考虑近-Hamiltonian系统

(1.1) $ \begin{eqnarray} \frac{{\rm d}x}{{\rm d}t} = \frac{\partial H(x, y)}{\partial y}+\varepsilon f(x, y), \ \ \frac{{\rm d}y}{{\rm d}t} = -\frac{\partial H(x, y)}{\partial x}+\varepsilon g(x, y), \end{eqnarray} $

其中$ 0 < |\varepsilon|\ll1 $, $ H(x, y) $是关于$ x $和$ y $的$ m+1 $次多项式, $ f(x, y) $和$ g(x, y) $是关于$ x $和$ y $的$ n $次多项式.假设Hamiltonian系统$ (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\} $.确定Abelian积分

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

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

很多数学工作者都致力于研究Abelian积分$ I(h) $的零点个数问题.当$ I(h) $的生成元满足一个Picard-Fuchs方程时,有很多优秀的工作.例如,如果$ m = 2 $且未扰动系统至少有一个中心,相应Hamiltonian函数$ H(x, y) $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. $

Horozov和Iliev^[2]证明$ Z(2, n)\leq5n+15 $.如果$ H(x, y) = \frac{1}{2}y^2+U(x) $,其中$ U(x) $使得系统$ (1.1)|_0 $有一个中心且$ \deg U(x) = 4 $,赵育林和张芷芬^[3]证明$ Z(3, n)\leq7n+5 $.对于

$ \begin{eqnarray*} H(x, y) = x^2+y^2+ax^4 +bx^2 y^2 +cy^4, \; \; ac(4ac- b^2)\neq 0, \end{eqnarray*} $

周鑫和李翠萍得到相应$ I(h) $的代数结构,并给出当$ a > 0 $, $ b = 0 $, $ c = 1 $时$ I(h) $的零点个数的上界^[4].对于

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

吴娟娟等^[5]证明$ Z(3, n)\leq2[\frac{n-1}{4}]+12[\frac{n-3}{4}]+23 $.

如果$ I(h) $的生成元满足两个不同的Picard-Fuchs方程时,相关的研究结果较少.对于

$ H(x, y) = \frac{1}{2}y^2+\frac{a}{2}x^2+\frac{b}{4}x^4+\frac{c}{6}x^6, \, c\neq0, $

赵丽琴等^[6]证明$ Z(5, n)\leq54n-13 $.杨纪华和赵丽琴得到Hamitonian函数

$ \begin{eqnarray*} H(x, y) = -x^2+rx^2y^2+x^4+y^4, \; \; r\geq0, \ r\neq2 \end{eqnarray*} $

的Abelian积分的在不同周期环域上零点个数的上界^[7].

受文献[6-7]的启发,本文研究一类Hamiltonian系统的Abelian积分的零点个数问题,该Abelian积分的生成元满足两个Picard-Fuchs方程.为此,我们对系统(1.1)作如下假设:

(H1)系统(1.1)的Abelian积分$ I(h) $可表示为

(1.2) $ \begin{eqnarray} I(h) = \sum\limits_{i = 1}^k\alpha_i(h)I_i(h)+\beta_1(h)J_1(h)+\beta_2(h)J_2(h), \ h\in \Sigma, \end{eqnarray} $

其中$ \alpha_i(h)\ (i = 1, 2, \cdots, k) $和$ \beta_j(h) $$ (j = 1, 2) $是关于$ h $的多项式,且$ \deg\alpha_i(h)\leq n_i $$ (i = 1, 2, \cdots, k) $, $ \deg\beta_i(h)\leq m_j $$ (j = 1, 2) $, $ n_1\geq n_2\geq\cdots\geq n_k $, $ m_1\geq m_2 $.

(H2)向量函数$ V_1(h) = \big(I_1(h), I_2(h), \cdots, I_k(h)\big)^T $和$ V_2(h) = \big(J_1(h), J_2(h)\big)^T $满足Picard-Fuchs方程

(1.3) $ \begin{equation} \begin{array}{ll} V_1(h) = (B_1h+C_1)V'_1(h)\triangleq A_1(h)V'_1(h), \\ V_2(h) = (B_2h+C_2)V'_2(h)\triangleq A_2(h)V'_2(h), \end{array} \end{equation} $

其中$ B_1 $和$ C_1 $是$ k\times k $阶常数矩阵, $ B_2 $和$ C_2 $是$ 2\times2 $阶常数矩阵.

