由于分段光滑微分系统在实际中有着广泛而重要的应用,它们周期轨(二维微分系统称为极限环,所谓极限环是指系统的一条孤立的周期轨)的个数受到学者们的普遍关注.目前,关于二维分段光滑微分系统的极限环分支的研究文献较多,见文献[1-5].据我们所知,相关的研究方法主要有两种:由韩茂安等建立的Melnikov函数法^{[1, 6]}和由Llibre等建立的平均法^[7].但是,文献[7]没有给出极限环个数的上界估计.韩茂安在文献[8]给出了相应的上界估计法.最近,杨纪华和赵丽琴把Picard-Fuchs方程法推广到二维分段光滑微分系统的极限环分支研究中^[9].但是,高维分段光滑微分系统的周期轨分支的研究相对较少,而且主要研究方法是平均法,见文献[10-12]. 2017年,田焕欢和韩茂安给出高维分段光滑可积微分系统的一阶Melnikov向量函数公式^[13].该公式可以用来研究高维分段光滑微分系统的周期轨的个数.

本文主要应用文献[13]中给出的一阶Melnikov向量函数公式研究如下$ n $ -维分段光滑扰动微分系统

(1.1) $ \begin{equation} \left\{\begin{array}{ll} \dot{x}_1 = x_2+\varepsilon g^+_1({\bf x}), \\ \dot{x}_2 = -x_1+\varepsilon g^+_2({\bf x}), \\ \dot{x}_3 = \varepsilon g^+_3({\bf x}), \\ \cdots\\ \dot{x}_n = \varepsilon g^+_n({\bf x}), \\ \end{array}\right.x_1\geq0, \quad \quad \left\{\begin{array}{ll} \dot{x}_1 = x_2+\varepsilon g^-_1({\bf x}), \\ \dot{x}_2 = -x_1+\varepsilon g^-_2({\bf x}), \\ \dot{x}_3 = \varepsilon g^-_3({\bf x}), \\ \cdots\\ \dot{x}_n = \varepsilon g^-_n({\bf x}), \\ \end{array}\right.x_1<0, \end{equation} $

其中$ {\bf x} = (x_1, x_2, \cdots, x_n)^T $, $ 0< \varepsilon\ll1 $,

$ \left\{\begin{array}{ll} { } g^+_i({\bf x}) = \sum\limits_{k_1+k_2+\cdots+k_n = 0}^ma^i_{k_1k_2\cdots k_n}x_1^{k_1}x_2^{k_2}\cdots x_n^{k_n}, \\ { } g^-_i({\bf x}) = \sum\limits_{k_1+k_2+\cdots+k_n = 0}^mb^i_{k_1k_2\cdots k_n}x_1^{k_1}x_2^{k_2}\cdots x_n^{k_n}, \end{array}\right.i = 1, 2, \cdots, n. $

当$ \varepsilon = 0 $,系统(1.1)在每个平面$ x_i = h_i $上有一个中心,其中$ i = 3, 4, \cdots, n $.本文将借鉴文献[9]中的方法,应用一阶Melnikov向量函数研究系统(1.1)的周期轨个数的上界.本文的主要结果如下.

定理1.1  用一阶Melnikov向量函数,当$ \varepsilon>0 $充分小时,系统(1.1)最多存在$ m^{n-1} $个周期轨.

注1.1  当$ n = 3, m = 1 $时,本文中定理1.1的结果与文献[13]中定理3的结果相同.

本小节主要介绍田焕欢和韩茂安在文献[13]中给出的下面$ n $ -维分段光滑微分系统的一阶Melnikov向量函数公式

(2.1) $ \begin{eqnarray} \left\{\begin{array}{ll} \dot{x} = H_y(x, y, {\bf z})+\varepsilon P(x, y, {\bf z}), \\ \dot{y} = -H_x(x, y, {\bf z})+\varepsilon Q(x, y, {\bf z}), \\ \dot{{\bf z}} = \varepsilon R(x, y, {\bf z}), \\ \end{array}\right. \end{eqnarray} $

其中$ 0< \varepsilon\ll1 $, $ {\bf z} = (z_1, z_2, \cdots, z_{n-2})^T\in{{\Bbb R}} ^{n-2} $, $ n\geq2 $,

