考虑扰动Hamilton系统

$\left\{ \begin{array}{*{35}{l}} \frac{dx}{dt}=\frac{\partial H}{\partial y}+\varepsilon f(x,y),\\ \frac{dy}{dt}=-\frac{\partial H}{\partial x}+\varepsilon g(x,y),\\ \end{array} \right.$

(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(1.1)的未扰动系统(1.1)$_{\varepsilon =0}$有连续闭轨线族$\{{{\Gamma }_{h}}\},\Sigma $为h的最大存在开区间,即

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

定义积分

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

称该积分为Abelian积分. 寻找Abelian积分I(h)的零点个数的上界Z(m,n)称为弱Hilbert 16问题^[1]. 关于弱Hilbert 16问题,Khovansky和Varchenko独立的证明了Z(m,n)的存在性,但是没有给出具体的表达式^[2-3]. 对于特殊形式的Hamiltonian H(x,y)=U(x)+V(y),其中U(x)和V(y)分 别是关于x和y的多项式,相关的研究很多. 例如,Petrov估计了Hamiltonian$H(x,y)={{y}^{2}}-x+{{x}^{3}},H(x,y)={{y}^{2}}+{{x}^{2}}-{{x}^{4}}$和H(x,y)=y²-2x²+x⁴相应Abelian积分 零点的个数^[4-6]. 当H$(x,y)=\frac{1}{2}{{y}^{2}}+U(x)$时,其中$\deg U(x)=4$且使得(1.1)₀至少有一个中心,赵育林和张芷芬证明了$Z(m,n)\le 7n+5$^[7]. 对于4次Hamiltonian H(x,y)=-x²+x⁴+y⁴和$H(x,y)={{x}^{2}}+{{y}^{2}}+a{{x}^{4}}+{{y}^{4}}$,周鑫和李翠萍得到了相应Abelian积分零点个数的上界^[8-9]. 对于Hamiltonian $H(x,y)=\frac{1}{2}{{y}^{2}}+\frac{a}{2}{{x}^{2}}+\frac{b}{4}{{x}^{4}}+\frac{c}{6}{{x}^{6}}$,齐明辉,赵丽琴和刘长剑证明了$Z(m,n)\le 54n-13$^[10].

但是当Hamiltonian H(x,y)中含有${{x}^{i}}{{y}^{j}}$时,其中i和j是正整数,目前研究结果很少.当m=2且未扰动系统含有周期环域时,相应Hamiltonian H(x,y)的规范型可化为

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

Horozov和Iliev应用Picard-Fuchs方程法证明了Z$(m,n)\le 5n+15$^[11]. 对原点是初等中心的Hamiltonian

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

吴娟娟,张永康和李翠萍证明了$Z(m,n)\le 2[\frac{n-1}{4}]+12[\frac{n-3}{4}]+23$^[12]. 对Hamiltonian

$H(x,y)=-{{x}^{2}}+a{{x}^{2}}{{y}^{2}}+{{x}^{4}}+{{y}^{4}},$

其中$a\ge 0,a\ne -2$,杨纪华,赵丽琴得到了相应Abelian积分零点个数的上界^[13].

受文献^{[7, 12]}的启发,本文研究如下Hamiltonian

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

(1.2)

其中a>0,b<-2,c<0,a^2>4bc. 与之相对应的向量场为

$\left\{ \begin{array}{*{35}{l}} \dot{x}=2y(a{{x}^{2}}+2c{{y}^{2}}),\\ \dot{y}=2x(1-2b{{x}^{2}}-a{{y}^{2}}). \\ \end{array} \right.$

(1.3)