(H3)存在$ Z(h) = a_2I_2(h)+a_3I_3(h)+\cdots+a_kI_k(h) $使得

(1.4) $ \begin{eqnarray} G_1(h)\left(\begin{array}{cc} I''_{1}(h)\\ I''_{2}(h)\\ \vdots\\ I''_{k-1}(h)\\ Z''(h) \end{array}\right) = \left(\begin{array}{cc} a_{11}(h) &a_{12}(h)\\ a_{21}(h) &a_{22}(h)\\ \vdots &\vdots\\ a_{k-11}(h) &a_{k-12}(h)\\ a_{k1}(h) &a_{k2}(h) \end{array}\right) \left(\begin{array}{cc} I'_{1}(h)\\ Z'(h) \end{array}\right), \end{eqnarray} $

其中$ a_i\ (i = 2, \cdots, k) $是常数, $ a_{ij}(h) $是多项式, $ G_1(h) = \det\big(A_1(h)\big) $, $ \deg a_{ij}(h)\leq\deg G_1(h) $$ (i = 1, 2, \cdots, k;j = 1, 2) $.

本文的主要结果是如下定理.

定理1.1  假设(H1)–(H3)成立,且当$ h\in\Sigma $时, $ I'_1(h)\neq0 $, $ J_1(h)\neq0 $,则系统(1.1)的Abelian积分$ I(h) $至多有$ (3k+4)n_1+(6k+23)m_1+13k+30 $个零点(计重数).

本文中,我们用$ \#\{\phi(h) = 0, h\in(a, b)\} $表示函数$ \phi(h) $在区间$ (a, b) $上的零点个数(计重数).设$ V $是定义在开区间$ {\Bbb I} $上的实解析函数组成的向量空间.

定义2.1^[8]  如果$ V $中任一非零函数至多有$ \dim(V)-1 $个零点(计重数),称$ V $为Chebyshev空间.

设$ S $是二阶线性解析微分方程

(2.1) $ \begin{equation} x''(t)+a_1(t)x'(t)+a_2(t)x(t) = 0 \end{equation} $

在开区间$ {\Bbb I} $上的解空间.

命题2.1^[8]  方程(2.1)的解空间$ S $是区间$ {\Bbb I} $上的Chebyshev空间当且仅当$ S $中存在一个在$ {\Bbb I} $上处处非零的元.

命题2.2^[8]  假设齐次方程(2.1)的解空间是Chebyshev空间, $ R(t) $是解析函数并且在区间$ {\Bbb I} $上有$ l $个零点(计重数).则非齐次方程

$ x''(t)+a_1(t)x'(t)+a_2(t)x(t) = R(t) $

的解$ x(t) $在$ {\Bbb I} $上至多有$ l+2 $个零点(计重数).

引理2.1   (1)假设当$ h\in\Sigma $时, $ I'_1(h)\neq0 $,令$ \omega_1(h) = \frac{Z'(h)}{I'_1(h)} $,则$ \omega_1(h) $满足Riccati方程

(2.2) $ \begin{equation} G_1(h)\omega'_1(h) = -a_{12}(h)\omega_1^2(h)+\big[a_{k2}(h)-a_{11}(h)\big]\omega_1(h)+a_{k1}(h). \end{equation} $

(2)假设当$ h\in\Sigma $时, $ J_1(h)\neq0 $,令$ \omega_2(h) = \frac{J_2(h)}{J_1(h)} $,则$ \omega_2(h) $满足Riccati方程

(2.3) $ \begin{equation} G_2(h)\omega'_2(h) = -b^*_{12}(h)\omega_2^2(h)+\big[b^*_{22}(h)-b^*_{11}(h)\big]\omega_2(h)+b^*_{21}(h), \end{equation} $

其中$ G_2(h) = \det\big(A_2(h)\big), \ (B_2h+C_2)^* = \left(\begin{array}{cc}b^*_{11}(h) &b^*_{12}(h)\\ b^*_{21}(h) &b^*_{22}(h) \end{array}\right) $是$ B_2h+C_2 $的伴随矩阵.

证   (2.2)式可由(1.4)式得到. (2.3)式可由(1.3)式得到.证毕.