$ \begin{eqnarray*} H(x, y, {\bf z}) = \left\{\begin{array}{ll}H^+, \ x\geq0, \\ H^-, \ x<0, \end{array}\right.{\qquad} P(x, y, {\bf z}) = \left\{\begin{array}{ll}P^+, \ x\geq0, \\ P^-, \ x<0, \end{array}\right.\\ Q(x, y, {\bf z}) = \left\{\begin{array}{ll}Q^+, \ x\geq0, \\ Q^-, \ x<0, \end{array}\right.{\qquad} R(x, y, {\bf z}) = \left\{\begin{array}{ll}R^+, \ x\geq0, \\ R^-, \ x<0, \end{array}\right. \end{eqnarray*} $

其中$ H^\pm, P^\pm, Q^\pm $和$ R^\pm = (R^\pm_1, R^\pm_2, \cdots, R^\pm_{n-2})^T $是$ C^\infty $函数.易知,当$ \varepsilon = 0 $时,系统(2.1)有$ n-1 $个首次积分$ H $, $ z_1 $, $ z_2 $, $ \cdots $, $ z_{n-2} $.系统(2.1)的前两个方程定义了一个平面Hamiltonian系统

(2.2) $ \begin{eqnarray} \left\{\begin{array}{ll}\dot{x} = H_y(x, y, {\bf z}), \\ \dot{y} = -H_x(x, y, {\bf z}).\end{array}\right. \end{eqnarray} $

对系统(2.2)做如下假设

(H1)   设$ G\subset {{\Bbb R}} ^{n-2} $是开集.对每个$ \hat{h}\in G $,存在依赖于$ \hat{h} $的开区间$ J_{\hat{h}} $使得

$ \widehat{AB}:{L}^+_{h_1, \hat{h}} = \{(x, y)|H^+(x, y, \hat{h}) = h_1, x\geq0\} $

和

$ \widehat{BA}:{L}^-_{h_1, \hat{h}} = \{(x, y)|H^-(x, y, \hat{h}) = h_1, x<0\} $

是两条具有公共端点$ A(h_1, \hat{h}) = (0, a(h_1, \hat{h})) $和$ B(h_1, \hat{h}) = (0, b(h_1, \hat{h})) $的曲线,且$ a(h_1, \hat{h})>b(h_1, \hat{h}) $.该两条曲线上不含有系统(2.2)的临界点.这样, $ {L}_{h_1, \hat{h}} = {L}^+_{h_1, \hat{h}}\cup {L}^-_{h_1, \hat{h}} $是系统(2.2)的周期轨.

(H2)   曲线$ {L}_{h_1, \hat{h}} $在点$ A $和$ B $与$ y $ -轴不相切.即对$ \hat{h}\in G $和$ h_1\in J_{\hat{h}} $,有

$ H^\pm_y(A(h_1, \hat{h}), \hat{h})\cdot H^\pm_y(B(h_1, \hat{h}), \hat{h})\neq0. $

定理2.1^[13]  假设系统(2.1)满足(H1)–(H2).则系统(2.1)的一阶Melnikov向量函数公式为

(2.3) $ \begin{eqnarray} M(h)& = &\left(\begin{array}{cc} M^+_1(h)+N_1(h)+N(h)\big(M^-_1(h)+N_2(h)\big)\\ M^+_2(h)+M^-_2(h)\\ \vdots\\ M^+_{n-1}(h)+M^-_{n-1}(h) \end{array}\right){}\\ &\triangleq&\big(M_1(h), \ M_2(h), \ \cdots, \ M_{n-1}(h)\big)^T, \ h = (h_1, \hat{h})^T, \end{eqnarray} $

其中

$ M^\pm_1(h) = \int_{{L}^\pm_{h_1, \hat{h}}}Q^\pm(x, y, \hat{h}){\rm d}x-P^\pm(x, y, \hat{h}){\rm d}y, $

$ N(h) = H^+_y(A^*(h))/H^-_y(A^*(h)), \ A^*(h) = (A(h_1, \hat{h}), \hat{h}), $

$ N_1(h) = \sum\limits_{k = 1}^{n-2}[M^+_{1, k}(h)+H^+_{z_k}(A(h))M^+_{k+1}(h)], $

