考虑如下近Hamilton系统

$\ \left\{\begin{array}{ll} \dot{x} = \frac{\partial H}{\partial y}+\varepsilon f(x,y),\\[3mm] \dot{y} = -\frac{\partial H}{\partial x}+\varepsilon g(x,y),\end{array}\right.$

(1.1)

其中$0＜\varepsilon \ll1$, $H(x,y)$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$n$次多项式.假设未扰动系统(1.1)$_{\varepsilon =0}$有一族由$H(x,y)=h\ (h\in \Sigma)$定义的闭轨线, 其中$\Sigma$是$\Gamma_h$存在的最大开区间.寻找Abelian积分

$I(h)=\oint_{\Gamma_h}g(x,y){\rm d}x-f(x,y){\rm d}y,h\in\Sigma$

的孤立零点个数的最小上界$Z(m,n)$称为弱Hilbert 16问题或Hilbert-Arnold问题^[1].

当$m$是有限数且$n$任意正整数时, 相关的研究很多^[4-8].例如, 当$m=2$时, 如果未扰动系统存在周期环域, 则相应的Hamilton函数的规范型为

$H(x,y)=\frac{1}{2}(x^2+y^2)-\frac{1}{3}x^3+axy^2+\frac{1}{3}by^3,$

其中$b^2=-4a(2a+1),$ $-\frac{1}{2}<a＜0$, Horozov和Iliev^[2]证明了$Z(2,n)\leq5n+15$.当$m=3$且

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

其中$\deg U(x)=4$且使得(1.1)$_{\varepsilon =0}$存在周期环域, 赵育林和张芷芬^[3]证明$H(x,y)$的规范型为下列四种之一

$H(x,y)=\frac{1}{2}y^2+\frac{1}{2}x^2-\frac{1}{3}x^3-\frac{1}{4a}x^4,\\ H(x,y)=\frac{1}{2}y^2+\frac{1}{2}x^2+\frac{1}{4}x^4,\\ H(x,y)=y^2+x^2-x^4,\\ H(x,y)=\frac{1}{2}y^2+\frac{1}{4}x^4,$

并且得到$Z(3,n)\leq 7n+5$.另外, Petrov^{[4, 5]}研究了Hamilton函数$H(x,y)=y^2-x+x^3$和$H(x,y)=y^2+x^2-x^4$.周鑫和李翠萍^{[6, 7]}研究了Hamilton函数$H(x,y)=-x^2+x^4+y^4$和$H(x,y)=x^2+y^2+ax^4+y^4$.对于Hamilton函数$H(x,y)=x^2+y^2-x^4+ax^2y^2+y^4\ (a>-2)$, 吴娟娟等^[8]证明了

$Z(3,n)\leq2\Big[\frac{n-1}{4}\Big]+12\Big[\frac{n-3}{4}\Big]+23. $

对于Hamilton函数$H(x,y)=\frac{1}{2}y^2+\frac{a}{2}x^2+\frac{b}{4}x^4+\frac{c}{6}x^6$, 赵丽琴等^[9]证明了$Z(5,n)\leq54n-13. $本文研究如下Hamilton函数

$H(x,y)=x^2+y^2+2xy+a(x^4+y^4),$

(1.2)

其中$a>0$.与系统(1.2)相对应的Hamilton系统为

$\left\{\begin{array}{ll} \dot{x} = 2(x+y+2ay^3),\\ \dot{y} = -2(x+y+2ax^3). \end{array}\right.$

(1.3)

系统(1.3)有一个奇点$O(0,0)$, 卵形线$H(x,y)=0$对应中心$O$.如图 1所示.

系统(1.3)有一个无界周期环域$\Gamma_h$, 其中

$\Gamma_h=\{(x,y)|H(x,y)=h,h\in(0,+\infty)\}.$

本文的主要结果如下.

定理1.1 对于Hamilton函数

$H(x,y)=x^2+y^2+2xy+a(x^4+y^4),\ a>0,$

系统(1.1)最多有$3[\frac{n-1}{4}]+12[\frac{n-3}{4}]+23$(计重数)个极限环, 其中$[p]$表示$p$的整数部分.