引理2.2  假设当$ h\in\Sigma $时, $ J_1(h)\neq0 $.令

(2.4) $ \begin{eqnarray} \Psi(h) = \big(\beta_1(h), \beta_2(h)\big)\left(\begin{array}{cc} J_1(h)\\ J_2(h) \end{array}\right)\triangleq\tau_2(h)V_2(h), \ h\in\Sigma, \end{eqnarray} $

则$ \Psi(h) $在$ \Sigma $上至多有$ 3m_1+2 $个零点(计重数).

证  令$ \chi_2(h) = \frac{\Psi(h)}{J_1(h)} $,所以$ \chi_2(h) = \beta_1(h)+\beta_2(h)\omega_2(h) $.由(2.3)式可得$ \chi_2(h) $满足

$ G_2(h)\beta_2(h)\chi'_2(h) = -b^*_{12}(h)\chi_2^2(h)+F_1(h)\chi_2(h)+F_0(h), $

其中

$ \begin{eqnarray*} F_1(h)& = &G_2(h)\beta'_2(h)+2b^*_{12}(h)\beta_1(h)+\big(b^*_{22}(h)- b^*_{11}(h)\big)\beta_2(h), \\ F_0(h)& = &G_2(h)\big(\beta'_1(h)\beta_2(h)-\beta_1(h)\beta'_2(h)\big)- b^*_{12}(h)\beta_1^2(h)\\ &&+b^*_{21}(h)\beta_2^2(h)-\beta_1(h)\beta_2(h)\big( b^*_{22}(h)-b^*_{11}(h)\big) \end{eqnarray*} $

且$ \deg F_0(h)\leq2m_1+1 $.由文献[3]中引理4.4可得

$ \#\{\chi_2(h) = 0\}\leq\#\{F_0(h) = 0\}+\#\{\beta_2(h) = 0\}+1. $

因此,当$ h\in \Sigma $

(2.5) $ \begin{eqnarray} \#\{\Psi(h) = 0\} = \#\{\chi_2(h) = 0\}\leq3m_1+2. \end{eqnarray} $

证毕.

引理2.3  如果$ K = m_1+m_2+3 $,则当$ h\in\Sigma $时,存在$ K $, $ K-1 $和$ K-2 $次多项式$ P_2(h) $, $ P_1(h) $和$ P_0(h) $使得$ L(h)\Psi(h) = 0 $,其中

(2.6) $ \begin{eqnarray} L(h) = P_2(h)\frac{{\rm d}^2}{{\rm d}h^2}+P_1(h)\frac{\rm d}{{\rm d}h}+P_0(h). \end{eqnarray} $

证  由(1.3)式可得

$ V'_2(h) = (I-B_2)^{-1}(B_2h+C_2)V''_2(h), $

其中$ I $是$ 2\times2 $阶单位矩阵.所以

$ \begin{eqnarray*} \Psi(h)& = &\tau_2(h) V_2(h) = \tau_2(h)(B_2h+C_2)V'_2(h)\\ & = &\tau_2(h)(B_2h+C_2)(I-B_2)^{-1}(B_2h+C_2) V''_2(h)\\ &\triangleq&M_{m_1+2}(h)J''_1(h)+M_{m_2+2}(h)J''_2(h), \end{eqnarray*} $

其中$ M_{m_1+2}(h) $表示$ h $的至多$ m_1+2 $多项式,等等.对于$ \Psi'(h) $,我们得到

$ \begin{eqnarray*} \Psi'(h)& = &\tau'_2(h)V_2(h)+\tau_2(h)V'_2(h)\\ & = &\Big(\tau'_2(h)(B_2h+C_2)+\tau_2(h)\Big)(I-B_2)^{-1}(B_2h+C_2) V''_2(h)\\ &\triangleq&M_{m_1+1}(h)J''_1(h)+M_{m_2+1}(h)J''_2(h). \end{eqnarray*} $

同样的可得

$ \begin{eqnarray*} \Psi''(h) = M_{m_1}(h)J''_1(h)+M_{m_2}(h)J''_2(h). \end{eqnarray*} $

假设

(2.7) $ \begin{eqnarray} P_2(h) = \sum\limits_{k = 0}^{K}p_{2, k}h^k, \ \ P_1(h) = \sum\limits_{m = 0}^{K-1}p_{1, m}h^m, \ \ P_0(h) = \sum\limits_{l = 0}^{K-2}p_{0, l}h^l \end{eqnarray} $

