考虑近Hamilton系统

$ \begin{equation} \left\{ \begin{aligned} \dot{x}&=-\frac{\partial H}{\partial y}+\varepsilon f(x, y), \\ \dot{y}&=\frac{\partial H}{\partial x}+\varepsilon g(x, y), \\ \end{aligned} \right. \end{equation} $

(1.1)

其中$ 0<|\varepsilon |\ll 1 $, $ H(x, y) $是关于$ x $和$ y $的$ m+1 $次实多项式, $ f(x, y) $和$ g(x, y) $是关于$ x $和$ y $的次数不超过$ n $的实多项式.假设系统$ {\left( \text{1}.\text{1} \right)}_{\varepsilon =0} $有连续闭轨线族

$ \begin{align} {{\Gamma }_{h}}\subseteq\{(x, y)\in {{\mathbb{R}}^{2}}|H(x, y)=h, h\in \Sigma \}. \end{align} $

$ \Sigma $为$ h $的最大存在开区间.定义后继函数

$ \begin{align} {d(h, \varepsilon)}:={{P}_{\varepsilon }}(h)-h=\varepsilon I(h)+o({{\varepsilon }^{2}}), \end{align} $

其中$ {{P}_{\varepsilon }}(h) $为Poincar é映射下$ h $的像, 则$ \varepsilon $的系数$ I(h) $可表示为

$ \begin{eqnarray} I(h)=\oint_{\Gamma {}_{h}}{f(x, y)dx-g(x, y)dy, h\in \Sigma, } \end{eqnarray} $

(1.2)

式(1.2)称为Abelian积分.当$ I(h) $不恒等于零时, Abelian积分$ I(h) $的孤立零点的最大个数(计重数)就给出了在扰动前系统(1.1)的闭轨族所形成的区域内产生极限环个数(计重数)的上界^[1].寻找Abelian积分$ I(h) $的零点个数的上界$ B(n) $称为弱化的Hilbert第16问题^[2].对Hamilton函数

$ \begin{align} H(x, y)=\frac{1}{2}({{x}^{2}}+{{y}^{2}})-\frac{1}{3}{{x}^{3}}+ax{{y}^{2}}+\frac{1}{3}b{{y}^{3}}, \end{align} $

Horozov等^[3]用Picard-Fuchs方程法证明了$ B(n)\le 5n+15 $.对于Hamilton函数

$ \begin{align} H(x, y)=a{{x}^{2}}+b{{y}^{2}}-{{x}^{4}}+c{{x}^{2}}{{y}^{2}}+{{y}^{4}}, \end{align} $

杨纪华^[4]等证明了$ B(n)\le 90n+24 $. 对于Hamilton函数$ H(x, y)=\frac{1}{2}{{y}^{2}}+U(x) $, $ \deg U(x)=4 $时, 赵育林等^[5]证明了$ B(n)\le 7n+5 $.对于4次Hamilton函数

$ \begin{align} H(x, y)={{x}^{2}}+y{}^{2}+a{{x}^{4}}+b{{x}^{2}}{{y}^{2}}+c{{y}^{4}} \end{align} $

周鑫等^[6]得到相应Abelian积分$ I(h) $的代数结构, 并给出了$ a>0, b=0 $和$ c=0 $时$ I(h) $零点个数的上界.对于Hamilton函数

$ \begin{align} H(x, y)={{x}^{2}}+y{}^{2}-{{x}^{4}}+a{{x}^{2}}{{y}^{2}}+{{y}^{4}}, a>-2, \end{align} $

吴娟娟等^[7]证明了$ B(n)\le 2[\frac{n-1}{4}]+12[\frac{n-3}{4}]+23 $.对Hamilton函数

$ \begin{align} H(x, y)=\frac{1}{2}{{y}^{2}}+\frac{a}{2}{{x}^{2}}+\frac{b}{4}{{x}^{4}}+\frac{c}{6}{{x}^{6}}, \end{align} $

赵丽琴等^[8]证明了$ B(n)\le 54n-13 $.对Hamilton函数

$ \begin{align} H(x, y)={{x}^{2}}+c{{x}^{4}}+{{y}^{4}}, \end{align} $

张永康等^[9]证明了$ B(2n+2)=B(2n+1)\le 2[\frac{n}{2}]+3[\frac{n-1}{2}]+4 $

本文研究了如下Hamilton函数

$ \begin{eqnarray} H(x, y)={\frac{1}{4}x^4+\frac{1}{4}y^4-\frac{1}{8}x^8+\frac{1}{72}x^{12}=h, h\in \Sigma=\left(0, \frac{\sqrt{3}}{12}\right), } \end{eqnarray} $

(1.3)

相应的向量场

$ \begin{equation} \left\{ \begin{aligned} \dot{x}&=-{{y}^{3}}, \\ \dot{y}&=\frac{x^3}{6}\left(6-6x^4+x^8\right). \end{aligned} \right. \end{equation} $