易知,系统(1.3)有一个幂零中心O(0,0)和四个鞍点${{S}_{1}}(\sqrt{\frac{2c}{4bc-{{a}^{2}}}},\sqrt{\frac{a}{{{a}^{2}}-4bc}})$,${{S}_{2}}(\sqrt{\frac{2c}{4bc-{{a}^{2}}}},-\sqrt{\frac{a}{{{a}^{2}}-4bc}})$,${{S}_{3}}(-\sqrt{\frac{2c}{4bc-{{a}^{2}}}},\sqrt{\frac{a}{{{a}^{2}}-4bc}})$和${{S}_{4}}(-\sqrt{\frac{2c}{4bc-{{a}^{2}}}},-\sqrt{\frac{a}{{{a}^{2}}-4bc}})$. H(x,y)=0对应于幂零中心O,$H(x,y)=\frac{c}{{{a}^{2}}-4bc}$ 对应于过四个鞍点S_i(i=1,2,3,4)的多角环. 系统(1.3)只有一个围绕幂零中心O的有界周期环域,记为${{\Gamma }_{h}}$,其中

${{\Gamma }_{h}}=\{(x,y)|H(x,y)=h,h\in (\frac{c}{{{a}^{2}}-4bc},0)\}.$

记Z(n)为Abelian积分I(h)在$\Sigma $上零点个数(计重数),其中$n=\max\{\deg f(x,y),\deg g(x,y)\},\Sigma=(\frac{c}{a^2-4bc},0).$ 本文的主要结果为下列定理1.1.

定理1.1 对Hamiltonian (1.2),

$Z(n)\le 3n+3[\frac{n-1}{4}]+14,$

其中[p]表示p的整数部分. 而且Z(1)=Z(2)=0,Z(3)=Z(4)$\leq$12.

对$h\in\Sigma$,记$I_{i,j}(h)=\oint_{\Gamma_h}x^iy^j{d}x$, 其中i和j是自然数. 因为$\Gamma_h$关于坐标轴对称,因此$I_{i,2j}(h)\equiv0, I_{2i+1,2j+1}(h)\equiv0$. 所以我们只需考虑$I_{2i,2j+1}(h)$即可.

引理 2.1 假设 $h\in\Sigma,i+j\geq2$,则

${{I}_{2i,2j+1}}(h)=\bar{\alpha }(h){{I}_{0,1}}(h)+\bar{\beta }(h){{I}_{0,3}}(h)+\bar{\gamma }(h){{I}_{2,1}}(h)+\bar{\delta }(h){{I}_{2,3}}(h),$

(2.1)

其中$\bar{\alpha}(h),\bar{\beta}(h),\bar{\gamma}(h)$和$\bar{\delta}(h)$是关于h的多项式,$ \deg \bar{\alpha}(h)\leq[\frac{n-1}{4}],\deg \bar{\beta}(h),\deg \bar{\gamma}(h)\leq[\frac{n-3}{4}]$和$\deg \bar{\delta}(h)\leq[\frac{n-1}{4}]-1$.

证 由Green公式可得

$I(h)=\sum\limits_{2i+2j+1\leq n}c_{2i,2j+1}\oint_{\Gamma_h}x^{2i}y^{2j+1}{d}x=\sum\limits_{2i+2j+1\leq n}c_{2i,2j+1}I_{2i,2j+1}(h),$

其中$c_{2i,2j+1}$是常数.

对(1.2)式两端同时关于x求导可得

$-x+2bx^3+2cy^3\frac{\partial y}{\partial x}+axy^2+ax^2y\frac{\partial y}{\partial x}=0.$

(2.2)

(2.2)式两端同乘以$x^{2i-3}y^{2j+1}{d}x$,并沿着$\Gamma_h$积分可得

$I_{2i,2j+1}=\frac{1}{2b}\Big[\frac{1}{2}I_{2i-2,2j+1}+\frac{2(2i-3)c}{2j+5}I_{2i-4,2j+5}+\frac{2(i-j-2)a}{2j+3}I_{2i-2,2j+3}\Big].$

(2.3)

同样,(1.2)式两端同乘以$x^{2i}y^{2j-3}{d}x$,并沿着$\Gamma_h$积分可得

$I_{2i,2j+1}=\frac{1}{c}[hI_{2i,2j-3}+I_{2i+2,2j-3}-bI_{2i+4,2j-3}-aI_{2i+2,2j-1}]. $