是$ h $的多项式,并且系数$ p_{2, k}, \ p_{1, m} $和$ p_{0, l} $待定使得$ L(h)\Psi(h) = 0 $,

(2.8) $ \begin{eqnarray} 0\leq k\leq K, \ \ 0\leq m\leq K-1, \ \ 0\leq l\leq K-2. \end{eqnarray} $

直接计算可得

$ \begin{eqnarray*} L(h)\Psi(h)& = &P_2(h)\Psi''(h)+P_1(h)\Psi'(h)+P_0(h)\Psi(h)\\[0.2cm] & = &P_2(h)\Big[M_{m_1}(h)J''_1(h)+M_{m_2}(h)J''_2(h)\Big]\\ &&+P_1(h)\Big[M_{m_1+1}(h)J''_1(h)+M_{m_2+1}(h)J''_2(h)\Big]\\ &&+P_0(h)\Big[M_{m_1+2}(h)J''_1(h)+M_{m_2+2}(h)J''_2(h)\Big]\\[0.3cm] & = &\Big[P_2(h)M_{m_1}(h)+P_1(h)M_{m_1+1}(h)+P_0(h)M_{m_1+2}(h)\Big]J''_1(h)\\ &&+\Big[P_2(h)M_{m_1}(h)+P_1(h)M_{m_1+1}(h)+P_0(h)M_{m_1+1}(h)\Big]J''_2(h)\\ &\triangleq&S(h)J''_1(h)+T(h)J''_2(h), \end{eqnarray*} $

其中$ S(h) $和$ T(h) $是$ h $的多项式且$ \deg S(h)\leq K+m_1 $, $ \deg T(h)\leq K+m_2 $.令

$ S(h) = \sum\limits_{i = 0}^{K+m_1}x_ih^i, \ \ T(h) = \sum\limits_{j = 0}^{K+m_2}y_jh^j, $

其中$ x_i $和$ y_j $可以由$ p_{2, k} $, $ p_{1, m} $和$ p_{0, l} $线性表示, $ k $, $ m $和$ l $满足(2.8)式.令

(2.9) $ \begin{eqnarray} x_i = 0, \ \ y_j = 0, \ \ 0\leq i\leq K+m_1, \ \ 0\leq j\leq K+m_2, \end{eqnarray} $

则系统(2.9)是一个齐次线性方程组,该方程组有$ 2K+m_1+m_2+2 $个方程, $ 3K+3 $个变量.因为$ 3K-(2K+m_1+m_2+2)\geq1 $,所以存在$ p_{2, k} $, $ p_{1, m} $和$ p_{0, l} $使(2.9)式成立,即可得$ L(h)\Psi(h) = 0 $.证毕.

引理2.4  如果当$ h\in\Sigma $时, $ G_1(h)\neq0 $,则$ L(h)I(h) = R(h) $,其中$ L(h) $由(2.6)式给出,

(2.10) $ \begin{eqnarray} R(h) = \frac{1}{G_1(h)}\Big[Q_0(h)I'_{1}(h)+Q_1(h)Z'(h)\Big]+Q_2(h)I'_{2}(h)+\cdots+Q_{k-1}(h)I'_{k-1}(h), \end{eqnarray} $

$ Q_i(h)\ (i = 0, 1, \cdots, k-1) $是$ h $的多项式,且$ \deg Q_0(h), \deg Q_1(h)\leq K+n_1+k $, $ \deg Q_i(h)\leq K+n_i-1, i = 2, 3, \cdots, k-1 $.

证  记

$ \Phi(h) = \big(\alpha_1(h), \alpha_2(h), \cdots, \alpha_k(h)\big) \big(I_1(h), I_2(h), \cdots, I_k(h)\big)^T \triangleq\tau_1(h)V_1(h). $