(1.4)

在多项式$ f(x, y)\text{=}\sum\limits_{1\le 4i+4j+1\le n}{{{a}_{ij}}{{x}^{4i}}{{y}^{4j+1}}} $和$ g(x, y)=\sum\limits_{1\le 4i+4j+1\le n}{{{b}_{ij}}{{x}^{4i+1}}{{y}^{4j}}} $扰动下相应的Abelian积分$ I(h) $零点个数的上界(计重数). 设$ P(x, y)\text{=}-{{y}^{3}}, Q(x, y)\text{=}\frac{x^3}{6}\left(6-6x^4+x^8\right). $由 $ P(x, y)\text{=}0 $且$ Q(x, y)\text{=}0 $可得系统(1.4)有五个奇点$ \widetilde{O_1}(-\sqrt[4]{3+\sqrt{3}}, 0) $, $ O\left(0, 0\right) $, $ O_1\left(\sqrt[4]{3+\sqrt{3}}, 0\right) $, $ C^{\prime}\left(-\sqrt[4]{3-\sqrt{3}}, 0\right) $, $ C\left(\sqrt[4]{3-\sqrt{3}}, 0\right) $. 根据矩阵

$ \begin{bmatrix}\frac{\partial P}{\partial x} & \frac{\partial P}{\partial y} \\ \frac{\partial Q}{\partial x} & \frac{\partial Q}{\partial y}\end{bmatrix} $

在五点$ \widetilde{O_1}, O, O_1, C^{\prime}, C $处的特征根可得, 当$ h\in \Sigma=\left(0, \frac{\sqrt{3}}{12}\right) $时, 系统有三个中心点$ \widetilde{O_1}(-\sqrt[4]{3+\sqrt{3}}, 0) $, $ O\left(0, 0\right) $, $ O_1\left(\sqrt[4]{3+\sqrt{3}}, 0\right) $和两个对称的退化鞍点$ C^{\prime}\left(-\sqrt[4]{3-\sqrt{3}}, 0\right) $, $ C\left(\sqrt[4]{3-\sqrt{3}}, 0\right) $, 有三个分别围绕中心$ \widetilde{O_1}, O, O_1 $的周期环域$ \widetilde{\Gamma_h^1}, \Gamma_h, \Gamma_h^{1} $. 如图 1所示.

其中

$ \widetilde{\Gamma_h^1}=\left\lbrace\left(x, y\right)\left|H(x, y)={\frac{1}{4}x^4+\frac{1}{4}y^4-\frac{1}{8}x^8+\frac{1}{72}x^{12}=h, x<-\sqrt[4]{3-\sqrt{3}}, h\in \left(0, \frac{\sqrt{3}}{12}\right)}\right.\right\rbrace, $

$ \Gamma_h=\left\lbrace\left(x, y\right)\left|H(x, y)={\frac{1}{4}x^4+\frac{1}{4}y^4-\frac{1}{8}x^8+\frac{1}{72}x^{12}=h, -\sqrt[4]{3-\sqrt{3}}<x<\sqrt[4]{3-\sqrt{3}}, h\in \left(0, \frac{\sqrt{3}}{12}\right)}\right.\right\rbrace, $

$ \Gamma_h^{1}=\left\lbrace\left(x, y\right)\left|H(x, y)={\frac{1}{4}x^4+\frac{1}{4}y^4-\frac{1}{8}x^8+\frac{1}{72}x^{12}=h, x>\sqrt[4]{3-\sqrt{3}}, h\in \left(0, \frac{\sqrt{3}}{12}\right)}\right.\right\rbrace. $

下面对$ h\in \Sigma=\left(0, \frac{\sqrt{3}}{12}\right) $时在周期环域$ \Gamma_h $上的情形进行讨论, 本文的主要结果为

定理1.1  对Hamilton函数

$ \begin{eqnarray} H(x, y)&={\frac{1}{4}x^4+\frac{1}{4}y^4-\frac{1}{8}x^8+\frac{1}{72}x^{12}=h, h\in \Sigma=\left(0, \frac{\sqrt{3}}{12}\right), } \end{eqnarray} $

在周期环域$ \Gamma_h $上, $ B(n)\le 7[\frac{n-1}{4}]+5, $其中$ n=\max \{\deg f, \deg g\} $且$ n=4k+1, k\in{\mathbb{N}}_{+} $.

在周期环域$ \Gamma_h $上, 由分部积分可得

$ \begin{align} \oint_{{\Gamma }_{h}}{{{x}^{4i+1}}{{y}^{4j}}dy}=-\frac{4i+1}{4j+1}\oint_{{{\Gamma }_{h}}}{{{x}^{4i}}{{y}^{4j+1}}dx}, \end{align} $

(2.1)

所以Abelian积分(1.2)可表示为

$ \begin{align} I(h)=\oint_{{\Gamma }_{h}}{L(x, y)}dx, \end{align} $

(2.2)

其中$ L(x, y)=\sum\limits_{1\le 4i+4j+1\le n}{{{c}_{ij}}{{x}^{4i}}{{y}^{4j+1}}}, $ $ i $和$ j $是自然数, $ {{c}_{ij}}=-\left(\frac{4i+1}{4j+1}{{{a}_{ij}}+{{b}_{ij}}}\right) $.

