德国数学家Hilbert在1900年世界数学家大会上提出了23个著名的数学问题^[20], 其中, 第十六个问题的第二部分是: 平面实$ n $次多项式自治系统

$ \frac{{\rm d}x}{{\rm d}t} = P_n(x, y), \ \frac{{\rm d}y}{{\rm d}t} = Q_n(x, y) $

的极限环最大个数及分布如何? 一百多年来, 出现了大量这方面的工作, 但是, 这个问题是非常难的, 即使对于$ n = 2 $的情形都还没有被完全解决, 仍然是一个公开问题, 这个问题研究进展和部分有效的研究方法可以参见综述性文献[4, 6, 23-24, 27, 32-33]. 基于Hilbert第十六问题的研究难度, Arnold^[2]在1977年首次提出弱化形式的Hilbert第十六问题^{[1, 3]}, 研究如下系统的Abel积分零点个数, 即系统的Poincaré 分支意义下的极限环个数

(1.1) $ \begin{equation} \dot{x} = {H}_y + \varepsilon p(x, y), \ \ \dot{y} = -{H}_x + \varepsilon q(x, y), \end{equation} $

其中$ p $, $ q $和$ H $是关于$ x $和$ y $的多项式, 次数满足：$ \max\{deg (p), deg (q)\} = n $和$ deg (H) = n+1 $. $ |\varepsilon| $是充分小的正数. 当$ \varepsilon = 0 $时, 系统(1.1) 变为Hamiltonian系统

(1.2) $ \begin{equation} \dot{x} = {H}_y, \ \ \dot{y} = -{H}_x. \end{equation} $

假设闭曲线簇$ \{L_h\} \subset\{(x, y)| H(x, y) = h, \ h\in J\} $是系统(1.2)的周期轨道, $ J $为开区间即对应周期环域的Hamiltonian函数值区间, 端点值为中心处和同异宿轨道的Hamiltonian量. 对应于系统(1.1)的Abel积分为

(1.3) $ \begin{equation} I(h, \delta) = \oint_{L_h}q{\rm d}x-p{\rm d}y, \ \ h\in J. \end{equation} $

事实上, $ I(h, \delta) $是系统(1.2)后继函数或Poincaré映射的关于$ \varepsilon $渐近展开式的一阶近似, $ I(h, \delta) $的零点的最大个数提供了系统(1.2)的极限环分支的最大个数, 参见文献[2, 11, 15-16, 22].

然而, 弱化形式的Hilbert第十六问题依旧难度很大, 目前仅对$ n = 2 $的情况彻底解决, 见多位学者的系列工作^{[12, 21, 26, 42]}, 统一的证明见文献[4]. 学者们关注更加简单的形式的系统(1.2), 具有较强应用背景的多项式Liénard系统

(1.4) $ \begin{equation} \dot{x} = y, \ \dot{y} = -g(x)+ \varepsilon f(x)y, \end{equation} $

其中$ deg(g(x)) = m $和$ deg(f(x)) = n $, 上述系统称为类型$ (m, n) $的Liénard系统.

Dumortier和Li^[7-10]系统的研究了不同情况的类型为$ (3, 2) $系统(1.4), 得到Abel积分零点个数并得到其上确界及分支图.

对类型$ (4, 3) $系统(1.4)

(1.5) $ \begin{equation} \dot{x} = y, \ \dot{y} = -x(x-1)(x-\alpha)(x-\beta)+ \varepsilon f(x)y, \end{equation} $

其中$ \alpha $和$ \beta $是实参数. 根据参数的不同取值, 系统$ (1.5)_{\varepsilon = 0} $具有6种不同情况的周期环域相图^[41], 见图 1.

当$ \alpha = \beta = 0 $时, 系统$ (1.5)_{\varepsilon = 0} $周期环域边界分别是初等中心和连接幂零鞍点的同宿轨, 见图 1(a). 王和肖^[40]证明了此情况的Abel积分零点个数上界是4和下界是3.

当$ \alpha = \beta<0 $或者$ \alpha = 0, \beta>0 , \beta\neq 1 $时, 系统$ (1.5)_{\varepsilon = 0} $周期环域内边界是初等中心外边界是连接双曲鞍点的同宿轨, 环域外部是幂零尖点, 如图 1(b). 孙和吴^[36]研究了当$ \alpha = \beta = -3 $时的特殊情况, 得到其Abel积分零点个数上确界. 对于一般的$ \alpha = \beta<0 $未见相关结果.

当$ \alpha = \beta = 1 $时, 系统$ (1.5)_{\varepsilon = 0} $周期环域内外边界分别是幂零中心和连接双曲鞍点的同宿轨, 如图 1(c). 孙, 苏和韩^[35]以及王^[39]分别研究了此情况的Abel积分零点个数, 得到其上界是5或6和下界是3. 孙和黄^[34]证明了另一类此情况的Abel积分零点个数上界是6和下界是4.

当$ 0<\alpha = \beta<1 $或者$ \alpha = 0, \beta<0 $时和$ \frac{2}{5}<\alpha<1, \beta = 1 $时, 系统$ (1.5)_{\varepsilon = 0} $周期环域边界分别是初等中心和连接幂零鞍点的同宿轨, 环域中间有一个幂零尖点同宿轨, 如图(e) 和(f). 关于上界问题未见相关发表结果.

本文考虑图 1中(d) 的情形, 研究如下类型为$ (4, 3) $的Liénard系统

(1.6) $ \begin{equation} \dot{x} = y, \ \dot{y} = x(x+1)(x-\frac{7}{4})^2+\varepsilon (a_0+a_1x+a_2x^2+x^3)y, \end{equation} $

其中$ 0 <\varepsilon \ll 1 $, $ a_0 $, $ a_1 $和$ a_2 $是有界实参数, 对应的Hamiltonian函数是

(1.7) $ \begin{equation} H(x, y) = \frac{1}{2}y^2+\frac{49}{32}x^2-\frac{7}{48}x^3-\frac{5}{8}x^4+\frac{1}{5}x^5. \end{equation} $

$ L_h = H(x, y) = h, h\in(0, \frac{409}{480}) $为(1.7)式定义的闭轨线族, 见图 2, 即对应于系统$ (1.7)_{\varepsilon = 0} $的逆时针方向闭轨线族. $ {H}(x, y) = \frac{409}{480} $定义通过双曲鞍点$ (-1, 0) $的同宿环, 记为$ L_{\frac{409}{480}} $, 幂零尖点$ (\frac{7}{4}, 0) $在同宿环的外侧.

系统$ (1.6) $对应如下的Abel积分

(1.8) $ \begin{equation} I(h, \delta) = \oint_{L_h}(a_0+a_1x+a_2x^2+x^3)y {\rm d}x\equiv a_0I_0(h)+a_1I_1(h)+a_2I_2(h)+I_3(h), \end{equation} $

其中$ \ h\in(0, \frac{409}{480}) $, $ \delta = (a_0, a_1, a_2) $, $ I_i(h) = \oint_{L_h}x^iy{\rm d}x $$ (i = 0, 1, 2, 3) $.

本文第二部分给出了相关的定义和引理; 第三部分证明了$ \{I_0, I_1, I_2, I_3\} $是精度为$ 3 $的Chebyshev系统, 得到Abel积分$ I(h, \delta) $零点个数的上界是$ 6 $; 第四部分证明$ I(h, \delta) $零点个数下界是3.

关于系统(1.1)的Abel积分零点个数上界, 根据不同特点的系统, 学者们提出了辐角原理^{[23, 31]}、直接法^{[5, 13, 25, 29]}、几何方法^{[7-10, 28]}等很多有效的方法. 本文使用Abel积分生成元的Chebyshev系统判定理论来研究Abel积分零点个数上界, 相关的定义详见参考文献[13, 25].

定义2.1   设$ f_0(x), f_1(x), \cdots, f_{n-1}(x) $是定义在开区间$ J $上的解析函数,

(ⅰ) 如果任意的线性组合

$ k_0f_0(x) + k_1f_1(x)+ \cdots + k_{n-1}f_{n-1}(x) $

在$ J $上至多有$ n-1 $个孤立零点, 那么$ \{f_0(x), f_1(x), \cdots, f_{n-1}(x)\} $称为Chebyshev系统.

(ⅱ) 如果对每个$ i = 1, 2, \cdots, n $, 任意线性组合$ k_0f_0(x) + k_1f_1(x)+ \cdots + k_{i-1}f_{i-1}(x) $至多有$ i-1 $个零点, 考虑零点重数, 那么有序函数列$ \{f_0(x), f_1(x), \cdots, f_{n-1}(x)\} $称为完全Chebyshev系统.

(ⅲ) 对于任意的$ k = 1, 2, \cdots, n $, 系统$ \{f_0(x), f_1(x), \cdots, f_{n-1}(x)\} $是在区间$ J $上的一个完全Chebyshev系统当且仅当对任意的$ x\in J $时, 如下Wronskian行列式

$ W[f_0, f_1, f_2, \cdots, f_{k-1}](x)\neq0 $

成立, 这里的

$ W[f_0, f_1, \cdots, f_{k-1}](x) = \det(f_i^j)_{0\leq i, j\leq k-1} = \left| \begin{array}{cccc} f_0(x) & f_1(x) & \cdots & f_{k-1}(x) \\ f_0^{'}(x) & f_1^{'}(x) & \cdots & f_{k-1}^{'}(x) \\ \cdots & \cdots & \cdots & \cdots \\ f_0^{(k-1)}(x) & f_1^{(k-1)}(x) & \cdots & f_{k-1}^{(k-1)}(x) \end{array} \right|, $

其中$ f'(x) $和$ f^{(i)}(x) $分别代表$ f(x) $的一阶导数和$ i $阶导数.