$ N_2(h) = \sum\limits_{k = 1}^{n-2}[M^-_{1, k}(h)-H^-_{z_k}(A(h))M^-_{k+1}(h)], $

$ M^\pm_{1, k}(h) = \int_{{L}^\pm_{h_1, \hat{h}}}H^\pm_{z_k}(x, y, \hat{h})R^\pm_k(x, y, \hat{h}){\rm d}t, $

$ M^\pm_{k+1}(h) = \int_{{L}^\pm_{h_1, \hat{h}}}R^\pm_k(x, y, \hat{h}){\rm d}t, \ k = 1, 2, \cdots, n-2. $

并且,如果对某个$ h_0\in G $, $ M(h_0) = 0 $和$ {\rm{det}}DM(h_0)\neq0 $,则对充分小的$ \varepsilon>0 $,在$ L_{h_0} $附近存在系统(2.1)的唯一周期轨.

接着,作者又给出了计算公式(2.3)中$ M^\pm_{1, k}(h) $和$ M^\pm_{k+1}(h) $的方法.

引理2.1^[13]  记

$ \bar{T}^\pm_{k}(x, y, \hat{h}) = \int_0^yH^\pm_{z_k}(x, y, \hat{h})R^\pm_k(x, y, \hat{h}){\rm d}y, \ \bar{R}^\pm_k(x, y, \hat{h}) = \int_0^yR^\pm_k(x, y, \hat{h}){\rm d}y, $

$ \bar{M}^\pm_{1, k}(h) = \int_{{L}^\pm_{h_1, \hat{h}}}\bar{T}^\pm_{k}(x, y, \hat{h}){\rm d}x, \ \bar{M}^\pm_{k+1} = \int_{{L}^\pm_{h_1, \hat{h}}}\bar{R}^\pm_k(x, y, \hat{h}){\rm d}x, $

则

$ M^+_{1, k}(h) = \frac{\partial \bar{M}^+_{1, k}(h)}{\partial h_1}, \ M^-_{1, k}(h) = \frac{\partial \bar{M}^-_{1, k}(h)}{\partial h_1}\cdot N(h), $

$ M^+_{k+1}(h) = \frac{\partial \bar{M}^+_{k+1}(h)}{\partial h_1}, \ M^-_{k+1}(h) = \frac{\partial \bar{M}^-_{k+1}(h)}{\partial h_1}\cdot N(h). $

引理2.2^[13]  假设系统(2.1)满足(H1)–(H2)且$ H(x, y, {\bf z}) $中不含$ {\bf z} $.则公式(2.3)变为

(2.4) $ \begin{equation} M(h) = \left(\begin{array}{c} M^+_1(h)+N(h)M^-_1(h)\\ M^+_2(h)+M^-_2(h)\\ \vdots\\ M^+_{n-1}(h)+M^-_{n-1}(h) \end{array}\right), \ h = (h_1, \hat{h})^T. \end{equation} $

下面的引理是韩茂安在文献[14]中给出的,该引理可以估计实系数多项式方程组的根的个数的上界.

引理2.3^[14]  设$ f_1, \cdots, f_n\in {{\Bbb R}} [x_1, x_2, \cdots, x_n] $是实系数多项式,且映射$ f = (f_1, \cdots, f_n):{{\Bbb R}} ^n\rightarrow{{\Bbb R}} ^n $只有有限多个零点.则$ f $的零点个数不超过$ \deg f_1\times\cdots\times\deg f_n $,其中$ \deg f_k $表示多项式$ f_k $的次数.

本节首先计算系统(1.1)的一阶Melnikov向量函数$ M(h) $,然后给出定理1.1的证明.易知当$ \varepsilon = 0 $时,系统(1.1)有首次积分

(3.1) $ \begin{equation} H^\pm(x_1, x_2) = \frac{1}{2}(x_1^2+x_2^2), \ x_3, \ x_4, \ \cdots, \ x_n. \end{equation} $