对$ {h\in \Sigma} $, 记

$ \begin{align} {{I}_{4i, 4j+1}}(h)=\oint_{{\Gamma}_{h}}{{x}^{4i}}{{y}^{4j+1}}dx, \end{align} $

(2.3)

所以

$ \begin{align} I(h)=\sum\limits_{1\le 4i+4j+1\le n}{{{c}_{ij}}{{I}_{4i, 4j+1}}(h)}. \end{align} $

从而, Abelian积分$ I(h) $的代数结构可由下面两个引理给出.

引理2.1   当$ n=4i+4j+1=4k+1\ge 5 $时, 积分$ {{I}_{4i, 4j+1}}(h) $可以表示为$ {{I}_{l, m}}(h)(l+m=4k-7, 4k-3) $和$ h{{I}_{l, m}}(h)(l+m=4k-11, 4k-7 $或$ 4k-3) $的线性组合.

证  在式(1.3)两端同时对$ x $求导可得

$ \begin{align} {{x}^{3}}-{{x}^{7}}+\frac{1}{6}{{x}^{11}+{{y}^{3}}\frac{\partial y}{\partial x}}=0, \end{align} $

(2.4)

给式(2.4)两端同乘以$ {{x}^{4i-11}{y}^{4j+1}} $, 并沿$ {\Gamma }_{h} $关于$ x $积分可得

$ \begin{align} {{I}_{4i, 4j+1}}= 6{{I}_{4i-4, 4j+1}}-6{{I}_{4i-8, 4j+1}}+\frac{6(4i-11)}{4j+5}{{{I}_{4i-12, 4j+5}}} , \end{align} $

(2.5)

其中$ i\ge 3, $给式(1.3)两端同乘以$ {{x}^{4i}}{{y}^{4j-3}} $, 并沿$ {{\Gamma }_{h}} $关于$ x $积分可得

$ \begin{align} {{I}_{4i, 4j+1}}=4h{{I}_{4i, 4j-3}}-{{I}_{4i+4, 4j-3}}+\frac{1}{2}{{I}_{4i+8, 4j-3}}-\frac{1}{18}{{I}_{4i+12, 4j-3}}. \end{align} $

(2.6)

由式(2.5)可得

$ \begin{align} {{I}_{4i+12, 4j-3}}=6{{I}_{4i+8, 4j-3}}-6{{I}_{4i+4, 4j-3}}+\frac{6(4i+1)}{4j+1}{{{I}_{4i, 4j+1}}}. \end{align} $

(2.7)

把式(2.7)代入式(2.6)可得

$ \begin{align} {{I}_{4i, 4j+1}}=\frac{4j+1}{8(i+3j+1)}[24h{{I}_{4i, 4j-3}}-4{{I}_{4i+4, 4j-3}}+{{I}_{4i+8, 4j-3}}]. \end{align} $

(2.8)

由式(2.6)可得

$ \begin{align} {{I}_{4i-12, 4j+5}}=4h{{I}_{4i-12, 4j+1}}-{{I}_{4i-8, 4j+1}}+\frac{1}{2}{{I}_{4i-4, 4j+1}}-\frac{1}{18}{{I}_{4i, 4j+1}}. \end{align} $

(2.9)

把式(2.9)代入式(2.5)可得

$ \begin{align} &{{I}_{4i, 4j+1}}\\=&\frac{1}{4(i+3j+1)}[72(4i-11)h{{I}_{4i-12, 4j+1}}-36(2i+2j-3){{I}_{4i-8, 4j+1}}+9(4i+8j-1){{I}_{4i-4, 4j+1}}]. \end{align} $

(2.10)

在式(2.5)中取$ (i, j)=(3, k-1) $可得

$ \begin{align} {{I}_{12, 4k-3}}= 6{{I}_{8, 4k-3}}-6{{I}_{4, 4k-3}}+\frac{6}{4k+1}{{{I}_{0, 4k+1}}}. \end{align} $

(2.11)

在式(2.8)中取$ (i, j)=(2, k-1) $可得

$ \begin{align} {{I}_{8, 4k-3}}=\frac{4k-3}{24k}[24h{{I}_{8, 4k-7}}-4{{I}_{12, 4k-7}}+{{I}_{16, 4k-7}}]. \end{align} $

(2.12)