(2.4)

由(2.3)和(2.4)式可得

$I_{2i,2j+1}=\frac{2j+1}{4(i+j+1)c}\Big[2hI_{2i,2j-3}+I_{2i+2,2j-3}-\frac{2(i+j+1)a}{2j-1}I_{2i+2,2j-1}\Big], $

(2.5)

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

(2.6)

下面用数学归纳法证明(2.1)式成立. 当 n=5,7时,由(2.5)和(2.6)式可得

$\left\{ \begin{array}{*{35}{l}} {{I}_{0,5}}=\frac{5h}{6c}{{I}_{0,1}}+\frac{5}{12c}{{I}_{2,1}}-\frac{5a}{6c}{{I}_{2,3}},\ {{I}_{4,1}}=\frac{h}{6b}{{I}_{0,1}}+\frac{7}{12b}{{I}_{2,1}}-\frac{a}{6b}{{I}_{2,3}},\\ {{I}_{0,7}}=\frac{7h}{8c}{{I}_{0,3}}+\frac{7}{16c}{{I}_{2,3}}-\frac{7a}{10c}{{I}_{2,5}},\ \ {{I}_{2,5}}=\frac{5h}{8c}{{I}_{2,1}}+\frac{5}{16c}{{I}_{4,1}}-\frac{5a}{6c}{{I}_{4,3}},\\ {{I}_{4,3}}=\frac{h}{8b}{{I}_{0,3}}+\frac{9}{16b}{{I}_{2,3}}-\frac{3a}{10b}{{I}_{2,5}},\ \ {{I}_{6,1}}=\frac{3h}{8b}{{I}_{2,1}}+\frac{11}{16b}{{I}_{4,1}}-\frac{a}{6b}{{I}_{4,3}},\\ \end{array} \right.$

(2.7)

所以当n=5,7时,(2.1)式成立. 假设当$2i+2j+1\le 2k-1\ (k\ge 3)$时,(2.1)式成立. 则当$2i+2j+1=2k+1\ (k\ge 2)$时,在(2.5)和(2.6)式分别取 $(i,j)=(0,k),(1,k-1),(2,k-2),\cdots ,(k-2,2)$和(i,j)=(k-1,1),(k,0)可得

$A\left( \begin{matrix} {{I}_{0,2k+1}} \\ {{I}_{2,2k-1}} \\ \vdots \\ {{I}_{2k-4,5}} \\ {{I}_{2k-2,3}} \\ {{I}_{2k,1}} \\ \end{matrix} \right)\ =\frac{1}{4k+4}\left( \begin{matrix} \frac{1}{c}[2(2k+1)h{{I}_{0,2k-3}}+(2k+1){{I}_{2,2k-3}}] \\ \frac{1}{c}[2(2k-1)h{{I}_{2,2k-5}}+(2k-1){{I}_{4,2k-5}}] \\ \vdots \\ \frac{1}{c}[10h{{I}_{2k-4,1}}+5{{I}_{2k-2,1}}] \\ \frac{1}{b}[(4k-3){{I}_{2k-4,3}}+2(2k-5)h{{I}_{2k-6,3}}] \\ \frac{1}{b}[(4k-1){{I}_{2k-2,1}}+2(2k-3)h{{I}_{2k-4,1}}] \\ \end{matrix} \right),$

其中

$A=\left( \begin{matrix} 1 & ~~\frac{(2k+1)a}{2(2k-1)c}~~ & 0 & ~~0~~ & \cdots & ~~0~~ & 0 & ~~~0~~~ & 0 & ~~0 \\ 0 & 1 & \frac{(2k-1)a}{2(2k-3)c} & 0 & \cdots & 0 & 0 & 0 & 0 & ~~0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & ~~\vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 1 & \frac{5a}{6c} & ~~0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & \frac{3a}{10b} & 1 & ~~0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \frac{a}{6b} & ~~1 \\ \end{matrix} \right)$