考虑在平面某开子集上解析的具有如下形式的Hamiltonian函数

$ H(x, y) = A(x)+\frac{1}{2}y^{2}, $

假定它在原点$ O(0, 0) $有局部最小值$ H(0, 0) = 0 $, 环绕中心$ O(0, 0) $有一族顺时针方向的闭轨$ \{L_h\}\subset \{(x, y)|H(x, y) = h, 0<h<h_0\} $, 形成未扰系统的周期环域. 每条周期轨与$ x $ -轴有两个交点, 设区间$ (x_l, x_r) $是周期轨族在$ x $ -轴上的投影. 根据假设可知, $ x A'(x)>0, x\in (x_l, x_r)\backslash\{0\} $, 因此, $ A(x) $定义了一个对合$ z(x) $满足$ A(x) = A(z(x)) $, 当$ 0<x<x_r $时, $ x_l<z(x)<0 $. 设$ f_i(x) $是解析函数, 考虑沿着每条轨线$ L_h $的如下形式的Abel积分

(2.1) $ \begin{eqnarray} I_{i}(h) = \oint_{L_h} f_i(x)y^{2s-1}{\rm d}x, \ \ i = 0, 1, \cdots, n-1, \end{eqnarray} $

关于其线性组合的零点个数有如下结果(见文献[30]).

引理2.1   假定上述条件成立, 考虑$ (2.1) $式定义的$ \rm Abel $积分$ I_i(h) $, 并定义

$ l_i(x): = \frac{f_i(x)}{A'(x)}- \frac{f_i(z(x))}{A'(z(x))}. $

如果以下条件成立

(ⅰ) $ W[l_0, \cdots , l_i]\neq 0 $, 当$ x\in (0, x_r) $和所有的$ i = 0, 1, \cdots, n-2, $

(ⅱ) $ W[l_0, \cdots , l_{n-1}] $在$ (0, x_r) $上有$ k $个零点(考虑重数),

(ⅲ) $ s> n+k-2 $,

那么任何非平凡线性组合$ \{I_0, I_1, \cdots, I_{n-1}\} $在$ (0, h_0) $至多有$ n+k-1 $个零点(考虑重数). 此时称$ \{I_0, I_1, \cdots, I_{n-1}\} $在$ (0, h_0) $上是精度为$ k $的Chebyshev系统.

在研究一些系统时, $ s $的值不能满足条件(ⅲ), 引理2.1并不能直接应用, 此时可以使用如下引理2.2, 提高$ I_i(h) $中$ y $的次数, 克服上述困难, 参见文献[13]中的引理4.1.

引理2.2   设$ F(x) $是关于$ x $的一元函数, 并且满足$ \frac{F(x)}{A'(x)} $在$ x = 0 $处解析. 对于任意$ k \in N $, 沿着$ {A(x) + \frac{1}{2}y^2 = h} $的每条轨线$ L_h $, 如下等式成立

$ \oint_{L_h} F(x)y^{k-2} {\rm d}x = \oint_{L_h}G(x)y^k {\rm d}x, $

其中$ G(x) = \frac{1}{k}(\frac{F(x)}{A'(x)})'(x) $.

Abel积分零点个数的Chebyshev性质的代数判定方法最初的思想是来源于文献[25], 后来在文献[13]和文献[30]中得以推广和发展.

计算Abel积分零点个数的一个有效的方法是研究(1.2) 周期环域$ \{L_{h}\} $端点附近的渐近展开式(参见文献[14, 37]及[38]). 假如系统(1.2)的周期环域的内边界是初等中心$ C(0, 0) $满足$ {H}(C) = 0 $, 外边界是连接双曲鞍点$ S_1(x_1, y_1) $的同宿环$ L_{h_s} $, 满足$ {H}(x_1, y_1) = h_s $, 那么

(2.2) $ \begin{equation} M(h, \delta) = \oint_{L_h}q{\rm d}x-p{\rm d}y, \ \ 0<h<h_s. \end{equation} $

引理2.3^[19]   假设$ (x, y) $在初等中心$ C(0, 0) $附近

(2.3) $ \begin{equation} \begin{array}{l} { } {H}(x, y) = \frac{y^2}{2}+ \frac{x^2}{2}+\sum\limits_{i+j\geq 3}h_{ij}x^iy^j, \\ { } p(x, y, \delta) = \sum\limits_{i+j\geq 0}a_{ij}x^iy^j, \ q(x, y, \delta) = \sum\limits_{i+j\geq0}b_{ij}x^iy^j, \end{array} \end{equation} $

那么

(2.4) $ \begin{equation} M(h, \delta) = \sum\limits_{j\geq 0}b_{j}(\delta)h^{j+1}, \ 0<h\ll 1. \end{equation} $

引理2.4^[18-19]  假设$ (x, y) $在双曲鞍点$ S_1(x_1, y_1) $附近

(2.5) $ \begin{equation} \begin{array}{ll} { } {H}(x, y) = h_s+\frac{\lambda}{2}(y-y_1)^2-\frac{\lambda}{2}(x-x_1)^2+\sum\limits_{i+j\geq 3} \overline{h}_{ij}(x-x_1)^i(y-y_1)^j, \ \lambda\neq0, \\ p(x, y, \delta) = \sum\limits_{i+j> 0}\overline{a}_{ij}(x-x_1)^i(y-y_1)^j, \ q(x, y, \delta) = \sum\limits_{i+j> 0}\overline{b}_{ij}(x-x_1)^i(y-y_1)^j, \end{array} \end{equation} $

那么Melnikov函数$ M(h, \delta) $有如下展开式

(2.6) $ \begin{eqnarray} M(h, \delta)& = &{c}_0(\delta)+c_1(\delta)(h-h_s)\ln|h-h_s|+ {c}_2(\delta)(h-h_s)+{c}_3(\delta)(h-h_s)^2\ln|h-h_s|{}\\ &&+O(|h-h_s|^2), \ \ 0< h_s-h\ll 1, \end{eqnarray} $

其中

$ {c}_{0}(\delta) = M(h_s, \delta) = \oint_{L_{h_s}}q{\rm d}x-p{\rm d}y, \ \ {c}_{1}(\delta) = -\frac{\bar a_{10}+\bar b_{01}}{|\lambda|}, $

$ {c}_{2}(\delta) = \oint_{L_{h_s}}(p_{x}+q_{y}-\bar a_{10} -\bar b_{01}){\rm d}t+b{c}_{1}, $

$ {c}_{3}(\delta) = -\frac{1}{2}\{(- 3\bar a_{30}- \bar b_{21}+\bar a_{12}+3\bar b_{03})- [(2\bar b_{02}+\bar a_{11})(3\bar h_{03}-\bar h_{21})+(2\bar a_{20}+\bar b_{11})(3\bar h_{30}-\bar h_{12})]\}+\bar{b}{c}_1, $

$ b $和$ \bar{b} $是确定的常数.

引理2.5   如果展开式(2.4) 和(2.6) 满足

(2.7) $ \begin{equation} {\rm rank}\frac{\partial(c_0, c_1, \cdots, c_{m-1}, b_0, b_1, \cdots, b_{k-1})}{\partial\delta}(\delta_0) = m+k, \end{equation} $

并存在$ \delta_0\in\mathbb{R}^N $和正整数$ m $和$ k $使得如下条件成立

(2.8) $ \begin{equation} \begin{array}{ll} c_0(\delta_0) = c_1(\delta_0) = , \cdots, = c_{m-1}(\delta_0) = 0, \ c_m(\delta_0)\neq 0, \\ b_0(\delta_0) = b_1(\delta_0) = \cdots = b_{k-1}(\delta_0) = 0, \ b_{k}(\delta_0)\neq 0. \end{array} \end{equation} $

令$ N^* = m+k+\frac{1-sgn(M(h_1, \delta_0)M(h_2, \delta_0))}{2} $, 其中$ h_1 = h_s-\varepsilon_0 $, $ h_2 = 0+\varepsilon_0 $和$ \varepsilon_0 $是充分小的正参数. 那么存在$ \delta_0 $附近的参数$ \delta $使得函数$ M(h, \delta) $存在$ N^* $个零点, 其中$ m $个零点在$ h = h_s $附近, $ k $个零点在$ h = 0 $附近.

注2.1   引理2.5是文献[17]的转述, 较为清晰的证明见文献[36].

本部分我们利用引理2.2和引理2.3以及结合相关代数定理和分析技巧研究(1.8)中Abel积分$ I(h, \delta) $的生成元

(3.1) $ \begin{equation} I_i = \oint_{L_h}x^{i}y{\rm d}x, \ i = 0, 1, 2, 3, \end{equation} $

证明$ \{I_0, I_1, I_2, I_3\} $在$ (0, \frac{409}{480}) $上是构成精度为$ 3 $的Chebyshev系统.