由式(2.10)可得

$ \begin{align} &{{I}_{4i+8, 4j-3}}\\=&\frac{1}{4(i+3j)}[72(4i-3)h{{I}_{4i-4, 4j-3}}-36(2i+2j-1){{I}_{4i, 4j-3}}+9(4i+8j-1){{I}_{4i+4, 4j-3}}]. \end{align} $

(2.13)

在式(2.13)中取$ (i, j)=(1, k-1), (2, k-1), (2, k-2) $可得

$ \begin{align} {{I}_{12, 4k-7}}=&\frac{1}{4(3k-2)}[72h{{I}_{0, 4k-7}}-36(2k-1){{I}_{4, 4k-7}}+9(8k-5){{I}_{8, 4k-7}}]. \end{align} $

(2.14)

$ \begin{align} {{I}_{16, 4k-7}}=&\frac{1}{4(3k-1)}[360h{{I}_{4, 4k-7}}-36(2k+1){{I}_{8, 4k-7}}+9(8k-1){{I}_{12, 4k-7}}]. \end{align} $

(2.15)

$ \begin{align} {{I}_{16, 4k-11}}=&\frac{1}{4(3k-4)}[360h{{I}_{4, 4k-11}}-36(2k-1){{I}_{8, 4k-11}}+9(8k-9){{I}_{12, 4k-11}}]. \end{align} $

(2.16)

在式(2.8)中分别取$ (i, j)=(0, k), (1, k-1), (2, k-2), $在式(2.10)中取$ (i, j)=(3, k-3), (4, k-4), \cdots, (k-1, 1), (k, 0), $同时结合(2.11), (2.12), (2.14), (2.15)和(2.16)可得

$ \begin{align} AJ=B, \end{align} $

其中$ J=({{I}_{0, 4k+1}}, {{I}_{4, 4k-3}}, {{I}_{8, 4k-7}}, {{I}_{12, 4k-11}}, {{I}_{16, 4k-15}}, \cdots , {{I}_{4k-4, 5}}, {{I}_{4k, 1}}), $

$ A=\left[ \begin{array}{cccccccc} 1 & \frac{4k+1}{2\left(3k+1\right)} & -\frac{\left(4k+1\right)\left(4k-3\right)}{8\left(3k+1\right)}{[\frac{h}{k}+\frac{1056k^2-880k+151}{128k\left(3k-1\right)\left(3k-2\right)}]} & 0 & 0 & \cdots & 0 & 0\\ 0 & 1 & -\frac{4k-3}{8\left(3k-1\right)}{[\frac{9\left(8k-5\right)}{4\left(3k-2\right)}-4]} & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & -\frac{4k-7}{8\left(3k-3\right)}{[\frac{9\left(8k-9\right)}{4\left(3k-4\right)}-4]}& 0& \cdots & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 &\cdots & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 &\cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 &0& \cdots & 1 & 0 \\ 0 & 0 & 0 & 0 &0& \cdots & 0 & 1 \\ \end{array} \right], $

$ B=\left[ \begin{array}{c} \frac{4k+1}{8(3k+1)}\left\lbrace 24h{{I}_{0, 4k-3}}+\frac{(72k+21)(4k-3)}{16k(3k-1)(3k-2)}h{{I}_{0, 4k-7}} +[\frac{15(4k-3)}{4k(3k-1)}h-\frac{(216k-75)(2k-1)(4k-3)}{32k(3k-1)(3k-2)}]{{I}_{4, 4k-7}} \right\rbrace \\ \frac{4k-3}{8(3k-1)}{[{\frac{18}{3k-2}h{I}_{0, 4k-7}}+(24h-\frac{18k-9}{3k-2}){I}_{4, 4k-7}]} \\ \frac{4k-7}{8(3k-3)}{[{\frac{90}{3k-4}h{I}_{0, 4k-11}}+(24h-\frac{18k-9}{3k-4}){I}_{8, 4k-11}]} \\ \frac{1}{4(3k-5)}[72h{{I}_{0, 4k-11}}-36(2k-3){{I}_{4, 4k-11}}+9(8k-11){{I}_{8, 4k-11}}] \\ \frac{1}{4(3k-7)}[360h{{I}_{4, 4k-15}}-36(2k-3){{I}_{8, 4k-15}}+9(8k-15){{I}_{12, 4k-15}}] \\ \vdots\\ \frac{1}{4(k+3)}[72(4k-15)h{{I}_{4k-16, 5}}-36(2k-3){{I}_{4k-12, 5}}+9(4k+3){{I}_{4k-8, 5}}] \\ \frac{1}{4(k+1)}[72(4k-11)h{{I}_{4k-12, 1}}-36(2k-3){{I}_{4k-8, 1}}+9(4k-1){{I}_{4k-4, 1}}] \\ \end{array} \right]. $

显然, $ \rm{det} $$ A=1 $, 且$ B $中的元素仅含有$ {{I}_{l, m}}(h)(l+m=4k-7, 4k-3) $和$ h{{I}_{l, m}}(h)(l+m=4k-11 $, $ 4k-7 $或$ 4k-3) $. 所以, 引理2.1成立.