由(1.3)和(1.4)式可得

$ \begin{eqnarray*} \Phi(h)& = &\tau_1(h)V_1(h) = \tau_1(h)(B_1h+C_1)V'_1(h)\\ & = &\rho_{n_1+1}(h)I'_{1}(h) +\cdots+\rho_{n_{k-1}+1}(h)I'_{k-1}(h)+\rho_{n_k+1}(h)I'_{k}(h)\\ & = &\rho_{n_1+1}(h)I'_{1}(h) +\cdots+\rho_{n_{k-1}+1}(h)I'_{k-1}(h)\\ &&+\rho_{n_k+1}(h)\Big[Z'(h)-a_2I'_{2}(h)-\cdots-a_{k-1}I'_{k-1}(h)\Big]\\ & = &\rho_{n_1+1}(h)I'_{1}(h) +\cdots+\rho_{n_{k-1}+1}(h)I'_{k-1}(h)+\rho_{n_k+1}(h)Z'(h), \\ \Phi'(h)& = &\tau'_1(h)V_1(h)+\tau_1(h)V'_1(h) = [\tau'_1(h)(B_1h+C_1)+\tau_1(h)]V'_1(h)\\ & = &\rho_{n_1}(h)I'_{1}(h) +\cdots+\rho_{n_{k-1}}(h)I'_{k-1}(h)+\rho_{n_k}(h)I'_{k}(h)\\ & = &\rho_{n_1}(h)I'_{1}(h) +\cdots+\rho_{n_{k-1}}(h)I'_{k-1}(h)\\ &&+\rho_{n_k}(h)\Big[Z'(h)-a_2I'_{2}(h)-\cdots-a_{k-1}I'_{k-1}(h)\Big]\\ & = &\rho_{n_1}(h)I'_{1}(h) +\cdots+\rho_{n_{k-1}}(h)I'_{k-1}(h)+\rho_{n_k}(h)Z'(h), \\ \Phi''(h)& = &\rho'_{n_1}(h)I'_{1}(h)+\rho'_{n_2}(h)I'_{2}(h)+\cdots+ \rho'_{n_k}(h)Z'(h)\\ &&+\rho_{n_1}(h)I''_{1}(h)+\rho_{n_2}(h)I''_{2}(h)+\cdots+ \rho_{n_k}(h)Z''(h)\\ & = &\frac{1}{G_1(h)}\Big[\rho_{n_1+k}(h)I'_{1}(h)+\rho_{n_1+k}(h)Z'(h)\Big]\\ & &+\rho_{n_2-1}(h)I'_{2}(h)+\cdots+\rho_{n_{k-1}-1}(h)I'_{k-1}(h). \end{eqnarray*} $

由引理2.3可得

(2.11) $ \begin{eqnarray} L(h)I(h)& = &L(h)\big(\Phi(h)+\Psi(h)\big) = L(h)\Phi(h) \\ & = &P_2(h)\Phi''(h)+P_1(h)\Phi'(h)+P_0(h)\Phi(h). \end{eqnarray} $

把$ \Phi(h) $, $ \Phi'(h) $和$ \Phi''(h) $代入(2.11)式即可得证.证毕.

引理2.5  如果当$ h\in\Sigma $时, $ G_1(h)\neq0 $, $ I'_1(h)\neq0 $,则$ R(h) $至多有$ (3k+4)K+(3k+4)n_1+4k+1 $个零点(计重数).

证  由(1.4)和(2.10)式可得

(2.12) $ \begin{eqnarray} R^{(i)}(h)& = &\frac{1}{G^{i+1}_1(h)}\Big[X_{K+n_1+k+ki}(h)I'_{1}(h)+Y_{K+n_1+k+ki}(h)Z'(h)\Big] \\ & = &\frac{I'_{1}(h)}{G^{i+1}_1(h)}\Big[X_{K+n_1+k+ki}(h)+Y_{K+n_1+k+ki}(h)\omega_1(h)\Big], \end{eqnarray} $

其中$ i = K+n_2 $, $ X_{K+n_1+k+ki}(h) $和$ Y_{K+n_1+k+ki}(h) $是关于$ h $的二次数不超过$ K+n_1+k+ki $的多项式.所以$ R^{(i)}(h) $和$ X_{K+n_1+k+ki}(h)+Y_{K+n_1+k+ki}(h)\omega_1(h) $在$ \Sigma $上有相同的零点.