本小节, 我们将把$I(h)$表示为四个基本积分的线性组合.记

$I_{ij}(h)=\oint_{\Gamma_h}x^iy^j{\rm d}x.$

显然当$h\in(0,+\infty)$时, $\Gamma_h$关于原点对称, 所以

$I_{i,2j}(h)=I_{2i+1,2j+1}(h)\equiv0. $

我们只需考虑$I_{2i,2j+1}(h)$即可.

引理2.1 如果$h\in(0,+\infty)$且$i+j\geq2$, 则

$I_{2i,2j+1}(h)=a(h)I_{0,1}(h)+b(h)I_{0,3}(h)+c(h)I_{2,1}(h)+d(h)I_{2,3}(h),$

(2.1)

其中$a(h)$, $b(h)$, $c(h)$和$d(h)$是关于$h$的多项式,

$\deg a(h)\leq\Big[\frac{n-1}{4}\Big]=p,~ \deg b(h),$

$ \deg c(h)\leq\Big[\frac{n-3}{4}\Big]=q,~ \deg d(h)\leq \Big[\frac{n-1}{4}\Big]-1=p-1. $

证  (1.2)式两端同时关于$x$求导, 可得

$x+y\frac{\partial y}{\partial x}+y+x\frac{\partial y}{\partial x}+2ax^3+2ay^3\frac{\partial y}{\partial x}=0.$

(2.2)

(2.2)式两端同时乘以$x^{2i-3}y^{2j+1}{\rm d}x$并分部积分可得

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

(2.3)

类似的, (1.2)式两端同乘以$x^{2i}y^{2j-3}{\rm d}x$并在$\Gamma_h$上积分可得

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

(2.4)

再由(2.3)和(2.4)式可得

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

(2.5)

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

(2.6)

下面用数学归纳法进行证明.当$n=5, 7$时, 从(2.5)和(2.6)式可得

$\left\{\begin{array}{ll} I_{0,5}=\frac{5h}{6a}I_{0,1}-\frac{5}{12a}I_{2,1}-\frac{35}{36a}I_{2,3},\ \ I_{4,1}=\frac{h}{6a}I_{0,1}-\frac{1}{36a}I_{0,3}-\frac{7}{12a}I_{2,1},\\[3mm] I_{0,7}=\frac{7h}{8a}I_{0,3}-\frac{7}{16a}I_{2,3}-\frac{77}{80a}I_{0,5},\ \ I_{2,5}=\frac{5h}{8a}I_{2,1}-\frac{5}{16a}I_{4,1}-\frac{15}{16a}I_{2,3},\\[3mm] I_{4,3}=\frac{h}{8a}I_{0,3}-\frac{9}{16a}I_{2,3}-\frac{3}{80a}I_{0,5},\ \ I_{6,1}=\frac{3h}{8a}I_{2,1}-\frac{11}{16a}I_{4,1}-\frac{1}{16a}I_{2,3},\end{array}\right.$

(2.7)

即当$n=5,\ 7$时, 结论成立.假设当$2i+2j+1\leq 2k-1\ (k\geq3)$时, 结论成立.当$2i+2j+1=2k+1\ (k\geq2)$时, 在(2.5)式中取$(i,j)=(0,k)$, $(1,k-1)$, $(2,k-2)$, 在(2.6)式中取$(i,j)=(3,k-3)$, $(4,k-4)\cdots,(k,0)$可得