引理2.2  在(1.2)中假设$ n\ge 5 $, 对Hamilton函数(1.3), $ I(h) $可表示为

$ \begin{align} I(h)=\alpha (h){{I}_{0, 1}}(h)+\beta (h){{I}_{4, 1}}(h)+\gamma(h){{I}_{8, 1}}(h), \end{align} $

(2.17)

其中$ \alpha (h) $, $ \beta (h) $, $ \gamma(h) $是关于$ h $的多项式, $ \deg \alpha (h)\le \left[ \frac{n-1}{4} \right]=k, \deg\{ \beta (h), \gamma(h)\}\le \left[ \frac{n-1}{4} \right]-1=k-1. $且$ k $是$ \deg \alpha (h) $的最小上界, $ k-1 $是$ \deg \beta (h) $和$ \deg{\gamma(h)} $的最小上界.

证   下面用数学归纳法证明. 由式(2.8)和式(2.10)可得

$ \begin{align} {{I}_{0, 5}}(h)=&\frac{5}{32}[24h{{I}_{0, 1}}-4{{I}_{4, 1}}+{{I}_{8, 1}}], {{I}_{0, 9}}(h)=\frac{9}{56}[24h{{I}_{0, 5}}-4{{I}_{4, 5}}+{{I}_{8, 5}}], \\ {{I}_{0, 13}}(h)=&\frac{13}{80}[24h{{I}_{0, 9}}-4{{I}_{4, 9}}+{{I}_{8, 9}}], {{I}_{4, 5}}(h)=\frac{1}{8}[24h{{I}_{4, 1}}-4{{I}_{8, 1}}+{{I}_{12, 1}}], \\ {{I}_{4, 9}}(h)=&\frac{9}{64}[24h{{I}_{4, 5}}-4{{I}_{8, 5}}+{{I}_{12, 5}}], {{I}_{8, 5}}(h)=\frac{5}{48}[24h{{I}_{8, 1}}-4{{I}_{12, 1}}+{{I}_{16, 1}}], \\ {{I}_{8, 9}}(h)=&\frac{1}{8}[24h{{I}_{8, 5}}-4{{I}_{12, 5}}+{{I}_{16, 5}}], {{I}_{12, 1}}(h)=\frac{1}{16}[72h{{I}_{0, 1}}-108{{I}_{4, 1}}+99{{I}_{8, 1}}], \\ {{I}_{16, 1}}(h)=&\frac{1}{20}[360h{{I}_{4, 1}}-180{{I}_{8, 1}}+135{{I}_{12, 1}}], {{I}_{12, 5}}(h)=\frac{1}{28}[72h{{I}_{0, 5}}-180{{I}_{4, 5}}+171{{I}_{8, 5}}], \\ {{I}_{16, 5}}(h)=&\frac{1}{32}[360h{{I}_{4, 5}}-252{{I}_{8, 5}}+207{{I}_{12, 5}}]. \end{align} $

(2.18)

所以, 当$ k=1, 2, 3 $时结论成立.

假设当$ k\le d-1 $时, 即$ n=4k+1\le 4d-3 $时, $ {I}(h)=\oint_{{{\Gamma }_{h}}}{L(x, y)dx} $能表示成

$ \begin{align} {I}(h)={{\alpha }^{n}}(h){{I}_{0, 1}}(h)+{{\beta }^{n}}(h){{I}_{4, 1}}(h)+{{\gamma}^{n}}(h){{I}_{8, 1}}(h), \end{align} $

(2.19)

其中$ \deg {{\alpha }^{n}}(h)\le \left[ \frac{n-1}{4} \right]=k\le d-1, \deg\{ {{\beta }^{n}}(h), {{\gamma}^{n}}(h)\}\le \left[ \frac{n-1}{4} \right]-1=k-1\le d-2. $

当$ k=d $时, 由引理2.1和式(2.19)可得

$ \begin{align} I(h)=&\sum\limits_{l+m\leq4k-3}{{{A}_{lm}}{{I}_{lm}}}+h\sum\limits_{l+m\le 4k-3}{{{B}_{lm}}{{I}_{lm}}} \\ =&{{\alpha }^{4d-3}}(h){{I}_{0, 1}}(h)+{{\beta }^{4d-3}}(h){{I}_{4, 1}}(h)+{{\gamma }^{4d-3}}(h){{I}_{8, 1}}(h)\\ &+h({{{\alpha}_{1} }^{4d-3}}(h){{I}_{0, 1}}(h)+{{{\beta}_{1}}^{4d-3}}(h){{I}_{4, 1}}(h)+{{{\gamma}_{1}}^{4d-3}}(h){{I}_{8, 1}}(h))\\ =&\alpha (h){{I}_{0, 1}}(h)+\beta (h){{I}_{4, 1}}(h)+\gamma (h){{I}_{8, 1}}(h). \end{align} $