系统$ (1.6) $的Hamiltonian函数$ H(x, y) $定义的周期轨线形成的周期环域在$ x $轴上的投影是$ (-1, x_r) $, 其中

$ x_r = \frac{41(2245+110\sqrt{409})^{\frac{1}{3}}-(2245+110\sqrt{409})^{\frac{2}{3}}-45}{24(2245+110\sqrt{409})^{\frac{1}{3}}}\approx 0.9081569122. $

记$ H(x, y): = A(x)+\frac{1}{2}y^{2} $, 其中$ A(x) = \frac{49}{32}x^2-\frac{7}{48}x^3-\frac{5}{8}x^4+\frac{1}{5}x^5 $, 由引理2.2的可知, 此时$ n = 4, s = 1 $, 不满足$ s >n+k-2 $.

下面根据引理2.2中的方法提高$ I_i(h) $中$ y $的次数.

因为在每条轨线$ L_h $上有$ \frac{2A(x)+y^2}{2h} = 1 $成立, 所以

(3.2) $ \begin{equation} I_i(h) = \frac{1}{2h}\oint_{L_h}(2A(x)+y^2)x^{i}y{\rm d}x = \frac{1}{2h}(\oint_{L_h}2x^{i}A(x)y{\rm d}x+\oint_{L_h}x^{i}y^3{\rm d}x), \ \ i = 0, 1, 2, 3. \end{equation} $

令$ k = 3 $和$ F(x) = 2x^{i}A(x) $, 根据引理2.2, 得到

(3.3) $ \begin{equation} \oint_{L_h}2x^{i}A(x)y{\rm d}x = \oint_{L_h}G_i(x)y^3{\rm d}x, \end{equation} $

其中$ G_i(x) = \frac{\rm d}{3{\rm d}x}(\frac{2x^{i}A(x)}{A'(x)}) = \frac{ x^{i} g_i(x)}{45(x+1)^{2}(4\, x-7 )^{3}} $, $ g_i(x) = 384\, i{x}^{5}-1488\, i{x}^{4}+384\, {x}^{5}-52\, i{ x}^{3}-1248\, {x}^{4}+5250\, i{x}^{2}+592\, {x}^{3}-1715\, ix+910\, {x}^{2} -5145\, i-1960\, x-5145. $

根据$ (3.2) $式和$ (3.3) $式以及考虑到在每条轨线$ L_h $上$ \frac{2A(x)+y^2}{2h} = 1 $, 可得

(3.4) $ \begin{eqnarray} I_i(h)& = & \frac{1}{2h}\oint_{L_h}(x^{i}+G_i(x))y^3{\rm d}x = \frac{1}{4h^2}\oint_{L_h}(2A(x)+y^2)(x^{i}+G_i(x))y^3{\rm d}x {}\\ & = & \frac{1}{4h^2}\oint_{L_h}2A(x)(x^{i}+G_i(x))y^3{\rm d}x+ \frac{1}{4h^2}\oint_{L_h}(x^{i}+G_i(x))y^5{\rm d}x. \end{eqnarray} $

取$ k = 5 $和$ F(x) = 2A(x)(x^{i}+G_i(x)) $, 以及由引理2.2可知

(3.5) $ \begin{equation} \oint_{L_h}2A(x)(x^{i}+G_i(x))y^3{\rm d}x = \oint_{L_h}\widetilde{G}_i(x)y^5{\rm d}x, \end{equation} $

其中, $ \widetilde{G}_i(x) = \frac{\rm d}{5{\rm d}x}(\frac{2A(x)(x^{i}+G_i(x))}{A'(x)}) = \frac{x^i\widetilde{g}_i(x)}{3375\, \left( x+1 \right) ^{4} \left( 4\, x-7 \right) ^{6} } $, 而

$ \begin{eqnarray*} \widetilde{g}_ i(x)& = &147456\, {i}^{2}{x}^{10}-1142784\, {i}^{2}{x}^{9} +1400832\, i{x}^{10}+2174208\, {i}^{2}{x}^{8}-9888768\, i{x}^{9}\\ &&+1253376 \, {x}^{10}+4186752\, {i}^{2}{x}^{7}+17451264\, i{x}^{8}-8146944\, {x}^{9} -16938416\, {i}^{2}{x}^{6}\\ &&+25995648\, i{x}^{7}+14505984\, {x}^{8}+606480 \, {i}^{2}{x}^{5}-100110736\, i{x}^{6}+9629568\, {x}^{7}\\ &&+43052380\, {i}^{2 }{x}^{4}+7873740\, i{x}^{5}-51823968\, {x}^{6}-17472420\, {i}^{2}{x}^{3}+ 201083750\, i{x}^{4}\\ &&+16222920\, {x}^{5}-51081275\, {i}^{2}{x}^{2}- 82489785\, i{x}^{3}+84509810\, {x}^{4}+17647350\, {i}^{2}x\\ &&-222452650\, i{x }^{2}-58255120\, {x}^{3} +26471025\, {i}^{2}+88236750\, ix-58524375\, {x}^{ 2}\\ &&+132355125\, i+75631500\, x+105884100. \end{eqnarray*} $

由$ (3.4) $式和$ (3.5) $式以及考虑到在每条轨线$ L_h $上$ \frac{2A(x)+y^2}{2h} = 1 $, 可知

(3.6) $ \begin{eqnarray} I_i(h)& = &\frac{1}{4h^2} \oint_{L_h}(x^{i}+G_i(x)+\widetilde{G}_i(x))y^5{\rm d}x{}\\ & = &\frac{1}{8h^3}\oint_{L_h}(2A(x)+y^2)(x^{i}+G_i(x)+\widetilde{G}_i(x))y^5{\rm d}x {}\\ & = &\frac{1}{8h^3} \oint_{L_h}2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x))y^5{\rm d}x+ \frac{1}{8h^3}\oint_{L_h}(x^{i}+G_i(x)+\widetilde{G}_i(x))y^7{\rm d}x.{\quad} \end{eqnarray} $

令$ k = 7 $和$ F(x) = 2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)) $, 根据引理2.2, 得到

(3.7) $ \begin{equation} \oint_{L_h}2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x))y^5{\rm d}x = \oint_{L_h}\overline{G}_i(x)y^7{\rm d}x, \end{equation} $

其中$ \overline{G}_i(x) = \frac{\rm d}{7{\rm d}x}(\frac{2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x))}{A'(x)}) = \frac{x^i\overline{g}_i(x)}{354375\, \left( x+1 \right) ^{6} \left( 4\, x-7 \right) ^{9} } $, 其中

$ \begin{eqnarray*} \overline{g}_i(x)& = & 56623104\, {i}^{3}{x}^{15}-658243584\, {i}^{3}{x} ^{14}+1302331392\, {i}^{2}{x}^{15}+2527690752\, {i}^{3}{x}^{13}\\ &&- 14219476992\, {i}^{2}{x}^{14}+7743209472\, i{x}^{15}-793939968\, {i}^{3}{ x}^{12}+51375144960\, {i}^{2}{x}^{13}\\ &&-79557230592\, i{x}^{14}+6497501184 \, {x}^{15}-19099800576\, {i}^{3}{x}^{11}-18083450880\, {i}^{2}{x}^{12}\\ &&+ 271507341312\, i{x}^{13}-63350636544\, {x}^{14}+37835345664\, {i}^{3}{x}^ {10}-323024188416\, {i}^{2}{x}^{11}\\ &&-108468449280\, i{x}^{12}+ 209745838080\, {x}^{13}+40641774272\, {i}^{3}{x}^{9}+604787884800\, {i}^{ 2}{x}^{10}\\ &&-1411402994688\, i{x}^{11}-123970240512\, {x}^{12}- 178096151520\, {i}^{3}{x}^{8}+572452435392\, {i}^{2}{x}^{9}\\ &&+ 2523116354304\, i{x}^{10}-812050421760\, {x}^{11}+14837592000\, {i}^{3}{x }^{7}-2394192638880\, {i}^{2}{x}^{8}\\ &&+2114953350976\, i{x}^{9}+ 1557227612160\, {x}^{10}+395827117560\, {i}^{3}{x}^{6}+198334846080\, {i} ^{2}{x}^{7}\\ &&-8548783943280\, i{x}^{8}+803678789760\, {x}^{9}-182123533200 \, {i}^{3}{x}^{5}+4634586405960\, {i}^{2}{x}^{6}\\ &&+751071173860\, i{x}^{7}- 4313677838400\, {x}^{8}-500022535950\, {i}^{3}{x}^{4}-2014540802400\, {i} ^{2}{x}^{5}\\ &&+14561090677630\, i{x}^{6}+1220542215200\, {x}^{7}+ 268772081725\, {i}^{3}{x}^{3}-5354037079650\, {i}^{2}{x}^{4}\\ &&- 6187921606825\, i{x}^{5}+5985510026650\, {x}^{6}+371520835875\, {i}^{3}{x }^{2}+2832990861225\, {i}^{2}{x}^{3}\\ &&-14961600223675\, i{x}^{4}- 3791441669400\, {x}^{5}-136193423625\, {i}^{3}x+3873019786125\, {i}^{2}{x }^{2}\\ &&+8096663004500\, i{x}^{3}-5213740226975\, {x}^{4}-136193423625\, {i} ^{3}-1485156857625\, {i}^{2}x\\ &&+10295728700250\, i{x}^{2}+4384625286300\, { x}^{3}-1498127659875\, {i}^{2}-4591663996500\, ix\\ &&+2697926868000\, {x}^{2} 4630576403250\, i-3398350189500\, x-3268642167000. \end{eqnarray*} $