$\left(\begin{array}{cc} I_{0,2k+1}\\ I_{2,2k-1}\\ I_{4,2k-3}\\ I_{6,2k-5}\\ \vdots\\ I_{2k-4,5}\\ I_{2k-2,3}\\ I_{2k,1} \end{array}\right) =\frac{1}{2a(k+1)}\left(\begin{array}{cc} 2(2k+1)hI_{0,2k-3}-\frac{(2k+1)}{2}I_{2,2k-3}-\frac{(4k-1)(2k+1)}{4k-2}I_{0,2k-1}\\[3mm] 2(2k-1)hI_{2,2k-5}-\frac{(2k-1)}{2}I_{4,2k-5}-\frac{(4k-3)(2k-1)}{4k-6}I_{2,2k-3}\\[3mm] 2(2k-3)hI_{4,2k-7}-\frac{(2k-3)}{2}I_{6,2k-7}-\frac{(4k-5)(2k-3)}{4k-10}I_{4,2k-5}\\[3mm] 3hI_{2,2k-5}-\frac{(2k+5)}{2}I_{4,2k-5}-\frac{3(2k-5)}{2(2k-3)}I_{2,2k-3}\\[2mm] \vdots\\[2mm] (2k-7)hI_{2k-8,5}-\frac{(4k-5)}{2}I_{2k-6,5}-\frac{5(2k-7)}{14}I_{2k-8,7}\\[3mm] (2k-5)hI_{2k-6,3}-\frac{(4k-3)}{2}I_{2k-4,3}-\frac{3(2k-5)}{10}I_{2k-6,5}\\[3mm] (2k-3)hI_{2k-4,1}-\frac{(4k-1)}{2}I_{2k-2,1}-\frac{(2k-3)}{6}I_{2k-4,3} \end{array}\right).$

所以

$\begin{eqnarray*}I_{2i,2j+1}(h)&=a_{2k-1}I_{0,1}+b_{2k-1}I_{0,3}+c_{2k-1}I_{2,1}+d_{2k-1}I_{2,3}\\& +h(a_{2k-3}I_{0,1}+b_{2k-3}I_{0,3}+c_{2k-3}I_{2,1}+d_{2k-3}I_{2,3})\\ &:=a_{2k+1}I_{0,1}+b_{2k+1}I_{0,3}+c_{2k+1}I_{2,1}+d_{2k+1}I_{2,3},\end{eqnarray*}$

其中$a_{2k-s}$, $b_{2k-s}$, $c_{2k-s}$和$d_{2k-s}\ (s=1,3)$是关于$h$的多项式, 且$\deg a_{2k-s}\leq[\frac{2k-1-s}{4}]$, $\deg b_{2k-s}$, $\deg c_{2k-s}\leq[\frac{2k-3-s}{4}]$和$\deg d_{2k-s}\leq[\frac{2k-5-s}{4}]$.容易验证

$\deg a_{2k+1}=\max\Big\{\Big[\frac{2k-2}{4}\Big],\Big[\frac{2k-4}{4}\Big]+1\Big\}=\Big[\frac{k}{2}\Big].$

同理可得

$\deg b_{2k+1},\deg c_{2k+1}\leq\Big[\frac{k-1}{2}\Big],\deg d_{2k+1}\leq\Big[\frac{k}{2}\Big]-1.$

证毕.

由引理2.1, 可得Abelian积分$I(h)$的代数结构如下.

引理2.2 对于$h\in(0,+\infty)$, 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$\delta (h)$是关于$h$的多项式, 且

$\deg \alpha(h)\leq\Big[\frac{n-1}{4}\Big]=p,~ \deg \beta(h),$

$ \deg \gamma(h)\leq\Big[\frac{n-3}{4}\Big]=q,~ \deg \delta (h)\leq\Big[\frac{n-1}{4}\Big]-1=p-1. $

本小节将得到Abelian积分$I(h)$的四个生成元$I_{0,1}$, $I_{0,3}$, $I_{2,1}$和$I_{2,3}$满足的Picard-Fuchs方程.

引理3.1 向量函数$V=(I_{0,1},I_{0,3},I_{2,1},I_{2,3})^T$满足Picard-Fuchs方程

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

(3.1)

其中

$B=\left(\begin{array}{cccc} 2&0&0&~~0\\[2mm] -\frac{3}{4a}&1&0&~~0\\[3mm] -\frac{1}{4a}&0&1&~~0\\[3mm] \frac{1}{8 a^2}&~~ -\frac{1}{12a}~~& -\frac{1}{4a}&~ \frac{2}{3} \end{array}\right),\ \ C=\left(\begin{array}{cccc} 0&~ -\frac{1}{3}~~&-1&~~0\\[3mm] 0& \frac{3}{8a}& \frac{3}{8a}& ~~-\frac{1}{2}\\[3mm] 0& \frac{1}{24a}& \frac{3}{8a}& ~~-\frac{1}{6}\\[3mm] 0& -\frac{1}{24 a^2}& -\frac{1}{8 a^2}&~~ \frac{5}{12a} \end{array}\right).$