所以$ {L}_{h_1, \hat{h}} = {L}^+_{h_1, \hat{h}}\cup {L}^-_{h_1, \hat{h}} $是系统

$ \left\{\begin{array}{ll}\dot{x}_1 = x_2, \\ \dot{x}_2 = -x_1, \end{array}\right. x_1\geq0, \ \ \ \ \left\{\begin{array}{ll}\dot{x}_1 = x_2, \\ \dot{x}_2 = -x_1, \end{array}\right. x_1<0 $

的周期轨,其中

$ {L}^+_{h_1, \hat{h}} = \{(x_1, x_2)|H^+(x_1, x_2) = h_1, x_1\geq0, h_1\in(0, +\infty)\}, $

$ {L}^-_{h_1, \hat{h}} = \{(x_1, x_2)|H^-(x_1, x_2) = h_1, x_1<0, h_1\in(0, +\infty)\}. $

由于(3.1)式中的$ H^\pm(x_1, x_2) $中不含有$ x_3, x_4, \cdots, x_n $,所以系统(1.1)的一阶Melnikov向量函数公式具有(2.4)式的形式.显然, $ N(h) = 1 $.下面我们计算(2.4)式中的$ M^\pm_i(h) $, $ i = 1, 2, \cdots, n-1 $.

我们首先计算

$ M_1(h) = M^+_1(h)+M^-_1(h), $

其中$ h = (h_1, \hat{h})^T = (h_1, h_3, h_4, \cdots, h_n)^T $.记

$ I_{k_1, k_2}(h_1) = \int_{L^+_{h_1, \hat{h}}}x_1^{k_1}x_2^{k_2}{\rm d}x_2, \ \ \ J_{k_1, k_2}(h_1) = \int_{L^-_{h_1, \hat{h}}}x_1^{k_1}x_2^{k_2}{\rm d}x_2. $

易知

$ I_{k_1, 2j+1}(h_1) = J_{k_1, 2j+1}(h_1) = 0. $

所以,我们只考虑$ I_{k_1, 2j}(h_1) $和$ J_{k_1, 2j}(h_1) $.由格林公式可得

(3.2) $ \begin{eqnarray} \int_{L^\pm_{{h_1, \hat{h}}}}x^{k_1}_1x_2^{k_2}{\rm d}x_1 = -\frac{k_2}{k_1+1}\int_{L^\pm_{{h_1, \hat{h}}}}x_1^{k_1+1}x_2^{k_2-1}{\rm d}x_2. \end{eqnarray} $

因此,由定理2.1和(3.2)式可得

(3.3) $ \begin{eqnarray} M_1(h)& = &\int_{L^+_{h_1, \hat{h}}}g_1^+(x_1, x_2, h_3, \cdots, h_n){\rm d}x_2-g_2^+(x_1, x_2, h_3, \cdots, h_n){\rm d}x_1{}\\ &&+ \int_{L^-_{h_1, \hat{h}}}g_1^-(x_1, x_2, h_3, \cdots, h_n){\rm d}x_2-g_2^-(x_1, x_2, h_3, \cdots, h_n){\rm d}x_1{}\\ & = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^mh_3^{k_3}\cdots h_n^{k_n}\Big[a^1_{k_1k_2\cdots k_n}\int_{L_{h_1, \hat{h}}^+}x_1^{k_1}x_2^{k_2}{\rm d}x_2- a^2_{k_1k_2\cdots k_n}\int_{L_{h_1, \hat{h}}^+}x_1^{k_1}x_2^{k_2}{\rm d}x_1\Big]{}\\ &&+\sum\limits_{k_1+k_2+\cdots+k_n = 0}^mh_3^{k_3}\cdots h_n^{k_n}\Big[b^1_{k_1k_2\cdots k_n}\int_{L_{h_1, \hat{h}}^-}x_1^{k_1}x_2^{k_2}{\rm d}x_2- b^2_{k_1k_2\cdots k_n}\int_{L_{h_1, \hat{h}}^-}x_1^{k_1}x_2^{k_2}{\rm d}x_1\Big]{}\\ & = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^mh_3^{k_3}\cdots h_n^{k_n}{}\\ &&\times \Big[a^1_{k_1k_2\cdots k_n}\int_{L_{h_1, \hat{h}}^+}x_1^{k_1}x_2^{k_2}{\rm d}x_2+ \frac{k_2}{k_1+1}a^2_{k_1k_2\cdots k_n}\int_{L^+_{h_1, \hat{h}}}x_1^{k_1+1}x_2^{k_2-1}{\rm d}x_2\Big]{}\\ &&+\sum\limits_{k_1+k_2+\cdots+k_n = 0}^mh_3^{k_3}\cdots h_n^{k_n}{}\\ &&\times \Big[b^1_{k_1k_2\cdots k_n}\int_{L_{h_1, \hat{h}}^-}x_1^{k_1}x_2^{k_2}{\rm d}x_2+ \frac{k_2}{k_1+1}b^2_{k_1k_2\cdots k_n}\int_{L^-_{h_1, \hat{h}}}x_1^{k_1+1}x_2^{k_2-1}{\rm d}x_2\Big]{}\\ &: = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^mh_3^{k_3}\cdots h_n^{k_n}\Big[\alpha_{k_1k_2\cdots k_n}I_{k_1, k_2}(h_1)+ \beta_{k_1k_2\cdots k_n}J_{k_1, k_2}(h_1)\Big], \end{eqnarray} $