根据$ (3.6) $式和$ (3.7) $式以及考虑到在每条轨线$ L_h $上$ \frac{2A(x)+y^2}{2h} = 1 $, 得到

(3.8) $ \begin{eqnarray} I_i(h)& = & \frac{1}{8h^3} \oint_{L_h}(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x))y^7{\rm d}x {}\\ & = & \frac{1}{16h^4}\oint_{L_h}(2A(x)+y^2)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x))y^7{\rm d}x {}\\ & = & \frac{1}{16h^4}(\oint_{L_h}2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x))y^7{\rm d}x{}\\ &&+ \oint_{L_h}(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x))y^9{\rm d}x). \end{eqnarray} $

取$ k = 9 $, $ F(x) = 2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)) $, 根据引理2.2, 得到

(3.9) $ \begin{equation} \oint_{L_h}2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x))y^7{\rm d}x = \oint_{L_h}\widehat{G}_i(x)y^9{\rm d}x, \end{equation} $

其中$ \widehat{G}_i(x) = \frac{\rm d}{9{\rm d}x}(\frac{2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x))}{A'(x)}) = \frac{x^i\widehat{g}_i(x)}{47840625\, \left( x+1 \right) ^{8} \left( 4\, x-7 \right) ^{12}} $, 而$ \widehat{g}_i(x) $是一个项数较多的多项式, 详见附录A.

由$ (3.8) $和$ (3.9) $式, 以及考虑到在每条轨线$ L_h $上$ \frac{2A(x)+y^2}{2h} = 1 $, 得到

(3.10) $ \begin{eqnarray} I_i(h)& = & \frac{1}{16h^4}\oint_{L_h}(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))y^9{\rm d}x {}\\ & = & \frac{1}{32h^5}\oint_{L_h}(2A(x)+y^2)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))y^9{\rm d}x {}\\ & = & \frac{1}{32h^5}(\oint_{L_h}2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))y^9{\rm d}x{}\\ &&+ \oint_{L_h}(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))y^{11}{\rm d}x). \end{eqnarray} $

取$ k = 11 $, $ F(x) = 2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))y^9 $, 根据引理2.2, 得到

(3.11) $ \begin{equation} \oint_{L_h}2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))y^9{\rm d}x = \oint_{L_h}\check{G}_i(x)y^{11}{\rm d}x, \end{equation} $

其中$ \check{G}_i(x) = \frac{\rm d}{11{\rm d}x}(\frac{2A(x)(x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x))}{A'(x)}) = \frac{x^i\check{g}_i(x)}{7893703125\, \left( x+1 \right) ^{10} \left( 4\, x-7 \right) ^{15}, } $$ \check{g}_i(x) $是一个较长的多项式, 详见附录A.

把$ (3.11) $式带入$ (3.10) $式中得

$ 32h^5I_i(h) = \oint_{L_h}f_i(x)y^{11}{\rm d}x\equiv \widetilde{I}_i(h), $

其中$ f_i(x) = x^{i}+G_i(x)+\widetilde{G}_i(x)+\overline{G}_i(x)+\widehat{G}_i(x)+\check{G}_i(x) $. 显然如果$ \{I_0, I_1, I_2, I_{3}\} $在$ (0, \frac{409}{480}) $上是精度为$ k $的Chebyshev系统, 那么当且仅当$ \{\widetilde{I}_0, \widetilde{I}_1, \widetilde{I}_2, \widetilde{I}_{3}\} $也是精度为$ k $的Chebyshev系统.

下面证明$ \{\widetilde{I}_0, \widetilde{I}_1, \widetilde{I}_2, \widetilde{I}_{3}\} $是精度为$ 3 $的Chebyshev系统, 此时$ s = 6 $, $ k = 3 $, 那么$ s > n+k-2 $. 取

$ l_i(x) = (\frac{f_i}{A'(x)})(x)-(\frac{f_i}{A'(x)})(z(x)), $

这里$ z(x) $是满足$ A(x) = A(z(x)) $与$ A(-1) = A(x_r) $的一个解析对合, 由$ A(x) = A(z) $得到

$ -\frac{1}{480}(x-z)q(x, z) = 0, $

其中

$ \begin{eqnarray*} \label{qq} q(x, z)& = &96\, {x}^{4}+96\, {x}^{3}z+96\, {x}^ {2}{z}^{2}+96\, x{z}^{3}+96\, {z}^{4}-300\, {x}^{3}-300\, {x}^{2}z-300\, x{ z}^{2}\\ & &-300{z}^{3}-70\, {x}^{2} -70\, xz-70\, {z}^{2}+735\, x+735\, z. \end{eqnarray*} $

事实上$ q(x, z) = 0 $是定义解析对合$ z(x) $的因式, 所以$ \frac{{\rm d}z}{{\rm d}x} = - \frac{\partial q(x, z)}{\partial x } / \frac{\partial q(x, z)}{\partial z } $, 因此

(3.12) $ \begin{equation} \frac{\rm d}{{\rm d}x}l_i(x) = \frac{\rm d}{{\rm d}x}(\frac{f_i}{A'(x)})(x) - \frac{\rm d}{{\rm d}z}[(\frac{f_i}{A'(x)})(z(x))] \times \frac{{\rm d}z}{{\rm d}x}. \end{equation} $

如果我们取$ x \in (-1, 0) $, 那么$ 0< z(x)< x_r $. 根据Sturm定理和相关分析技巧, 同时借助符号计算系统Maple.18, 我们有以下引理.

引理3.1  如下结论成立

(ⅰ) $ W[l_1(x)]\neq 0, x\in (-1, 0); $

(ⅱ) $ W[l_1(x), l_2(x)]\neq 0, x\in (-1, 0); $

(ⅲ) $ W[l_1(x), l_2(x), l_0(x)]\neq 0, x\in (-1, 0); $

(ⅳ) $ W[l_1(x), l_2(x), l_0(x), l_3(x)] \neq 0, x\in(-1, 0)\backslash \{x_1, x_2, x_3\}, $

其中, $ x_1, x_2, x_3 $是$ W[l_1(x), l_2(x), l_0(x), l_3(x)] $在$ (-1, 0) $上的单根.

上面四个Wronskian行列式都是关于$ (x, z) $的二元函数, 而且$ (x, z) $满足$ q(x, z) = 0 $, 也即$ z $是由$ q(x, z) $决定的关于$ x $的对合, 限定$ x \in(-1, 0) $, 此时$ 0 < z< x_r. $因此, 研究上述四个Wronskian行列式是否在$ (-1, 0) $上有零点, 我们只需研究上述四个Wronskian行列式与$ q(x, z) $是否有满足

(3.13) $ \begin{equation} -1<x<0 < z< x_r \end{equation} $

的公共根$ (x, z) $.

证  根据(3.12)式, 借助于符号计算系统Maple.18, 得到

$ \begin{array}{llll} W[l_1] = -\frac{16(x-z) p_1(x, z)}{7893703125\left( x+1 \right) ^{11} \left( 4\, x-7 \right) ^{17} \left( z+1 \right) ^{11} \left( 4\, z-7 \right) ^{17} }, \\ W[l_1, l_2] = \frac{-2048(x-z)^3 p_2(x, z)}{62310549025634765625\left( 4\, z-7 \right) ^{34} \left( z+1 \right) ^{21} \left( 4\, x-7 \right) ^{34} \left( x+1 \right) ^{21}p(x, z) }, \\ W[l_1, l_2, l_0] = \frac{65536(x-z)^6 p_3(x, z)}{70265853652016979217529296875 {z}^{3}{x}^{3} \left( 4\, x-7 \right) ^{50} \left( x+1 \right) ^{30} \left( 4\, z-7 \right) ^{50} \left( z+1 \right) ^{30}p^3(x, z) }, \\ W[l_1, l_2, l_0, l_3] = \frac{ 8388608(x-z)^{10} p_4(x, z)}{2401115967765017711699008941650390625{x}^{4}{z}^{4} \left( 4\, x-7 \right) ^{66} \left( x+1 \right) ^{38} \left( 4\, z-7 \right) ^{66} \left( z+1 \right) ^{38} p^{6}(x, z) }, \end{array} $

其中$ p(x, z) $, $ p_1(x, z) $, $ p_2(x, z) $, $ p_3(x, z) $和$ p_4(x, z) $是次数分别为$ 3 $, $ 52 $, $ 104 $, $ 157 $和$ 208 $的表达式较长的多项式. 计算$ p(x, z) $和$ q(x, z) $关于$ z $的结式得

$ \begin{eqnarray*} R(p, q, z)& = &9393093476352000\, {x}^{12}-70448201072640000\, {x}^{11}+120251165442048000\, {x}^{10}\\ &&+ 299722849910784000\, {x}^{9} -993818796097536000\, {x}^{8}-337425938251776000\, {x}^{7}\\ & &+ 2985665866629120000\, {x}^{6}-238540595134464000\, {x}^{5}- 4784259428831232000\, {x}^{4}\\ &&+691607350038528000\, {x}^{3}+ 4099730011776000000\, {x}^{2}-274628165575680000\, x\\ &&-1441797869272320000. \end{eqnarray*} $

由Sturm定理得, $ q(x, z) $和$ p(x, z) $没有满足(3.13) 的公共根, 也就是在$ (-1, 0) $上有$ R(p, q, z)\neq 0 $. 所以在$ (-1, 0) $上$ p(x, z)\neq 0 $. 因此, 要证明上述四个Wronskian行列式与$ q(x, z) $是否有满足(3.13)式的公共根, 我们只需证明$ p_i(x, z) $和$ q(x, z) $是否有满足(3.13)式的公共根, 其中$ i = 1, 2, 3, 4 $.