证 由(1.2)式可得

$\frac{\partial y}{\partial h}=\frac{1}{2(x+y+2ay^3)},$

即

$I'_{i,j}=\frac{j}{2}\oint_{\Gamma_h}\frac{x^iy^{j-1}}{x+y+2ay^3}{\rm d}x.$

(3.2)

因此

$I_{i,j}=\frac{2}{j+1}I'_{i+1,j+1}+\frac{2}{j+2}I'_{i,j+2}+\frac{4a}{j+4}I'_{i,j+4}.$

(3.3)

(3.3)式两端同乘以$h$, 可得

$hI'_{i,j}=I'_{i+2,j}+\frac{j}{j+2}I'_{i,j+2}+aI'_{i+4,j}+\frac{aj}{j+4}I'_{i,j+4}+\frac{2j}{j+1}I'_{i+1,j+1}.$

(3.4)

另一方面

$I_{i,j}=\oint_{\Gamma_h}x^iy^j{\rm d}x=-\frac{j}{i+1}\oint_{\Gamma_h}x^{i+1}y^{j-1}{\rm d}y\\ =\frac{j}{i+1}\oint_{\Gamma_h}\frac{x^{i+1}y^{j-1}(x+y+2ax^3)}{x+y+2ay^3}{\rm d}x\\ =\frac{2j}{i+1}\Big[\frac{1}{j}I'_{i+2,j}+\frac{2}{j+1}I'_{i+1,j+1}+\frac{4a}{j}I'_{i+4,j}\Big].$

(3.5)

由(3.4)-(3.6)式可得

$I_{i,j}=\frac{2}{i+j+1}\Big[2hI'_{i,j}-I'_{i+2,j}-\frac{2j}{j+1}I'_{i+1,j+1}-\frac{j}{j+2}I'_{i,j+2}\Big].$

(3.6)

再由(3.7)式可得

$I_{0,1}=2hI'_{0,1}-I'_{2,1}-\frac{1}{3}I'_{0,3},\ \ I_{0,3}=hI'_{0,3}-\frac{1}{2}I'_{2,3}-\frac{3}{10}I'_{0,5},\\ I_{2,1}=hI'_{2,1}-\frac{1}{2}I'_{4,1}-\frac{1}{6}I'_{2,3},\ \ I_{2,3}=\frac{2}{3}hI'_{2,3}-\frac{1}{3}I'_{4,3}-\frac{1}{5}I'_{2,5}.$

注意到(2.5)和(2.6)式, 即可得结论成立.证毕.

取

$Z=\frac{1}{12a}I_{0,3}+\frac{1}{4a}I_{2,1}+\frac{1}{3}I_{2,3}.$

(3.7)

我们可以得到下面的引理3.2.

引理3.2 函数$I_{0,1}$, $I_{0,3}$, $I_{2,1}$和$Z$满足

$G(h)\left(\begin{array}{cc} I"_{0,1}\\ I"_{0,3}\\ I"_{2,1}\\ Z" \end{array}\right)= \left(\begin{array}{cc} a_{11}(h)&a_{12}(h)\\ a_{21}(h)&a_{22}(h)\\ a_{31}(h)&a_{32}(h)\\ a_{41}(h)&a_{42}(h) \end{array}\right) \left(\begin{array}{cc} I'_{0,1}\\ Z' \end{array}\right),$

(3.8)