是(k+1)×(k+1)阶矩阵. 直接计算得$\det A=1-\frac{a^2}{4bc}\neq0$. 因此

$\begin{align} & {{I}_{2i,2j+1}}(h)={{{\bar{\alpha }}}_{2k-1}}{{I}_{0,1}}+{{{\bar{\beta }}}_{2k-1}}{{I}_{0,3}}+{{{\bar{\gamma }}}_{2k-1}}{{I}_{2,1}}+{{{\bar{\delta }}}_{2k-1}}{{I}_{2,3}} \\ & +h({{{\bar{\alpha }}}_{2k-3}}{{I}_{0,1}}+{{{\bar{\beta }}}_{2k-3}}{{I}_{0,3}}+{{{\bar{\gamma }}}_{2k-3}}{{I}_{2,1}}+{{{\bar{\delta }}}_{2k-3}}{{I}_{2,3}}) \\ & :={{{\bar{\alpha }}}_{2k+1}}{{I}_{0,1}}+{{{\bar{\beta }}}_{2k+1}}{{I}_{0,3}}+{{{\bar{\gamma }}}_{2k+1}}{{I}_{2,1}}+{{{\bar{\delta }}}_{2k+1}}{{I}_{2,3}},\\ \end{align}$

其中${{\bar{\alpha }}_{2k-s}},{{\bar{\beta }}_{2k-s}},{{\bar{\gamma }}_{2k-s}}$和$\bar{\delta}_{2k-s}\ (s=1,3)$是h的多项式,并满足$\deg \bar{\alpha}_{2k-s}\leq[\frac{2k-1-s}{4}],\deg \bar{\beta}_{2k-s},\deg \bar{\gamma}_{2k-s}\leq[\frac{2k-3-s}{4}] $和$\deg \bar{\delta}_{2k-s}\leq[\frac{2k-1-s}{4}]-1$. 容易得到

$\deg {{\bar{\alpha }}_{2k+1}}=\max \{[\frac{2k-2}{4}],\ [\frac{2k}{4}]\}=[\frac{k}{2}].$

同理,可得

$\deg {{\bar{\beta }}_{2k+1}},\ \deg {{\bar{\gamma }}_{2k+1}}\le [\frac{k-1}{2}],\ \deg {{\bar{\delta }}_{2k+1}}\le [\frac{k}{2}]-1.$

引理2.1证毕.

由引理2.1,我们可得Abelian积分I(h)的代数结构如下.

引理 2.2 当$h\in\Sigma$时,Abelian 积分I(h)可表示为

$I(h)=\alpha(h)I_{0,1}+\beta(h)I_{0,3}+\gamma(h)I_{2,1}+\delta(h)I_{2,3},$

(2.8)

其中$\alpha(h),\beta(h),\gamma(h)$和$\delta(h)$是h的多项式, 并且$\deg \alpha(h)\leq[\frac{n-1}{4}],\deg \beta(h),\deg \gamma(h)\leq[\frac{n-3}{4}]$和$\deg \delta(h)\leq[\frac{n-1}{4}]-1$.

引理 3.1 记$V=(I_{0,1},I_{0,3},I_{2,1},I_{2,3})^T$,则V满足如下的Picard-Fuchs方程

$(Bh+C)V'=V$,

(3.1)

其中

$\begin{align} & B=\left( \begin{matrix} 2 & 0 & 0 & ~~0 \\ 0 & 1 & 0 & ~~0 \\ \frac{1}{4b} & 0 & 1 & ~~0 \\ \frac{a}{8({{a}^{2}}-4)b} & ~~-\frac{1}{3({{a}^{2}}-4)}~~ & \frac{a}{2({{a}^{2}}-4)} & ~~\frac{2}{3} \\ \end{matrix} \right),\\ & C=\left( \begin{matrix} 0 & ~~0~~ & 1 & ~~0 \\ 0 & 0 & 0 & ~~\frac{1}{2} \\ 0 & 0 & \frac{3}{8b} & ~~-\frac{a}{12b} \\ 0 & 0 & \frac{3a}{16({{a}^{2}}-4)b} & ~~-\frac{{{a}^{2}}+20b}{24({{a}^{2}}-4)b} \\ \end{matrix} \right). \\ \end{align}$