(ⅰ) 首先计算$ q(x, z) $和$ p_1(x, z) $关于$ z $的结式, 得到

$ R(q, p_1, z) = 19342813113834066795298816\cdot \left( x+1 \right) ^{10} \left( 4\, x-7 \right) ^{32}\xi_{1r}(x), $

其中$ \xi_{1r}(x) $是关于$ x $的次数为$ 166 $的项数较多的多项式. 应用Sturm定理可得, 在$ x\in(-1, 0) $上, $ \xi_{1r}(x)\neq 0 $. 因此, $ p_1(x, z) $和$ q(x, z) $没有满足不等式(3.13)的公共根. 所以当$ x\in (-1, 0) $时$ W[l_1] \neq0 $.

(ⅱ) 为了探讨$ q(x, z) $和$ p_2(x, z) $是否有满足条件的公共根, 我们把$ q(x, z) $和$ p_2(x, z) $代入到文献[37]中的附录A的程序中

> with(RegularChains):

> with(ChainTools):

> with(SemiAlgebraicSetTools):

> sys : = [p_2(x, z), q(x, z)]:

> R : = PolynomialRing([x, z]):

> dec : = Triangularize(sys, R);

  [regular_chain, regular_chain, regular_chain]

> L : = map(Equations, dec, R);

$ [[p^r_1(x, z), p^r_2(z)], [4z-7, 4x-7], [z+1, x+1]], $

其中$ p^r_1(x, z) = p_{11}(z)x+p_{12}(z) $, $ p_{11}(z) $是次数为$ 278 $的多项式, $ p_{12}(z) $是次数为$ 279 $的多项式, $ p^r_{2}(z) $是次数为$ 334 $的多项式. 在Maple程序中, 正则链$ L[1][1] $和L[1][2]分别表示$ p^r_1 $和$ p^r_2 $. 显然两个正则链$ [4z-7, 4x-7] $和$ [z+1, x+1] $没有满足条件(3.13)的零点. 正则链$ [p^r_1, p^r_2] $是无平方因子而且是零维的(因为变量个数等于方程个数), 调整正则链$ [p^r_1, p^r_2] $中的多项式顺序然后实根隔离得到

> C : = Chain([L[1][2], L[1][1]], Empty(R), R);

        regular_chain

> RL : = RealRootIsolate(C, R, 'abserr' = 1/10^5);

        [box, box, box, box, box, box]

> map(BoxValues, RL, R);

$ \begin{eqnarray*} &&[x = [-\frac{166047}{131072}, -\frac{83023}{65536}], z = [-\frac{12847550852447264786699110431560920661}{21267647932558653966460912964485513216}, -\frac{25695101704894529573398220863121841321}{42535295865117307932921825928971026432}]] \\ &&[x = [\frac{57225}{65536}, \frac{114451}{131072}], z = [-\frac{130579124207821927758551173057129720387771624925469323192599402428584276639511331} {118571099379011784113736688648896417641748464297615937576404566024103044751294464}, \\ &&-\frac{65289562103910963879275586528564860193885812462734661596299701214292138319755665} {59285549689505892056868344324448208820874232148807968788202283012051522375647232}]] \\ && [x = [-\frac{19795}{32768}, -\frac{79179}{131072}], z = [-\frac{1067307572670673929555921262095838065161724660961973744789276451429}{842498333348457493583344221469363458551160763204392890034487820288}, \\ && -\frac{266826893167668482388980315523959516290431165240493436197319112857}{210624583337114373395836055367340864637790190801098222508621955072}]] \\ && [x = [\frac{257575}{524288}, \frac{128789}{262144}], z = [-\frac{69277551632659850252385127838921853737544978589010398489912099249}{52656145834278593348959013841835216159447547700274555627155488768}, \\ && -\frac{138555103265319700504770255677843707475089957178020796979824198497}{105312291668557186697918027683670432318895095400549111254310977536}]] \\ &&[x = [-\frac{144347}{131072}, -\frac{72173}{65536}], z = [\frac{152130703691490297053410098983442873154101}{174224571863520493293247799005065324265472}, \frac{76065351845745148526705049491721436577051}{87112285931760246646623899502532662132736}]] \\ &&[x = [-\frac{172447}{131072}, -\frac{86223}{65536}], z = [\frac{8850210479459909}{18014398509481984}, \frac{17700420958919819}{36028797018963968}]], \end{eqnarray*} $

上面得到的6个区间对是$ p_2(x, z) $和$ q(x, z) $在整个$ x-z $平面上所有公共根存在的区间. 显然$ p_2(x, z) $和$ q(x, z) $没有满足(3.13)的公共根. 因此, 当$ x\in (0, 1) $时$ W[l_1, l_2] \neq0 $.

(ⅲ) 把$ q(x, z) $和$ p_3(x, z) $代入到上面的算法程序中, 三角分解得到正则链

$ [[p^r_3(x, z), p^r_4(z)], [4z-7, 4x-7], [z+1, x+1]], $

其中$ p^r_3(x, z) = p_{31}(z)x+p_{32}(z) $, $ p_{31}(z) $和$ p_{32}(z) $是次数分别为$ 422 $和$ 423 $的多项式, $ p^r_{4}(z) $是次数为$ 510 $的多项式. 对正则链$ [p^r_3, p^r_4] $中的多项式调整顺序后实根隔离, 得到如下8个区间对

$ \begin{eqnarray*} &&[x = [-\frac{77597}{65536}, -\frac{155193}{131072}], \\ && z = [-\frac{2703326281672012770259650847849313005164772322865213216210682038740972915} {3533694129556768659166595001485837031654967793751237916243212402585239552}, \\ && -\frac{1351663140836006385129825423924656502582386161432606608105341019370486457} {1766847064778384329583297500742918515827483896875618958121606201292619776}]] \\ && [x = [\frac{59141}{65536}, \frac{118283}{131072}], \\ && z = [-\{215294173585539734989103573476748400 19440611856050418918488074352999942056371944710023467967 3759989845\\ &&99064594469977255503\}/ \{20657999024695268717247353376024094994637646342633788102645274852325180976134729557037\\ && 162826241102651487225375781979947008\}, -\{4305883471710794699782071469534968003888122371 2100837836976148705999\\ &&88411274388 9420046935934751997969198129188939954511005\}/ \{4131599804939053743449470675204818998927529268526757\\ &&62052905497046503619 52269459114074325652482205302974450751563959894016\}]] \\ &&[x = [-\frac{6267}{8192}, -\frac{100271}{131072}], \\ && z = [-\frac{331491328330134392938385464656700375289768559784477057232401020096904427949097850062708631569695075057} {279968092772225526319680285071055534765205687154331191862498637620473983897520118172609686658950889472}, \\ && -\frac{662982656660268785876770929313400750579537119568954114464802040193808855898195700125417263139390150113} {559936185544451052639360570142111069530411374308662383724997275240947967795040236345219373317901778944}]] \\ && [x = [\frac{51355}{65536}, \frac{102711}{131072}], \\ && z = [-\frac{84916938191726551604338967723381304089558526360923681212110853332024021426300089301562980642632420015235} {71671831749689734737838152978190216899892655911508785116799651230841339877765150252188079784691427704832}, \\ && -\frac{169833876383453103208677935446762608179117052721847362424221706664048042852600178603125961285264840030469} {143343663499379469475676305956380433799785311823017570233599302461682679755530300504376159569382855409664}]]\\ && [x = [\frac{23339}{65536}, \frac{186715}{524288}], \\ && z = [-\frac{20974137332438701558251887341142213087197039654791097735294519552970247067658496803489} {15541351137805832567355695254588151253139254712417116170014499277911234281641667985408}, \\ && -\frac{41948274664877403116503774682284426174394079309582195470589039105940494135316993606977} {31082702275611665134711390509176302506278509424834232340028998555822468563283335970816}]] \\ &&[x = [-\frac{273203}{262144}, -\frac{546403}{524288}], \\ &&z = [\frac{23202132131872376501887846649475732460328971517181097691790951} {25711008708143844408671393477458601640355247900524685364822016}, \\ && \frac{2900266516484047062735980831184466557541121439647637211473869} {3213876088517980551083924184682325205044405987565585670602752}]] \\ &&[x = [-\frac{155295}{131072}, -\frac{77647}{65536}], \\ && z = [\frac{279603400387855417412049014047917306142228651} {356811923176489970264571492362373784095686656}, \frac{1118413601551421669648196056191669224568914605} {1427247692705959881058285969449495136382746624}]] \\ &&[x = [-\frac{176891}{131072}, -\frac{88445}{65536}], z = [\frac{400966712053531}{1125899906842624}, \frac{100241678013383}{281474976710656}]], \end{eqnarray*} $

容易看到$ q(x, z) $和$ p_3(x, z) $没有满足(3.13)式的公共根. 故, 当$ x\in (-1, 0) $时$ W[l_1, l_2, l_0]\neq0 $.