其中

$\begin{eqnarray*}G(h)=&&\frac{4}{3}h\Big(h+\frac{1}{4 a}\Big)^2\Big(h+\frac{1}{2a}\Big),\\ &&a_{11}(h)=-\frac{2}{3}h\Big(h+\frac{1}{4 a}\Big)\Big(h+\frac{1}{2a}\Big),\ a_{12}(h)=\frac{1}{3}\Big(h+\frac{1}{4 a}\Big),\\ &&a_{21}(h)=\frac{1}{2 a}h\Big(h+\frac{1}{4 a}\Big)\Big(h+\frac{1}{2a}\Big),\ a_{22}(h)=h\Big(h+\frac{1}{4a}\Big),\\ &&a_{31}(h)=\frac{1}{6a}h\Big(h+\frac{1}{4 a}\Big)\Big(h+\frac{1}{2a}\Big),\ a_{32}(h)=\frac{1}{3}h\Big(h+\frac{1}{4 a}\Big),\\ &&a_{41}(h)=\frac{1}{12 a^2}h\Big(h+\frac{1}{4 a}\Big)\Big(h+\frac{1}{2a}\Big),\ a_{42}(h)=\frac{2}{3}h\Big(h+\frac{1}{4 a}\Big)\Big(h+\frac{1}{2a}\Big).\end{eqnarray*}$

证 对(3.1)式两端同时关于$h$求导可得

$(Bh+C)V"=(I-B)V',$

(3.9)

其中$I$是$4\times4$单位矩阵.把(3.7)式代入(3.9)式即可得(3.8)式.证毕.

容易验证, 当$h\in(0,+\infty)$时, $I'_{0,1}\neq0$.所以可以得到如下引理.

引理3.3 当$h\in(0,+\infty)$时, 令

$\omega(h)=\frac{Z'(h)}{I'_{0,1}(h)},$

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)式的第一个和第四个方程可得(3.10)式.证毕.

在下面的讨论中, 记$\#\varphi (h)$为函数$\varphi (h)$在$(0,+\infty)$上零点的个数(计重数). $B(n)$为Abelian积分$I(h)$在$(0,+\infty)$上零点个数(计重数), 其中$n=\max\{\deg f,\deg g\}$.由(2.8), (3.1)和(3.7)式可得

$\begin{array}{l} I(h)=\alpha_1(h)I'_{0,1}+\beta_1(h)I'_{0,3}+\gamma_1(h)I'_{2,1}+\delta _1(h)Z',\\ I'(h)=\alpha_2(h)I'_{0,1}+\beta_2(h)I'_{0,3}+\gamma_2(h)I'_{2,1}+\delta _2(h)Z',\end{array}$

(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$\deg \alpha_s(h)\leq p-s+2$, $\deg \beta_s(h)$, $\deg \gamma_s(h)\leq q-s+2$, $\deg \delta _s(h)\leq p-s+1 \ (s=1,2)$.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$\deg \alpha_3(h)\leq p+q+1$, $\deg \beta_3(h)\leq 2q+1$, $\deg \delta _3(h)\leq p+q$.

引理4.1^{[10, 引理5.1]} $\#I(h)\leq\#\gamma_1(h)+\#F_1(h)+1.$

引理4.2 设$S$是$\beta_3(h)$在$(0,+\infty)$上零点组成的集合, 则当$h\in(0,+\infty)\setminus S$时, 有

$\Big(\frac{F_1(h)}{\beta_3(h)}\Big)'=\frac{F_2(h)}{G(h)\beta^2_3(h)},$

其中$F_2(h)=\alpha_4(h)I'_{0,1}+\delta _4(h)Z'$, 且$\deg \alpha_4(h)\leq p+3q+5$, $\deg \delta _4(h)\leq p+3q+4$.

证 由(3.8)式, 当$h\in(0,+\infty)\setminus S$时, 有

$\Big(\frac{F_1(h)}{\beta_3(h)}\Big)'=\Big(\frac{\alpha_3}{\beta_3}\Big)'I'_{0,1}+\Big(\frac{\alpha_3}{\beta_3}\Big)I"_{0,1}+ +I"_{0,3}+\Big(\frac{\delta _3}{\beta_3}\Big)'Z'+\Big(\frac{\delta _3}{\beta_3}\Big)Z"\\ =\frac{1}{G(h)\beta^2_3(h)}(\alpha_4(h)I'_{0,1}+\delta _4(h)Z'),$

(4.2)

其中

$\begin{array}{l} \alpha_4(h)=G(h)(\alpha'_3\beta_3-\alpha_3\beta'_3)+a_{11}(h)\alpha_3\beta_3+a_{21}(h)\beta^2_3+a_{41}(h)\delta _3\beta_3,\\ \delta _4(h)=G(h)(\delta '_3\beta_3-\delta _3\beta'_3)+a_{12}(h)\alpha_3\beta_3+a_{22}(h)\beta^2_3+a_{42}(h)\delta _3\beta_3,\end{array}$

(4.3)

这意味着$\deg \alpha_4(h)\leq p+3q+5$, $\deg \delta _4(h)\leq p+3q+4$.证毕.