其中$ \deg \{{{{\alpha }^{4d-3}}(h), {{{\alpha}_{1} }^{4d-3}}(h)}\}\le d-1, \deg \{{{{\beta}^{4d-3}}(h), {{\beta}_{1}}^{4d-3}(h), {{\gamma}^{4d-3}}(h), {{{\gamma}_{1}}^{4d-3}}(h)}\}\le d-2 $, $ {A}_{lm} $和$ {B}_{lm} $是任意常数.因此,

$ \begin{array}{l} \deg \alpha (h)\le \max \{\deg {{\alpha }^{4d-3}}(h), \deg {{\alpha_1 }^{4d-3}}(h)+1\}\le d=k, \nonumber\\ \deg \beta (h)\le \max \{\deg {{\beta }^{4d-3}}(h), \deg {{\beta_1 }^{4d-3}}(h)+1\}\le d-1=k-1, \nonumber\\ \deg \gamma (h)\le \max \{\deg {{\gamma }^{4d-3}}(h), \deg {{\gamma_1 }^{4d-3}}(h)+1\}\le d-1=k-1.\nonumber\\ \end{array} $

所以, 对任意的$ k\in{\mathbb{N}}_{+} $, 引理2.2成立.

本节将证明$ {I}_{0, 1} $, $ {I}_{4, 1} $, $ {I}_{8, 1} $满足的Picard-Fuchs方程和相关的Riccati方程.

引理3.1    对Hamilton函数(1.3), 向量$ V=({I}_{0, 1}, {I}_{4, 1}, {I}_{8, 1})^T $满足Picard-Fuchs方程

$ \begin{align} RV=(Bh+C)V', \end{align} $

(3.1)

其中

$ B=\left[ \begin{array}{ccc} 24& 0 & 0 \\ \frac{9}{2}& 24& 0 \\ \frac{99}{8}& 18& 24 \end{array} \right], C=\left[ \begin{array}{ccc} 0& -4 & 0 \\ 0&-\frac{27}{4}& \frac{35}{16} \\ 0&-\frac{297}{16}& \frac{513}{64} \end{array} \right], R=\left[ \begin{array}{ccc} 8& 0 & 0 \\ -\frac{9}{2}&16& 0 \\ -\frac{99}{8}&-18& 24 \end{array} \right]. $

证    在式(1.3)两端关于$ h $求导可得

$ \begin{align} \frac{\partial y}{\partial h}=\frac{1}{{y}^{3}}, \end{align} $

(3.2)

进而由式(2.3)可得$ {{{I}'}_{4i, 4j+1}}=(4j+1)\oint_{{{\Gamma }_{h}}}{{{x}^{4i}}{{y}^{4j-3}}\text{d}x}, $即可得

$ \begin{align} {{I}_{4i, 4j-3}}=\frac{1}{4j+1}{{{I}'}_{4i, 4j+1}}. \end{align} $

(3.3)

把(3.3)代入(2.8)可得

$ \begin{align} {{I}_{4i, 4j+1}}=\frac{1}{8(i+3j+1)}[24h{{{I}'}_{4i, 4j+1}}-4{{{I}'}_{4i+4, 4j+1}}+{{{I}'}_{4i+8, 4j+1}}]. \end{align} $

(3.4)

在式(3.4)中分别取$ (i, j)=(0, 0), (1, 0) $和$ (2, 0) $可得

$ \begin{equation} \left\{ \begin{aligned} {{I}_{0, 1}}=\frac{1}{8}(24h{{{I}'}_{0, 1}}-4{{{I}'}_{4, 1}}+{{{I}'}_{8, 1}}), \\ {{I}_{4, 1}}=\frac{1}{16}(24h{{{I}'}_{4, 1}}-4{{{I}'}_{8, 1}}+{{{I}'}_{12, 1}}), \\ {{I}_{8, 1}}=\frac{1}{24}(24h{{{I}'}_{8, 1}}-4{{{I}'}_{12, 1}}+{{{I}'}_{16, 1}}). \end{aligned} \right. \end{equation} $

(3.5)

由式(2.18)可得

$ \begin{equation} \left\{\begin{array}{l} {{{I}'}_{12, 1}}=\frac{1}{16}(72{{I}_{0, 1}}+72h{{{I}'}_{0, 1}}-108{{{I}'}_{4, 1}}+99{{{I}'}_{8, 1}}), \\ {{{I}'}_{16, 1}}=\frac{1}{20}\left(\frac{1215}{2}{{I}_{0, 1}}+360{{I}_{4, 1}}+\frac{1215}{2}h{{{I}'}_{0, 1}}+360h{{{I}'}_{4, 1}}-\frac{3645}{4}{{{I}'}_{4, 1}} +\frac{10485}{4}{{{I}'}_{8, 1}}\right). \end{array} \right. \end{equation} $

(3.6)

把式(3.6)代入式(3.5)即可得结论成立.证毕.