假设$ X_{K+n_1+k+ki}(h) $和$ Y_{K+n_1+k+ki}(h) $有公因子$ \theta_l(h) $,其中$ \theta_l(h) $是$ h $的多项式且$ \deg \theta_l(h)\leq K+n_1+k+ki $.令

$ X_{K+n_1+k+ki}(h) = \theta_l(h)\bar{X}(h), \ \ Y_{K+n_1+k+ki}(h) = \theta_l(h)\bar{Y}(h), $

则$ \bar{X}(h) $和$ \bar{Y}(h) $是$ h $的次数不超过$ K+n_1+k+ki-l $多项式.所以

$ X_{K+n_1+k+ki}(h)+Y_{K+n_1+k+ki}(h)\omega_1(h) = \theta_l(h)[\bar{X}(h)+\bar{Y}(h)\omega_1(h)], $

且

$ \#\{X_{K+n_1+k+ki}(h)+Y_{K+n_1+k+ki}(h)\omega_1(h) = 0\} = l+\#\{\bar{X}(h)+\bar{Y}(h)\omega_1(h) = 0\}. $

令$ \chi_1(h) = \bar{X}(h)+\bar{Y}(h)\omega_1(h) $,由(2.2)式可得

(2.13) $ \begin{eqnarray} G_1(h)\bar{Y}(h)\chi'_1(h) = -a_{12}\chi_1^2(h)+\bar{F}_1(h)\chi_1(h)+\bar{F}_0(h), \end{eqnarray} $

其中

$ \begin{eqnarray*} \bar{F}_1(h)& = &G_1(h)\bar{Y}'(h)+2a_{12}(h)\bar{X}(h) +\big(a_{k2}(h)-a_{11}(h)\big)\bar{Y}(h), \\ \bar{F}_0(h)& = &G_1(h)(\bar{X}'(h)\bar{Y}(h) -\bar{X}(h)\bar{Y}'(h))- a_{12}(h)\bar{X}^2(h)\\ & &+a_{k1}(h)\bar{Y}^2(h)-\big( a_{k2}(h)-a_{11}(h)\big)\bar{X}(h)\bar{Y}(h), \end{eqnarray*} $

且$ \deg \bar{F}_0(h)\leq2K+2n_1+(2i+3)k-2l $.由文献[3,引理4.4]可得

$ \#\{\chi_1(h) = 0\}\leq\#\{\bar{F}_0(h) = 0\}+\#\{\bar{Y}(h) = 0\}+1. $

因此,当$ h\in \Sigma $时

(2.14) $ \begin{eqnarray} \#\{R(h) = 0\}\leq \#\{\chi_1(h) = 0\}+i \leq (3k+4)K+(3k+4)n_1+4k+1. \end{eqnarray} $

证毕.

定理1.1的证明  由引理2.2可得, $ \Psi(h) $至多有$ 3m_1+2 $个零点.假设

$ P_2(\tilde{h}_i) = 0, \ \Psi(\bar{h}_j) = 0, \ \tilde{h}_i, \bar{h}_j\in\Sigma, \ 1\leq i\leq K, \ 1\leq j\leq3m_1+2. $

记$ \tilde{h}_i $和$ \bar{h}_j $为$ h_m^* $,并重排它们使得$ h_m^* < h_{m+1}^* $, $ m = 1, 2, \cdots, K+3m_1+2 $.令

$ \Delta_s = (h_s^*, h_{s+1}^*), \ s = 0, 1, \cdots, K+3m_1+2, $

其中$ h_0^* $和$ h_{K+3m_1+3}^* $表示区间$ \Sigma $的左右端点.则$ P_2(h)\neq0 $, $ \Psi(h)\neq0 $, $ h\in\Delta_s $.由引理2.3可得$ L(h)\Psi(h) = 0 $.所以,由命题2.1,方程

$ L(h) = P_2(h)\Big(\frac{{\rm d}^2}{{\rm d}h^2}+\frac{P_1(h)}{P_2(h)}\frac{\rm d}{{\rm d}h}+\frac{P_0(h)}{P_2(h)}\Big) $

的解空间是区间$ \Delta_s $上的Chebyshev空间.由命题2.2知, $ I(h) $在$ \Delta_s $上至多有$ 2+l_s $个零点,其中$ l_s $是$ R(h) $在$ \Delta_s $上的零点个数.综上可得

$ \begin{eqnarray*} \#\{I(h) = 0\}&\leq&\#\{R(h) = 0\}+2\times {\rm区间} \Delta_s {\rm的个数} + {\rm区间} \Delta_s {\rm的端点的个数} \\ &\leq&(3k+7)K+(3k+4)n_1+9m_1+4k+9\\ & \leq&(3k+4)n_1+(6k+23)m_1+13k+30. \end{eqnarray*} $

证毕.