引理4.3 ^{[11,中引理4.4]}  $\#F_1(h)\leq\#\beta_3(h)+\#F_2(h)+2.$

引理4.4 当$h\in(0,+\infty)$时, 令$\chi(h)=\frac{F_2(h)}{I'_{01}(h)}$, 则$\chi(h)$满足

$G(h)\delta _4(h)\chi'(h)=-a_{12}(h)\chi^2(h)+R_1(h)\chi(h)+R_0(h),$

(4.4)

其中$\deg R_0(h)\leq2p+6q+12$,

$R_0(h)=G(h)(\alpha'_4\delta _4-\alpha_4\delta '_4)-a_{12}(h)\alpha_4^2-\alpha_4\delta _4(a_{42}-a_{11})+a_{41}(h)\delta _4^2.$

证 因为$\chi(h)=\alpha_4(h)+\delta _4(h)\omega(h)$, $\chi'(h)=\alpha'_4(h)+\delta '_4(h)\omega(h)+\delta _4(h)\omega'(h)$.注意到(3.10)式, 即可得结论成立.证毕.

引理4.5^{[3, 引理4.5]} $\#\chi(h)\leq\#R_0(h)+\#\delta _4(h)+1.$

定理1.1的证明 如果$n\geq5$, 由引理4.1, 4.3和4.5可得

$\begin{eqnarray*}\#I(h)&\leq&\#\gamma_1(h)+\#\beta_3(h)+\#\delta _4(h)+\#R_0(h)+5\\ &\leq&3p+12q+23=3\Big[\frac{n-1}{4}\Big]+12\Big[\frac{n-3}{4}\Big]+23.\end{eqnarray*}$

因此

$B(n)\leq3\Big[\frac{n-1}{4}\Big]+12\Big[\frac{n-3}{4}\Big]+23.$

当$n=1$时, $I(h)=c_0I_{01}(h)$, 其中$c_0$是非零常数.又$I_{01}\neq0$, 所以$B(1)=0$.

当$n=3$时, 函数$I(h)$可表示为

$I(h)=c_1I_{0,1}+c_2I_{0,3}+c_3I_{2,1},$

其中$c_1$, $c_2$和$c_3$是常数.由(3.8)式可得

$G(h)I"(h)=c_4(h)I_{0,1}+c_5(h)Z':=W(h),$

其中$c_4(h)=(c_1a_{11}(h)+c_2a_{21}(h)+c_3a_{31}(h))I'_{0,1}$, $c_5(h)=(c_1a_{12}(h)+c_2a_{22}(h)+c_3a_{32}(h))I'_{0,1}$, $\deg c_4(h)\leq3$和$\deg c_5(h)\leq2$.

类似于引理4.4的证明, 我们得到

$G(h)c_5(h)\nu'(h)=-a_{12}(h)\nu^2(h)+c_6(h)\nu(h)+c_7(h),$

其中$\nu(h)=\frac{W(h)}{I'_{0,1}}$, $c_7(h)=G(h)(c'_4c_5-c_4c'_5)-a_{12}c_4^2-c_4c_5(a_{41}-a_{11})+a_{41}c_5^2$, 且$\deg c_7(h)\leq8$.

因为$G(h)$在$(0,+\infty)$上是非零常数.由引理4.5可得

$\#I"(h)=\#W(h)=\#\nu\leq\#c_5(h)+\#c_7(h)+1\leq11.$

再由$I(0)=0$, 可得$\#I(h)\leq12$.因此, $B(3)\leq12$.再由Poincare-Pontryagin定理^[12]可知定理1.1成立.