证 由(1.2)式可得

$\frac{\partial y}{\partial h}=\frac{1}{2y(2cy^2+ax^2)},$

所以

$I'_{i,j}=\frac{j}{2}\oint_{\Gamma_h}\frac{x^iy^{j-2}}{2cy^2+ax^2}{d}x.$

(3.2)

因此,

$I_{i,j}=\frac{2a}{j+2}I'_{i+2,j+2}+\frac{4c}{j+4}I'_{i,j+4}.$

(3.3)

(3.2)式两端同乘以 h可得

$hI'_{i,j}=-I'_{i+2,j}+bI'_{i+4,j}+\frac{cj}{j+4}I'_{i,j+4}+\frac{aj}{j+2}I'_{i+2,j+2}.$

(3.4)

另一方面,

$\begin{align} & {{I}_{i,j}}=\oint_{{{\Gamma }_{h}}}{{{x}^{i}}}{{y}^{j}}dx=-\frac{j}{i+1}\oint_{{{\Gamma }_{h}}}{{{x}^{i+1}}}{{y}^{j-1}}dx \\ & =-\frac{j}{i+1}\oint_{{{\Gamma }_{h}}}{\frac{{{x}^{i+2}}{{y}^{j-2}}(1-2b{{x}^{2}}-a{{y}^{2}})}{2c{{y}^{2}}+a{{x}^{2}}}}dx \\ & =-\frac{2j}{i+1}[\frac{1}{j}{{{{I}'}}_{i+2,j}}-\frac{2b}{j}{{{{I}'}}_{i+4,j}}-\frac{a}{j+2}{{{{I}'}}_{i+2,j+2}}]. \\ \end{align}$

(3.5)

由(3.3)--(3.5)式得

${{I}_{i,j}}=\frac{2}{i+j+1}(2h{{{I}'}_{i,j}}+{{{I}'}_{i+2,j}}).$

(3.6)

由(3.6)式可得

$\begin{align} & {{I}_{0,1}}=2h{{{{I}'}}_{0,1}}+{{{{I}'}}_{2,1}},\ \ {{I}_{0,3}}=h{{{{I}'}}_{0,3}}+\frac{1}{2}{{{{I}'}}_{2,3}},\\ & {{I}_{2,1}}=2h{{{{I}'}}_{2,1}}+\frac{1}{2}{{{{I}'}}_{4,1}},\ \ {{I}_{2,3}}=\frac{2}{3}h{{{{I}'}}_{2,3}}+\frac{1}{3}{{{{I}'}}_{4,3}}. \\ \end{align}$

注意到(2.5)和(2.6)式,即可得引理3.1的结论成立.

取

$Z=\frac{1}{3({{a}^{2}}-4)}{{I}_{0,3}}-\frac{a}{2({{a}^{2}}-4)}{{I}_{2,1}}+\frac{1}{3}{{I}_{2,3}},$

(3.7)

我们可得如下引理.

引理 3.2 $I_{0,1},I_{0,3},I_{2,1}$和Z满足

$G(h)\left( \begin{array}{*{35}{l}} {{{{I}''}}_{0,1}} \\ {{{{I}''}}_{0,3}} \\ {{{{I}''}}_{2,1}} \\ {{Z}''} \\ \end{array} \right)=\left( \begin{matrix} {{a}_{11}}(h)~~ & {{a}_{12}}(h) \\ {{a}_{21}}(h)~~ & {{a}_{22}}(h) \\ {{a}_{31}}(h)~~ & {{a}_{32}}(h) \\ {{a}_{41}}(h)~~ & {{a}_{42}}(h) \\ \end{matrix} \right)\left( \begin{array}{*{35}{l}} {{{{I}'}}_{0,1}} \\ {{Z}'} \\ \end{array} \right),$