其中$ \alpha_{k_1k_2\cdots k_n} $和$ \beta_{k_1k_2\cdots k_n} $是常数.

引理3.1

(3.4) $ \begin{eqnarray} I_{k_1, k_2}(h_1) = \left\{\begin{array}{ll}\delta_{k_1k_2}^1h_1^{[\frac{k_1+k_2}{2}]}I_{0, 0}(h_1), \ \ \mbox{当$k_1+k_2$是偶数}, \\ \delta_{k_1k_2}^2h_1^{[\frac{k_1+k_2}{2}]}I_{1, 0}(h_1), \ \ \mbox{当$k_1+k_2$是奇数}, \end{array}\right. \end{eqnarray} $

(3.5) $ \begin{eqnarray} J_{k_1, k_2}(h_1) = \left\{\begin{array}{ll}\delta_{k_1k_2}^3h_1^{[\frac{k_1+k_2}{2}]}J_{0, 0}(h_1), \ \ \mbox{当$k_1+k_2$是偶数}, \\ \delta_{k_1k_2}^4h_1^{[\frac{k_1+k_2}{2}]}J_{1, 0}(h_1), \ \ \mbox{当$k_1+k_2$是奇数}, \end{array}\right. \end{eqnarray} $

其中$ \delta_{k_1k_2}^i $, $ i = 1, 2, 3, 4 $是常数.

证  我们只证明(3.4)式. (3.5)式可以类似的证明.对方程

$ H^+(x_1, x_2) = \frac{1}{2}(x^2_1+x^2_2) = h_1 $

两边同时关于$ x_2 $求偏导数可得

(3.6) $ \begin{eqnarray} x_2+x_1\frac{\partial x_1}{\partial x_2} = 0. \end{eqnarray} $

(3.6)式两边同乘以$ x_1^{k_1}x_2^{k_2-1}{\rm d}x_2 $,沿着$ L^+_{h_1, \hat{h}} $积分,并注意到(3.2)式可得

(3.7) $ \begin{eqnarray} I_{k_1, k_2}(h_1) = \frac{k_2-1}{k_1+2}I_{k_1+2, k_2-2}(h_1). \end{eqnarray} $

同样, $ H^+(x_1, x_2) = h_1 $两端同乘以$ x_1^{k_1-2}x_2^{k_2}{\rm d}x_2 $并沿着$ L^+_{h_1, \hat{h}} $积分可得

(3.8) $ \begin{eqnarray} I_{k_1, k_2}(h_1) = 2h_1I_{k_1-2, k_2}(h_1)-I_{k_1-2, k_2+2}(h_1). \end{eqnarray} $

再由(3.7)和(3.8)式可得

(3.9) $ \begin{eqnarray} I_{k_1, k_2}(h_1) = \frac{2(k_2-1)}{k_1+k_2+1}h_1I_{k_1, k_2-2}(h_1) \end{eqnarray} $

和

(3.10) $ \begin{eqnarray} I_{k_1, k_2}(h_1) = \frac{2k_1}{k_1+k_2+1}h_1I_{k_1-2, k_2}(h_1). \end{eqnarray} $