(ⅳ) 把$ q(x, z) $和$ p_4(x, z) $代入到上面的算法程序中, 三角分解得到正则链

$ [[p^r_5(x, z), p^r_6(z)], [4z-7, 4x-7], [z+1, x+1]], $

其中$ p^r_5(x, z) = p_{51}(z)x+p_{52}(z) $, 多项式$ p_{51}(z) $, $ p_{52}(z) $和$ p^r_{6}(z) $的次数分别为$ 559 $, $ 561 $和$ 680 $. 对正则链$ [p^r_5, p^r_6] $中的多项式实根隔离后得到如下10个区间对

$ \begin{eqnarray*} &&[x = [-\frac{170419}{131072}, -\frac{85209}{65536}], \\ && z = [-\frac{795241159813809950861337563313481171541693640180033213} {1532495540865888858358347027150309183618739122183602176}, -\frac{198810289953452487715334390828370292885423410045008303} {383123885216472214589586756787577295904684780545900544}]], \\ && [x = [-\frac{386707}{524288}, -\frac{193355}{262144}], \\ && z = [-\frac{11084672095528422217674276908486659546374460919926434879704809176904510203981245734731} {15541351137805832567355695254588151253139254712417116170014499277911234281641667985408}, \\ && -\frac{5542336047764211108837138454243329773187230459963217439852404588452255101990622867365} {7770675568902916283677847627294075626569627356208558085007249638955617140820833992704}]], \\ && [x = [-\frac{36505}{32768}, -\frac{146019}{131072}], \\ && z = [-\frac{34223032128351225372005064978750581793387152805269983891981269105943951459879954679722522858760183742306201078933941} {39402006196394479212279040100143613805079739270465446667948293404245721771497210611414266254884915640806627990306816}, \\ && -\frac{8555758032087806343001266244687645448346788201317495972995317276485987864969988669930630714690045935576550269733485} {9850501549098619803069760025035903451269934817616361666987073351061430442874302652853566563721228910201656997576704}]], \\ && [x = [-\frac{115617}{131072}, -\frac{57809}{65536}], \\ &&z = [-(9 7703193195996882242806710742613782378168037765115696352113660977200499236156219447913\\ &&08313841688486091387769814942917806859151)/(1083074099265943304522818040680892071654858232568678349675968586177586448\\ &&3615725089999900023844295226942934417817982702456930304), -(4885159659799844112140335537130689118908401888255784817605 \\ &&683048860024961807810972395654156920844243045693884907471458903429575)/(541537049632971652261409020340446035827429116\\ &&2843391748379842930887932241807862544999950011922147613471467208908991351228465152)]], \\ &&[x = [-\frac{118995}{131072}, -\frac{29749}{32768}], \\ &&z = [-(122458510361161212198731295080012702412664729663402164251962476568353033943639847700767438\\ &&522794140854461370679357361007604656374979760409981138514092054132578371227800048711)/(123665200736552267030251260509\\ &&823595 017565674550605919957031528046448612553265933585158200530621522494798835713008069669675682517153375604983773077\\ &&550946583958 303386074349568), -(6122925518058060609936564754000635120633236483170108212598123828417651697181992385038\\ &&3719261397070427230 685339678680503802328187489880204990569257046027066289185613900024355)/(61832600368276133515125630\\ &&25491179750878283727530295 9978515764023224306276632966792579100265310761247399417856504034834837841258576687802491886\\ &&538775473291979151693037174784)]], \\ &&[x = [-\frac{28461}{32768}, -\frac{113843}{131072}], \\ &&z = [(2591136222918880268063082955851190872512344341156985661847699980486727902005860274400385907 \\ &&5642909920831003428327747556768484783493618037)/(232588391774594204975783618524161450993165235419941779290076863737804\\ &&57219628733546438113622840434097944400691400517693873107252115668992), (64778405572972006701577073896279771812808608528\\ &&92464154619249951216819755014650686000964768910727480207750857081936889192121195873404509)/(58147097943648551243945904\\ &&63104036274829130885498544482251921593445114304907183386609528405710108524486100172850129423468276813028917248)]]\\ &&[x = [-\frac{4251}{8192}, -\frac{68015}{131072}], \\ && z = [-\frac{6403770300638479429419474158201605523331732644167924760238227379010131108852367918376032409023522422277348256939895} {4925250774549309901534880012517951725634967408808180833493536675530715221437151326426783281860614455100828498788352}, \\ && -\frac{3201885150319239714709737079100802761665866322083962380119113689505065554426183959188016204511761211138674128469947} {2462625387274654950767440006258975862817483704404090416746768337765357610718575663213391640930307227550414249394176}]], \\ && [x = [-\frac{519175}{524288}, -\frac{129793}{131072}], \\ &&z = [\frac{225751695451877482375298799607089930803532000364657801511247335503669450181546020019557} {248661618204893321077691124073410420050228075398673858720231988446579748506266687766528}, \\ && \frac{112875847725938741187649399803544965401766000182328900755623667751834725090773010009779} {124330809102446660538845562036705210025114037699336929360115994223289874253133343883264}]]\\ && [x = [-\frac{472957}{524288}, -\frac{236477}{262144}], \\ &&z = [\frac{6383714842460013288044212925318720379033757062475770275889058374306380652117} {7237005577332262213973186563042994240829374041602535252466099000494570602496}, \\ && \frac{3191857421230006644022106462659360189516878531237885137944529187153190326059} {3618502788666131106986593281521497120414687020801267626233049500247285301248}]]\\ && [x = [-\frac{46743}{65536}, -\frac{93485}{131072}], \\ && z = [\frac{282588614619387647412772690288705924986687827864180031} {383123885216472214589586756787577295904684780545900544}, \frac{4415447103427931990824573285761030077916997310377813} {5986310706507378352962293074805895248510699696029696}]]. \end{eqnarray*} $

由此我们知道上述10个区间对中的最后三个区间对有满足不等式(3.13)根, 假设$ (x_1, z_1) $, $ (x_2, z_2) $和$ (x_3, z_3) $是$ p_4(x, z) $和$ q(x, z) $的公共根. 把$ W[l_1(x), l_2(x), l_0(x), l_3(x)] $简记为$ W_4(x, z(x)) $, 并对它求导得到

$ \begin{array}{rl} \frac{{\rm d} W_4}{{\rm d} x}& = \frac{\partial W_4}{\partial x}+\frac{\partial W_4}{\partial z} \times \frac{{\rm d}z}{{\rm d}x} \\ & = \frac{ 8388608(x-z)^{9} p_5(x, z)}{2401115967765017711699008941650390625{x}^{5}{z}^{5} \left( 4\, x-7 \right) ^{67} \left( x+1 \right) ^{39} \left( 4\, z-7 \right) ^{66} \left( z+1 \right) ^{38} p^{7}(x, z)}, \end{array} $

其中$ p_5(x, z) $是含有$ 14383 $项以及次数是$ 214 $的多项式. 同样根据Sturm定理和利用上述符号计算的方法可得: $ p_5(x, z) $和$ q(x, z) $没有公共根. 因此, $ (x_1, z_1) $, $ (x_2, z_2) $和$ (x_3, z_3) $是$ W_4(x, z(x)) $的三个单根.

根据引理2.1, 得到如下定理.

定理3.1  $ \{I_0, I_1, I_2, I_3\} $是精度为$ 3 $的Chebyshev系统, 即Abel积分$ I(h, \delta) $至多有$ 6 $个零点.