(3.8)

其中

$\begin{align} & G(h)=\frac{4}{3}{{h}^{2}}(h+\frac{1}{4b})(h-\frac{1}{{{a}^{2}}-4}),\\ & {{a}_{11}}(h)=-\frac{2}{3}h(h+\frac{1}{8b})(h-\frac{1}{{{a}^{2}}-4}),\ \ {{a}_{12}}(h)=-\frac{a}{12b}h,\\ & {{a}_{21}}(h)=0,\ \ {{a}_{22}}(h)=-h(h+\frac{1}{4b}),\\ & {{a}_{31}}(h)=-\frac{1}{6b}{{h}^{2}}(h-\frac{1}{{{a}^{2}}-4}),\ \ {{a}_{32}}(h)=\frac{a}{6b}{{h}^{2}},\\ & {{a}_{41}}(h)=\frac{a}{12({{a}^{2}}-4)b}{{h}^{2}}(h-\frac{1}{{{a}^{2}}-4}),\\ & {{a}_{42}}(h)=\frac{2}{3}h({{h}^{2}}+\frac{({{a}^{2}}-4b-8)h-1}{8b({{a}^{2}}-4)}). \\ \end{align}$

证 对(3.1)式两端关于h求导得

$\text{(Bh+C)V }\!\!'\!\!\text{ }\!\!'\!\!\text{ =(I-B)V }\!\!'\!\!\text{ ,}$

(3.9)

其中I是4×4阶单位矩阵. 把(3.7)式代入(3.9)式即可得(3.8)式成立.

易知当$h\in\Sigma$时,$I'_{0,1}(h)\neq0$. 所以可得如下引理.

引理 3.3当$h\in\Sigma$时,令$\omega(h)=\frac{Z'}{I'_{0,1}}$,则$omega(h)$满足如下的Riccati方程

$G(h)\omega'(h)=-a_{12}(h)\omega^2(h)+(a_{42}(h)-a_{11}(h))\omega(h)+a_{41}(h)$.

(3.10)

证 由(3.8)式中的第一个和第四个方程即可得(3.10)式成立.

由(2.8),(3.1)和(3.7)式可得

$\begin{array}{*{35}{l}} I(h)={{\alpha }_{1}}(h){{{{I}'}}_{0,1}}+{{\beta }_{1}}(h){{{{I}'}}_{0,3}}+{{\gamma }_{1}}(h){{{{I}'}}_{2,1}}+{{\delta }_{1}}(h){Z}',\\ {I}'(h)={{\alpha }_{2}}(h){{{{I}'}}_{0,1}}+{{\beta }_{2}}(h){{{{I}'}}_{0,3}}+{{\gamma }_{2}}(h){{{{I}'}}_{2,1}}+{{\delta }_{2}}(h){Z}',\\ \end{array}$

(4.1)

其中$\alpha_s(h),\beta_s(h),\gamma_s(h)$和$\delta_s(h)$是关于h的多项式, 且满足$\deg \alpha_s(h)\leq [\frac{n-1}{4}]-s+2,\deg \beta_s(h), \deg \gamma_s(h)\leq [\frac{n-3}{4}]-s+2,\deg \delta_s(h)\leq [\frac{n-1}{4}]-s+1 \ (s=1,2)$. 从(4.1)式中消去$I'_{2,1}$可得

${{\gamma }_{1}}(h){I}'(h)={{\gamma }_{2}}(h)I(h)+{{F}_{1}}(h),$

其中${{F}_{1}}(h)={{\alpha }_{3}}(h){{{{I}'}}_{0,1}}+{{\beta }_{3}}(h){{{{I}'}}_{0,3}}+{{\delta }_{3}}(h){Z}'$,且$\deg {{\alpha }_{3}}(h)\le [\frac{n-1}{4}]+[\frac{n-3}{4}]+1,\deg {{\beta }_{3}}(h)\le 2[\frac{n-3}{4}]+1,\deg {{\delta }_{3}}(h)\le [\frac{n-1}{4}]+[\frac{n-3}{4}]$.