例3.1  在文献[7]中,作者研究了如下近-Hamiltonian系统

(3.1) $ \begin{eqnarray} \left\{\begin{array}{ll} \dot{x} = 2y(rx^2+2y^2)+\varepsilon f(x, y), \\ \dot{y} = 2x(1-2x^2-ry^2)+\varepsilon g(x, y), \\ \end{array}\right.\ r\geq0, \ r\neq2, \end{eqnarray} $

其中$ f(x, y) $和$ g(x, y) $是关于$ x $和$ y $的$ n $次多项式.系统(3.1)的Abelian积分如下

(3.2) $ \begin{eqnarray} I(h) = \big[\alpha(h)I_{01}+\beta(h)I_{03}+\gamma(h)I_{21}+\delta(h)I_{23}\big]+\big[\xi(h)I_{11}+\eta(h)I_{13}\big], h\in(-\frac{1}{4}, 0), \end{eqnarray} $

其中$ \alpha(h), \ \beta(h), \ \gamma(h), \ \delta(h), \ \xi(h) $和$ \eta(h) $是$ h $的多项式,且$ \deg \alpha(h)\leq[\frac{n-1}{4}] $, $ \deg \beta(h), $$ \deg \gamma(h)\leq[\frac{n-3}{4}] $, $ \deg \delta(h)\leq[\frac{n-1}{4}]-1 $, $ \deg \xi(h)\leq[\frac{n-2}{4}] $, $ \deg \eta(h)\leq[\frac{n}{4}]-1 $.作者证明$ I(h) $至多有$ 40n+6[\frac{n-1}{4}]+6[\frac{n-2}{4}]+11 $个零点(计重数).

容易验证, (H1)–(H3)满足,且$ I'_{01}(h)\neq0 $, $ I_{11}(h)\neq0 $, $ h\in(-\frac{1}{4}, 0) $.比较(3.2)式和条件(H1)得

$ k = 4, \ n_1 = [\frac{n-1}{4}], \ m_1 = [\frac{n-2}{4}]. $

所以,由定理1.1得$ I(h) $至多有$ 16[\frac{n-1}{4}]+47[\frac{n-2}{4}]+82 $个零点(计重数).

例3.2  在文献[6]中,作者研究了如下系统

(3.3) $ \begin{eqnarray} \left\{\begin{array}{ll} \dot{x} = y+\varepsilon f(x, y), \\ \dot{y} = -(ax+bx^3+cx^5)+\varepsilon g(x, y), \\ \end{array}\right.\ a, b, c\in\mathbb{R} , \ c\neq0. \end{eqnarray} $

系统(3.3)的Abelian积分$ I(h) $可表示为

$ \begin{eqnarray*} I(h) = &\big[{\alpha}(h)I_{01}+{\beta}(h)I_{21}+{\gamma}(h)I_{41}\big] +\big[{\xi}(h)I_{11}+{\eta}(h)I_{31}\big], \end{eqnarray*} $

其中$ {\alpha}(h), \ {\beta}(h), \ {\gamma}(h), \ {\xi}(h) $和$ {\eta}(h) $是$ h $的多项式,且$ \deg {\alpha}(h)\leq[\frac{n-1}{2}] $, $ \deg {\beta}(h), $$ \deg {\gamma}(h)\leq[\frac{n-3}{2}] $, $ \deg {\xi}(h)\leq[\frac{n-2}{2}] $, $ \deg {\eta}(h)\leq[\frac{n-4}{2}] $.作者得到$ I(h) $至多有$ 54n-13 $个零点(计重数).

容易验证,定理1.1的条件成立,且$ k = 3, \ n_1 = [\frac{n-1}{2}], \ m_1 = [\frac{n-2}{2}] $.所以, $ I(h) $至多有$ 13[\frac{n-1}{2}]+41[\frac{n-2}{2}]+69 $个零点(计重数).

注3.1  由上面对例3.1和例3.2的讨论可知,应用本文定理1.1得到的$ I(h) $的零点个数的上界比原来文献得到的小,这主要是因为我们改善了引理2.3的结果.