引理3.2    对系统(1.4), $ {I}_{0, 1} $, $ {I}_{4, 1} $和$ {I}_{8, 1} $满足

$ \begin{equation} D(h)\left[ \begin{array}{c} {I}''_{0, 1} \\ {I}''_{4, 1}\\ {I}''_{8, 1} \end{array}\right] =\left[ \begin{array}{cc} a_{11}(h)&a_{12}(h) \\ a_{21}(h)& a_{22}(h) \\ a_{31}(h)&a_{32}(h) \end{array} \right]\left[ \begin{array}{c} {I}'_{0, 1} \\ {I}'_{4, 1} \end{array} \right], \end{equation} $

(3.7)

其中

$ \begin{array}{l} D(h)=288h(48h^2-1), \\ a_{11}(h)=-9216h^2-288h+144, a_{12}(h)=-48h-36, \\ a_{21}(h)=144h(1+12h)-324h(16h+5)+1188h, a_{22}(h)=-288h(16h+5)+1728h, \\ a_{31}(h)=3456h^2+324h(12h-11)-3564h(4h-1), a_{32}(h)=288h(12h-11)-5184h(4h-1). \end{array} $

证    在式(3.1)两端对$ h $求导可得

$ \begin{align} (R-B)V'=(Bh+C)V'', \end{align} $

(3.8)

其中

$ R-B=\left[ \begin{array}{ccc} -16& 0 & 0 \\ -9& -8& 0 \\ -\frac{99}{4}& -36& 0 \end{array} \right]. $

由(3.8)可得(3.7). 证毕.

注    由(3.8)推导(3.7)时，由于三阶矩阵$ (Bh+C) $的逆矩阵计算量大，系数复杂，本文借助数学软件Maple进行相关计算，程序如下：LinearAlgebra: -MatrixInverse(rtable$ (1..3, 1..3, [[-4*h, -4, 1], [(1/7)*k*h, -4*h-9/7*k, 27/28*k-1], [-(4/7*k)*h+(3/28*(k^2))*h, k*h+36/7*k-27/28*(k^2), -4*h+81/112*\\(k^2)-48/7*k]], \text{subtype=Matrix})).$

由式(3.3)可得$ {{I}'_{0, 1}}(h)= \oint_{{{\Gamma }_{h}}}{y^{-3}\text{d}x}=3 \int\!\!\! \int_{\operatorname{int}({{\Gamma }_{h}})}{y^{-4}}{dxdy}\ne 0, $所以可得如下引理.

引理3.3    对Hamilton函数(1.3), $ \omega(h)=\frac{{I}'_{4, 1}(h)}{{I}'_{0, 1}(h)} $满足Riccati方程

$ \begin{align} D(h)\omega'(h)=-a_{12}(h)\omega^2(h)+[a_{32}(h)-a_{11}(h)]\omega(h)+a_{31}(h). \end{align} $

(3.9)

证    由于$ {\omega }'(h)=\frac{{{I}''_{4, 1}}(h){{I}'_{0, 1}}(h)-{{I}'_{4, 1}}(h){{I}''_{0, 1}}(h)}{{{I}'_{0, 1}}^{2}(h)} $, 结合式(3.7)即可得式(3.9).证毕.

由(2.17)和(3.1)可得

$ \begin{align} I'(h)={\alpha_1} (h){{I}'_{0, 1}}(h)+{\beta_1} (h){{I}'_{4, 1}}(h)+{\gamma_1}(h){{I}'_{8, 1}}(h), \end{align} $

(4.1)

其中$ \deg {{\alpha}_{1} }{(h)}\le k, \deg \{{{\beta}_{1}}(h), {{\gamma}_{1}}(h)\}\le k-1 $.

引理4.1   设$ \Omega=\{h|{{\gamma}_{1}}(h)=0, {h\in \Sigma}\}, $则当$ h\in \Sigma\backslash\Omega $时, 等式

$ \begin{align} \left(\frac{{I}'(h)}{{\gamma_1}(h)}\right)'=\frac{M(h)}{D(h){\gamma_1}^2(h)} \end{align} $

(4.2)

成立, 其中

$ \begin{align} M(h)={{{\alpha}_2}(h)}{{I}'_{0, 1}}(h)+{{{\beta_2} }(h)}{{I}'_{4, 1}}(h), \end{align} $

(4.3)

$ \deg {{{\alpha}_{2} }{(h)}\le 2k+1, \deg {{\beta}_{2}}(h)}\le 2k $.