引理 4.1^[14] $\#I(h)\le \#{{\gamma }_{1}}(h)+\#{{F}_{1}}(h)+1$.

引理 4.2 设S是${{\beta }_{3}}(h)$和G(h)在$\Sigma $上的零点组成的集合,则在$\Sigma \backslash S$上,

$(\frac{{{F}_{1}}(h)}{{{\beta }_{3}}(h)}{)}'=\frac{{{F}_{2}}(h)}{G(h)\beta _{3}^{2}(h)},$

其中${{F}_{2}}(h)={{\alpha }_{4}}(h){{{{I}'}}_{0,1}}+{{\delta }_{4}}(h){Z}',\deg {{\alpha }_{4}}(h)\le [\frac{n-1}{4}]+3[\frac{n-3}{4}]+5,\deg {{\delta }_{4}}(h)\le [\frac{n-1}{4}]+3[\frac{n-3}{4}]+4$.

证 由(3.8)式和F₁(h)的定义知,在$\Sigma \backslash S$上

$\begin{align} & (\frac{{{F}_{1}}(h)}{{{\beta }_{3}}(h)}{)}'=(\frac{{{\alpha }_{3}}}{{{\beta }_{3}}}{)}'{{{{I}'}}_{0,1}}+(\frac{{{\alpha }_{3}}}{{{\beta }_{3}}}){{{{I}''}}_{0,1}}++{{{{I}''}}_{0,3}}+(\frac{{{\delta }_{3}}}{{{\beta }_{3}}}{)}'{Z}'+(\frac{{{\delta }_{3}}}{{{\beta }_{3}}}){Z}'' \\ & =\frac{1}{G(h)\beta _{3}^{2}(h)}({{\alpha }_{4}}(h){{{{I}'}}_{0,1}}+{{\delta }_{4}}(h){Z}'),\\ \end{align}$

(4.2)

其中

$\begin{array}{*{35}{l}} {{\alpha }_{4}}(h)=G(h)(\alpha _{3}^{'}{{\beta }_{3}}-{{\alpha }_{3}}\beta _{3}^{'})+{{a}_{11}}(h){{\alpha }_{3}}{{\beta }_{3}}+{{a}_{21}}(h)\beta _{3}^{2}+{{a}_{41}}(h){{\delta }_{3}}{{\beta }_{3}},\\ {{\delta }_{4}}(h)=G(h)(\delta _{3}^{'}{{\beta }_{3}}-{{\delta }_{3}}\beta _{3}^{'})+{{a}_{12}}(h){{\alpha }_{3}}{{\beta }_{3}}+{{a}_{22}}(h)\beta _{3}^{2}+{{a}_{42}}(h){{\delta }_{3}}{{\beta }_{3}},\\ \end{array}$

(4.3)

所以$\deg {{\alpha }_{4}}(h)\le [\frac{n-1}{4}]+3[\frac{n-3}{4}]+5,\deg {{\delta }_{4}}(h)\le [\frac{n-1}{4}]+3[\frac{n-3}{4}]+4$.

因为G(h)在$\Sigma$上最多有两个零点. 不失一般性,下面假设G(h)在\Sigma上恰好有两个零点.

引理 4.3^[7] $\#{{F}_{1}}(h)\le \#{{\beta }_{3}}(h)+\#{{F}_{2}}(h)+2$.