根据引理2.3, Abel积分$ I(h, \delta) $在$ h = 0 $附近的展开式是

$ I(h, \delta) = b_0(\delta)h+b_1(\delta)h^2+b_2(\delta)h^3+b_3(\delta)h^4+O(h^5). $

为了获得系数$ b_i $的值, 令$ x = \frac{4}{7}u, y = v $, 和时间变换$ t = \frac{7}{4}t $, 系统(1.6) 变为

(4.1) $ \begin{equation} \dot{u} = v, \ \dot{v} = \frac{1 }{9604}{u \left( 4\, u+7 \right) \left( 16\, u-49 \right) ^{2}} + \varepsilon ( \frac{4a_0}{7}+\frac{16a_1}{49} u + \frac{64a_2}{343} u^2 +\frac{256}{2401} u^3)v. \end{equation} $

那么系统的Hamiltonian函数就变为(2.3) 式的形式, 而积分$ I(h, \delta) $保持不变. 对系统(4.1) 应用文献[19]中的计算方法得

$ b_0(\delta) = \frac{7}{2}\pi a_0, \ b_1(\delta) = \frac{370}{1029} \pi a_0 + \frac{4}{49} \pi a_1+\frac{4}{7} \pi a_2, $

$ b_2(\delta) = \frac{411128}{3176523} \pi a_{{0}}+\frac{544}{151263} \pi a_{{1}}+ \frac{29920}{151263} \pi a_{{2}}+ \frac{320}{7203} \pi , $

$ b_3(\delta) = \frac{471730096}{7004233215} \pi a_0- \frac{14138272}{ 1089547389} \pi a_1+ \frac{15153248}{155649627} \pi a_2 +\frac{94592}{7411887} \pi . $

令尺度变换为$ x = \frac{4}{11}u-1, \ y = v $和时间变换$ x = \frac{11}{4}t $, 系统(1.6) 变为

(4.2) $ \begin{equation} \begin{array}{l} \dot{u} = v, \\ { } \dot{v} = -\frac{1}{161051}u(4u-11)(16u-121)^2+\varepsilon \frac{4}{11}(a_0+a_1(\frac{4}{11}u-1) \\ {\qquad} { } +a_2(\frac{4}{11}u-1)^2+(\frac{4}{11}u-1)^3)v, \end{array} \end{equation} $

其Hamiltonian函数就变为引理2.4中(2.5) 的形式, 而积分$ I(h, \delta) $保持不变. 对系统(4.2) 应用引理2.4得: 当$ 0<-(h-\frac{409}{480}) \ll 1 $时有

$ I(h, \delta) = c_0(\delta)+c_1(\delta)(h-\frac{409}{480})\ln|h-\frac{409}{480}|+ c_2(\delta)(h-\frac{409}{480})+c_3(\delta)(h-\frac{409}{480})^2\ln|h- \frac{409}{480}|+h.o.t., $

其中

$ \begin{eqnarray*} &&c_0 = 3.5303228958816917001a_0-0.69635879526150535901a_1+0.70279708246537375570a_2\\ &&{\qquad} -0.014376738278135976678;\\ && c_1 = \frac{4}{11}(-a_0+a_1-a_2+1), {\qquad} c_3 = \frac{32}{1331}a_2-\frac{96}{1331}; \\ && c_2 = 4.9419383738406755526a_1-2.8816960225630379282a_2+5.0510463136722214575. \end{eqnarray*} $

容易得到$ \det \frac{\partial (b_0, b_1, b_2)}{\partial (a_0, a_1, a_2)}\neq 0. $解方程$ b_0 = b_1 = b_2 = 0 $得到$ a_0 = 0, a_1 = \frac{245}{136}, a_2 = -\frac{35}{136} $, 令$ \delta_0 = \{0, \frac{245}{136}, -\frac{35}{136}, 1\} $带入展开式中首个非零系数得到$ b_3(\delta_0) = - \frac{214016}{6000099} \pi $, $ c_0(\delta_0) = -0.2077883334 $, 因此

$ \frac{1-sgn(I(h_1, \delta_0)I(h_2, \delta_0))}{2} = \frac{1-sgn(b_3(\delta_0)c_0(\delta_0))}{2} = 0. $

根据引理2.5可以知道$ I(h, \delta) $在$ h = 0 $附近有3个零点.

命题4.1  存在$ \delta_0\in R^4 $使得Abel积分$ I(h, \delta) $在$ (0, \frac{409}{480}) $上有$ 3 $个零点.

