}(c_1(λ_2^H))\approx-263.3673<0,$$ $$αpha'(λ_1^H)=A'(λ_1^H)/2\approx-1.3643<0,~~ αpha'(λ_2^H)=A'(λ_1^H)/2\approx-1.2662<0.$$ 这样,在点$λ=λ_1^H$ 和 $λ=λ_2^H$,分支方向是次临界的.