引理 4.4 当$h\in\Sigma$时,令$\bar{\omega }(h)=\frac{{{F}_{2}}(h)}{{{{{I}'}}_{0,1}}(h)}$, 则$\bar{\omega}(h)$满足

$G(h){{\delta }_{4}}(h){\bar{\omega }}'(h)=-{{a}_{12}}(h){{{\bar{\omega }}}^{2}}(h)+{{R}_{1}}(h)\bar{\omega }(h)+{{R}_{0}}(h),$

(4.4)

其中$\deg {{R}_{0}}(h)\le 2[\frac{n-1}{4}]+6[\frac{n-3}{4}]+12$,

${{R}_{0}}=G(h)({{{{\alpha }'}}_{4}}{{\delta }_{4}}-{{\alpha }_{4}}{{{{\delta }'}}_{4}})-{{a}_{12}}(h)\alpha _{4}^{2}-{{\alpha }_{4}}{{\delta }_{4}}({{a}_{42}}-{{a}_{11}})+{{a}_{41}}(h)\delta _{4}^{2}.$

证 因为$\bar{\omega }(h)={{\alpha }_{4}}(h)+{{\delta }_{4}}(h)\omega (h),{\bar{\omega }}'(h)={{{{\alpha }'}}_{4}}(h)+{{{{\delta }'}}_{4}}(h)\omega (h)+{{\delta }_{4}}(h){\omega }'(h)$. 注意到(3.10)式,即可得结论成立.

引理 4.5^[7] $\#\bar{\omega }(h)\le \#{{R}_{0}}(h)+\#{{\delta }_{4}}(h)+1$ .

定理1.1的证明 如果$n\geq5$,由引理4.1,4.3和4.5可得

$\#I(h)\le \#{{\gamma }_{1}}(h)+\#{{\beta }_{3}}(h)+\#{{\delta }_{4}}(h)+\#{{R}_{0}}(h)+5\le 3n+3[\frac{n-1}{4}]+14.$

因此,

$Z(n)\le 3n+3[\frac{n-1}{4}]+14.$

当n=1时,$\text{I(h)=}{{\text{c}}_{0}}{{\text{I}}_{0,1}}\text{(h)}$,其中c₀是非零常数. 因为${{I}_{0,1}}\ne 0$,所以Z(1)=Z(2)=0.

当n=3时,$\text{I(h)=}{{\text{c}}_{1}}{{\text{I}}_{0,1}}\text{+}{{\text{c}}_{2}}{{\text{I}}_{0,3}}\text{+}{{\text{c}}_{3}}{{\text{I}}_{2,1}}\text{,}$其中c₁,c₂和c₃是常数. 由(3.8)式可知

$\text{G(h)I }\!\!'\!\!\text{ }\!\!'\!\!\text{ (h)=}{{\text{c}}_{4}}\text{(h)I}_{0,1}^{'}\text{+}{{\text{c}}_{5}}\text{(h)Z }\!\!'\!\!\text{ :=W(h),}$

其中${{c}_{4}}(h)={{c}_{1}}{{a}_{11}}(h)+{{c}_{2}}{{a}_{21}}(h)+{{c}_{3}}{{a}_{31}}(h),{{c}_{5}}(h)={{c}_{1}}{{a}_{12}}(h)+{{c}_{2}}{{a}_{22}}(h)+{{c}_{3}}{{a}_{32}}(h),\deg {{c}_{4}}(h)\le 3$和$\deg {{c}_{5}}(h)\le 2$.

与引理4.4的证明相似,可得

$G(h){{c}_{5}}(h){\nu }'(h)=-{{a}_{12}}(h){{\nu }^{2}}(h)+{{c}_{6}}(h)\nu (h)+{{c}_{7}}(h),$

其中$\nu (h)=\frac{W(h)}{{{{{I}'}}_{0,1}}},{{c}_{7}}(h)=G(h)({{{{c}'}}_{4}}{{c}_{5}}-{{c}_{4}}{{{{c}'}}_{5}})-{{a}_{12}}c_{4}^{2}-{{c}_{4}}{{c}_{5}}({{a}_{41}}-{{a}_{11}})+{{a}_{41}}c_{5}^{2}$,且$\deg {{c}_{7}}(h)\le 8$.

由引理4.5可得

$\#{I}''(h)=\#W(h)=\#\nu \le \#{{c}_{5}}(h)+\#{{c}_{7}}(h)+1\le 11.$

又I(0)=0,所以$\#I(h)\leq12$. 因此,$Z(3)=Z(4)\leq12$.