证   由(4.1)和(3.7)可得

$ \begin{align} &\left(\frac{{I}'(h)}{{\gamma_1}(h)}\right)'\\ =&\left(\frac{{{\alpha}_{1} }(h)}{{\gamma_1}(h)}\right)'{{I}'_{0, 1}}(h) +\left(\frac{{{\beta}_{1} }(h)}{{\gamma_1}(h)}\right)'{{{I}'_{4, 1}}(h)}+\frac{1}{{\gamma_1}(h)}[{{{\alpha}_{1} }(h)}{{{I}''_{0, 1}}(h)} +{{{\beta}_{1} }(h)}{{{I}''_{4, 1}}(h)}+{{{\gamma}_{1} }(h)}{{{I}''_{8, 1}}(h)}]\\ =&\frac{1}{D(h){\gamma^2_1}(h)}[{{{\alpha}_2}(h)}{{I}'_{0, 1}}(h)+{{{\beta_2} }(h)}{{I}'_{4, 1}}(h)], \end{align} $

其中

$ \begin{array}{c} {{\alpha}_2}(h)=D(h)[{\alpha'_1(h)}{\gamma}_{1}{(h)}-{\alpha_1(h)}{\gamma'}_1(h)]+a_{11}(h){\alpha_1(h)}{\gamma}_{1}(h) +a_{21}(h){\beta_1(h)}{\gamma}_{1}(h)+a_{31}(h){\gamma}^2_{1}(h), \\ {{\beta}_2}(h)=D(h)[{\beta'_1}(h){\gamma}_{1}{(h)}-{\beta_1(h)}{\gamma'}_{1}(h)]+a_{12}(h){\alpha_1(h)}{\gamma}_{1}(h) +a_{22}(h){\beta_1(h)}{\gamma}_{1}(h)+a_{32}(h){\gamma}^2_{1}(h).\\ \end{array} $

并且$ \deg {{{\alpha}_{2} }{(h)}\le 2k+1, \deg {{\beta}_{2}}(h)}\le 2k. $证毕.

引理4.2^[3]   设$ I'(h), M(h), {\gamma}_{1}(h) $在$ \Sigma $中的零点个数分别为$ \#I'(h), \#M(h), \#{\gamma}_{1}(h) $, 则

$ \begin{align} \#I'(h)\le\#M(h)+\#{\gamma_1} (h)+1. \end{align} $

(4.4)

定理4.3   对Hamilton函数(1.3), $ W(h)=\frac{M(h)}{{I}'_{0, 1}(h)} $满足Riccati方程

$ \begin{align} D(h){\beta_2(h)}{W'(h)}={R_0(h)}{W^2(h)+{R_1(h)}{W(h)}+R_2(h)}, \end{align} $

(4.5)

其中

$ \begin{align*} {R_0}(h)=&-a_{12}(h), {R_1}(h)=D(h){\beta'}_2(h)+[a_{32}(h)-a_{11}(h)]{{\beta_2}(h)}+{2a_{12}(h)}{{\alpha}_{2}(h)}, \\ {R_2}(h)=&D(h)[{{\alpha'}_{2}}(h){{\beta_2}(h)}-{{\alpha}_2}(h){\beta'_2}(h)]-[{a_{32}}(h)-a_{11}(h)] {\alpha_2(h)}{\beta}_{2}(h)-a_{12}(h){\alpha^2_2}(h)\\ &+a_{31}(h){\beta}^2_{2}(h), \deg{R_2(h)}\le 4k+3. \end{align*} $

证   由式(4.3)可得

$ \begin{align} W(h)=\frac{M(h)}{{I}'_{0, 1}(h)}={{{\alpha}_2}(h)}+{{{\beta_2} }(h)}\omega(h), \end{align} $

(4.6)

所以

$ \begin{align} W'(h)={{{\alpha'}_2}(h)}+{{{\beta'_2} }(h)}\omega(h)+{{{\beta_2} }(h)}\omega'(h), \end{align} $

同时结合式(3.9)即可得证.

引理4.4^[5]  假设$ \Sigma_0=(a, b)\subset\Sigma $, 则当$ h\in\Sigma_0 $时, $ \#W(h)\le \#\beta_2 (h)+\#R_2(h)+1. $其中表达式$ \#\psi(h) $表示$ \psi(h) $在$ {\Sigma }_{0} $上的零点个数（计重数）.

定理1.1的证明   分别由引理4.1和引理4.3可得$ \deg {{\beta}_{2}}(h)\le 2k, \deg{R_2(h)}\le 4k+3. $从而由引理4.4可得

$ \begin{align} \#W(h)\le 6k+4. \end{align} $

由式(4.1)可得$ \deg {{\gamma}_{1}}(h)\le k-1 $. 从而由引理4.2可得

$ \begin{align} \#I'(h)\le 7k+4=7\left[\frac{n-1}{4}\right]+4. \end{align} $

因此, 对$ h\in\Sigma=\left(0, \frac{\sqrt{3}}{12}\right) $且在周期环域$ \Gamma_h $上时, $ B(n)\le 7\left[\frac{n-1}{4}\right]+5. $定理1.1证毕.