$ \begin{eqnarray*} \widehat{g}_i(x)& = & 21743271936\, {i}^{4}{x}^{20}-337020715008\, {i}^ {4}{x}^{19}+902345785344\, {i}^{3}{x}^{20}+1947155300352\, {i}^{4}{x}^{ 18}\\ &&-13340856287232\, {i}^{3}{x}^{19}+12225154646016\, {i}^{2}{x}^{20}- 3734576824320\, {i}^{4}{x}^{17}\\ &&+73523231981568\, {i}^{3}{x}^{18}- 172570913538048\, {i}^{2}{x}^{19}+57502800543744\, i{x}^{20}\\ &&- 9837268107264\, {i}^{4}{x}^{16} -136003068887040\, {i}^{3}{x}^{17}+ 908274535759872\, {i}^{2}{x}^{18}\\ &&-776385312325632\, i{x}^{19}+ 46158248411136\, {x}^{20}+57098499194880\, {i}^{4}{x}^{15}\\ &&- 320659129368576\, {i}^{3}{x}^{16}-1622710045900800\, {i}^{2}{x}^{17}+ 3911906485075968\, i{x}^{18}\\ &&-600057229344768\, {x}^{19} -44815874629632\, {i}^{4}{x}^{14}+1804428195840000\, {i}^{3}{x}^{15}\\ &&-3419438491828224\, {i }^{2}{x}^{16}-6781855228231680\, i{x}^{17} +2935235882778624\, {x}^{18}\\ &&- 242748635172864\, {i}^{4}{x}^{13}-1376418726739968\, {i}^{3}{x}^{14}+ 18717493627699200\, {i}^{2}{x}^{15}\\ &&-12645385026011136\, i{x}^{16}- 5184562798264320\, {x}^{17}+504067880386816\, {i}^{4}{x}^{12}\\ &&- 6749200188650496\, {i}^{3}{x}^{13} -13811287692214272\, {i}^{2}{x}^{14}+ 67986151994327040\, i{x}^{15}\\ &&-7120817429151744\, {x}^{16}+ 385930447203840\, {i}^{4}{x}^{11} +13334347974124544\, {i}^{3}{x}^{12}\\ &&- 62240170436834304\, {i}^{2}{x}^{13}-48819499532894208\, i{x}^{14}+ 42357616705142784\, {x}^{15}\\ &&-1859066034259840\, {i}^{4}{x}^{10}+ 9567311208176960\, {i}^{3}{x}^{11}+116894051834859776\, {i}^{2}{x}^{12}\\ &&- 202605696645421056\, i{x}^{13} -34623158043893760\, {x}^{14}+ 232638482711040\, {i}^{4}{x}^{9}\\ &&-43706677545342560\, {i}^{3}{x}^{10}+ 80183761064984640\, {i}^{2}{x}^{11} +364265736011574784\, i{x}^{12}\\ &&- 105856976777748480\, {x}^{13}+3825663033082800\, {i}^{4}{x}^{8}+ 4826507910708960\, {i}^{3}{x}^{9}\\ &&-344750925701251440\, {i}^{2}{x}^{10}+ 237257188014468160\, i{x}^{11}+205625342088176640\, {x}^{12}\\ &&- 1942599151516800\, {i}^{4}{x}^{7} +81486171059542200\, {i}^{3}{x}^{8}+ 32313042212163440\, {i}^{2}{x}^{9}\\ &&-973210836556168560\, i{x}^{10}+ 86893815127257600\, {x}^{11} -4968404400849400\, {i}^{4}{x}^{6}\\ &&- 38628679019830200\, {i}^{3}{x}^{7}+585542516667623900\, {i}^{2}{x}^{8}+ 85682970107683060\, i{x}^{9}\\ &&-473260803748353600\, {x}^{10}+ 3336656112741000\, {i}^{4}{x}^{5}-98108883434572850\, {i}^{3}{x}^{6}\\ &&- 259038587930044300\, {i}^{2}{x}^{7} +1506386751569733250\, i{x}^{8}+ 111801539547923400\, {x}^{9}\\ &&+4271894088030625\, {i}^{4}{x}^{4}+ 62849843078829625\, {i}^{3}{x}^{5} -648945857027027175\, {i}^{2}{x}^{6}\\ &&- 640282776626374175\, i{x}^{7}+633854025029053150\, {x}^{8}- 2727924010003500\, {i}^{4}{x}^{3}\\ &&+80295985523665625\, {i}^{3}{x}^{4}+ 396831165120894875\, {i}^{2}{x}^{5}-1526032367853908250\, i{x}^{6} \\ &&- 371974171717943200\, {x}^{7} -2392918453091250\, {i}^{4}{x}^{2}- 50339692179451500\, {i}^{3}{x}^{3}\\ &&+497435031616126625\, {i}^{2}{x}^{4}+ 915334619977916250\, i{x}^{5} -545943480733507325\, {x}^{6}\\ &&+ 934286886067500\, {i}^{4}x-44226090453746250\, {i}^{3}{x}^{2}- 309030626097669000\, {i}^{2}{x}^{3}\\ &&+1070530847541066625\, i{x}^{4}+ 455011579920135000\, {x}^{5}+700715164550625\, {i}^{4}\\ &&+17584613891341875 \, {i}^{3}x -267329985430803750\, {i}^{2}{x}^{2}-686730944873631000\, i{x} ^{3}\\ &&+328318416576255000\, {x}^{4}+13313588126461875\, {i}^{3} + 112081058939311875\, {i}^{2}x\\ &&-550762119336791250\, i{x}^{2}- 328451324063310000\, {x}^{3}+85487250075176250\, {i}^{2}\\ &&+ 272010953400795000\, ix-134365707879952500\, {x}^{2}+207411688706985000 \, i\\ &&+182986760114077500\, x+134537311593720000. \\ \check{g}_i(x)& = & 8349416423424\, {i}^{5}{x}^{25}-161769943203840 \, {i}^{5}{x}^{24}+542712067522560\, {i}^{4}{x}^{25}\\ &&+1248063809126400\, { i}^{5}{x}^{23}-10123667413401600\, {i}^{4}{x}^{24}+12837227751014400\, { i}^{3}{x}^{25}\\ &&-4199767332618240\, {i}^{5}{x}^{22}+75179102180474880\, {i }^{4}{x}^{23}-230691766586572800\, {i}^{3}{x}^{24}\\ &&+132400870934446080\, {i}^{2}{x}^{25}-128361179381760\, {i}^{5}{x}^{21}-244190968251678720\, { i}^{4}{x}^{22}\\ &&+1649988951877877760\, {i}^{3}{x}^{23}- 2294276127562137600\, {i}^{2}{x}^{24}+536630039496622080\, i{x}^{25}\\ &&+ 47446563348283392\, {i}^{5}{x}^{20}+4491468047646720\, {i}^{4}{x}^{21}- 5175446121228533760\, {i}^{3}{x}^{22}\\ &&+15819641797937725440\, {i}^{2}{x}^ {23}-8980570662858915840\, i{x}^{24}+416532033662091264\, {x}^{25}\\ &&- 122872252807249920\, {i}^{5}{x}^{19}+2476864648909946880\, {i}^{4}{x}^{ 20}+312782726501498880\, {i}^{3}{x}^{21}\\ &&-47954437723788410880\, {i}^{2}{ x}^{22}+59807588168025243648\, i{x}^{23}-6768645547008983040\, {x}^{24}\\ &&- 84757548745359360\, {i}^{5}{x}^{18}-6200405368638013440\, {i}^{4}{x}^{19 }+47251544271838248960\, {i}^{3}{x}^{20}\\ &&+4603116763868037120\, {i}^{2}{x }^{21}-175632368857572704256\, i{x}^{22}+43936786630985121792\, {x}^{23} \\ &&+892954894024704000\, {i}^{5}{x}^{17}-3896410598754877440\, {i}^{4}{x}^{ 18}-114279991958793093120\, {i}^{3}{x}^{19}\\ &&+395509894676710686720\, {i}^ {2}{x}^{20}+23334842404769366016\, i{x}^{21}-127631663668918222848\, {x} ^{22}\\ &&-871827308773232640\, {i}^{5}{x}^{16}+40614681601515356160\, {i}^{4 }{x}^{17}-66482306153915351040\, {i}^{3}{x}^{18}\\ &&-924000042752999424000 \, {i}^{2}{x}^{19}+1310764280856928321536\, i{x}^{20}+ 34871358205168975872\, {x}^{21}\\ &&-2805700280400581632\, {i}^{5}{x}^{15}- 38104928217383485440\, {i}^{4}{x}^{16}+679271893748789084160\, {i}^{3}{x }^{17}\\ &&-506616612223111004160\, {i}^{2}{x}^{18}-2969190885812535558144\, i{x}^{19}+822603245951813419008\, {x}^{20}\\ &&+6128803009406784000\, {i}^{5} {x}^{14}-116064868062049617920\, {i}^{4}{x}^{15}-608757156006774435840 \, {i}^{3}{x}^{16}\\ &&+5022814319768594841600\, {i}^{2}{x}^{17}- 1532017135907750412288\, i{x}^{18}-1909040785622528163840\, {x}^{19}\\ &&+ 3630160137132435200\, {i}^{5}{x}^{13}+241830764938912396800\, {i}^{4}{x} ^{14}-1782924007474252528640\, {i}^{3}{x}^{15}\\ &&-4280826400914870435840\, {i}^{2}{x}^{16}+14841434990018486108160\, i{x}^{17}- 661477003362874294272\, {x}^{18}\\ &&-19466038509919545600\, {i}^{5}{x}^{12}+ 139403544167385648640\, {i}^{4}{x}^{13}+3529130837913598310400\, {i}^{3} {x}^{14}\\ &&-12219308244819234334720\, {i}^{2}{x}^{15}- 12114093162629403095040\, i{x}^{16}+8425431873064317911040\, {x}^{17}\\ &&+ 3207823901935352000\, {i}^{5}{x}^{11}-703083765791662241280\, {i}^{4}{x} ^{12}+2018932504191269036160\, {i}^{3}{x}^{13}\\ &&+22925891709048381849600 \, {i}^{2}{x}^{14}-33573408175727656700928\, i{x}^{15}- 7545076079242242170880\, {x}^{16}\\ &&+38025927523737468000\, {i}^{5}{x}^{10} +100690818541640252800\, {i}^{4}{x}^{11}-9466961819797455709120\, {i}^{3 }{x}^{12}\\ &&+13168356691159101434240\, {i}^{2}{x}^{13}+ 60165113456335838361600\, i{x}^{14}-16610142161518865326080\, {x}^{15}\\ &&- 21022762577859182000\, {i}^{5}{x}^{9}+1272307274477934660000\, {i}^{4}{x }^{10}+1122394607286237006400\, {i}^{3}{x}^{11}\\ &&-57117090557956188473280 \, {i}^{2}{x}^{12}+34147307908236387343360\, i{x}^{13}+ 32134927138792357824000\, {x}^{14}\\ &&-50013209205512490000\, {i}^{5}{x}^{8} -655610034367880183200\, {i}^{4}{x}^{9}+15929074610803259245200\, {i}^{3 }{x}^{10}\\ &&+5359242848618029548800\, {i}^{2}{x}^{11}- 139535653041444336771920\, i{x}^{12}+12671091684173454537600\, {x}^{13}\\ &&+ 38951062522221482500\, {i}^{5}{x}^{7}-1571432562057929088000\, {i}^{4}{x }^{8}-7597594434027750502000\, {i}^{3}{x}^{9}\\ &&+89614016600183289555600\, {i}^{2}{x}^{10}+11208608243147020241900\, i{x}^{11}- 66526305925131451476000\, {x}^{12}\\ &&+46328800242150011250\, {i}^{5}{x}^{6} +1157772289758053291000\, {i}^{4}{x}^{7}-18433890463459926745500\, {i}^{ 3}{x}^{8}\\ &&-39495758281644704362400\, {i}^{2}{x}^{9}+ 203996129764714944131650\, i{x}^{10}+13390754363639789878000\, {x}^{11}\\ &&- 39811707617263596875\, {i}^{5}{x}^{5}+1386772626982184940750\, {i}^{4}{x }^{6}+12786953862584806605750\, {i}^{3}{x}^{7}\\ &&-96996027153274178695500 \, {i}^{2}{x}^{8}-85362590995600371931275\, i{x}^{9}+ 86611590975170822478750\, {x}^{10}\\ &&-30954574367417465625\, {i}^{5}{x}^{4} -1144310851147657506250\, {i}^{4}{x}^{5}+15375875050019135932000\, {i}^{ 3}{x}^{6}\\ &&+63358998946119555180250\, {i}^{2}{x}^{7}- 205455131633102371281625\, i{x}^{8}-47353003054336518889000\, {x}^{9}\\ &&+ 23007593141817243750\, {i}^{5}{x}^{3}-896010683537711135625\, {i}^{4}{x} ^{4}-12146269739197091141875\, {i}^{3}{x}^{5}\\ &&+75851002106313306631500\, {i}^{2}{x}^{6}+129492346309664282217000\, i{x}^{7}- 76582020887237366475625\, {x}^{8}\\ &&+14388018045439500000\, {i}^{5}{x}^{2}+ 653162453481485430000\, {i}^{4}{x}^{3}-9511021432558201408125\, {i}^{3}{ x}^{4}\\ &&- 57482026222957278897500\, {i}^{2}{x}^{5}+149005398507077198723000\, i{x}^{6}+62083240086558015563000\, {x}^{7}\\ &&- 6008632536021609375\, {i}^{5}x+410777915197297725000\, {i}^{4}{x}^{2}+ 6821701070771446138125\, {i}^{3}{x}^{3} \\ &&-44298909686217952250625\, {i}^{2 }{x}^{4}-111456853481215282950000\, i{x}^{5}+47558535343250977794000\, { x}^{6}\\ &&-3605179521612965625\, {i}^{5}-172705266606792543750\, {i}^{4}x+ 4283746247670752953125\, {i}^{3}{x}^{2}\\ &&+31621075675553700826875\, {i}^{2 }{x}^{3}-80431918853967979515000\, i{x}^{4}-49495636624672283535000\, {x }^{5}\\ &&-104550206126776003125\, {i}^{4}-1842418410759540337500\, {i}^{3}x+ 19464380587208907590625\, {i}^{2}{x}^{2}\\ &&+59828798211511486027500\, i{x}^ {3}-21896627044362914812500\, {x}^{4}-1124816010743245275000\, {i}^{3}\\ &&- 8894321230249816003125\, {i}^{2}x+33882016694795354925000\, i{x}^{2}+ 26369801704410533302500\, {x}^{3}\\ &&-5465452154765255887500\, {i}^{2}-18497661099750181912500\, ix +7441384832978272312500\, {x}^{2}\\ &&-11363525852124067650000\, i-11608678059593749312500\, x- 6921944681496894000000. \end{eqnarray*} $