下面我们用数学归纳法证明(3.4)式.由(3.9)和(3.10)式可得

(3.11) $ \begin{eqnarray} \left\{\begin{array}{ll} { } I_{0, 2}(h_1) = \frac{2}{3}h_1I_{0, 0}(h_1), \ \ I_{2, 0}(h_1) = \frac{4}{3}h_1I_{0, 0}(h_1), \\ { } I_{1, 2}(h_1) = \frac{1}{2}h_1I_{1, 0}(h_1), \ \ I_{3, 0}(h_1) = \frac{3}{2}h_1I_{1, 0}(h_1). \end{array}\right. \end{eqnarray} $

所以当$ k_1+k_2 = 2, 3 $时, (3.4)式成立.假设当$ k_1+k_2\leq l-1 $时, (3.4)式成立,其中$ l\geq4 $是偶数.在(3.9)式中取$ (k_1, k_2) = (0, l), (2, l-2), (4, l-4), \cdots, (l-2, 2) $,在(3.10)式中取$ (k_1, k_2) = (l, 0) $,我们得到

$ \begin{eqnarray*} \left(\begin{array}{c} I_{0, l}(h_1)\\ I_{2, l-2}(h_1)\\ I_{4, l-4}(h_1)\\ \vdots\\ I_{l-2, 2}(h_1)\\ I_{l, 0}(h_1) \end{array}\right)\ \ = \frac{2}{l+1}\left(\begin{array}{c} (l-1)h_1I_{0, l-2}(h_1)\\ (l-3)h_1I_{2, l-4}(h_1)\\ (l-5)h_1I_{4, l-6}(h_1)\\ \vdots\\ h_1I_{l-2, 0}(h_1)\\ lh_1I_{l-2, 0}(h_1) \end{array}\right). \end{eqnarray*} $

因此,当$ k_1+k_2 = l $时,我们得到

$ \begin{eqnarray*} I_{k_1, k_2}(h_1) = &h_1\gamma^1_{k_1k_2}h_1^{[\frac{l-2}{2}]}I_{0, 0}(h_1) = \delta^1_{k_1k_2}h_1^{[\frac{l}{2}]}I_{0, 0}(h_1). \end{eqnarray*} $

如果$ k_1+k_2 = l $是奇数,同样可以证明(3.4)式成立.证毕.

引理3.2  $ M_1(h) $可表示为

(3.12) $ \begin{eqnarray} M_1(h) = h_2\sum\limits_{k_1+k_2+\cdots+k_n = 0}^m\lambda^1_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}, \end{eqnarray} $

其中$ h_2 = \sqrt{h_1} $, $ \lambda^1_{k_1k_2\cdots k_n} $是常数.

证  由(3.3)式和引理3.1可得

(3.13) $ \begin{eqnarray} M_1(h)& = &\sum\limits_{{k_1+k_2+\cdots+k_n = 0, \atop k_1+k_2 = 0\, mod\, 2}}^m\alpha_{k_1k_2\cdots k_n}\delta^1_{k_1k_2}h_3^{k_3}\cdots h_n^{k_n}h_1^{[\frac{k_1+k_2}{2}]}I_{0, 0}(h_1){}\\ &&+\sum\limits_{{k_1+k_2+\cdots+k_n = 0, \atop k_1+k_2 = 1\, mod\, 2}}^m\alpha_{k_1k_2\cdots k_n}\delta^2_{k_1k_2}h_3^{k_3}\cdots h_n^{k_n}h_1^{[\frac{k_1+k_2}{2}]}I_{1, 0}(h_1){}\\ &&+\sum\limits_{{k_1+k_2+\cdots+k_n = 0, \atop k_1+k_2 = 0\, mod\, 2}}^m\beta_{k_1k_2\cdots k_n}\delta^3_{k_1k_2}h_3^{k_3}\cdots h_n^{k_n}h_1^{[\frac{k_1+k_2}{2}]}J_{0, 0}(h_1){}\\ &&+\sum\limits_{{k_1+k_2+\cdots+k_n = 0, \atop k_1+k_2 = 1\, mod\, 2}}^m\beta_{k_1k_2\cdots k_n}\delta^4_{k_1k_2}h_3^{k_3}\cdots h_n^{k_n}h_1^{[\frac{k_1+k_2}{2}]}J_{1, 0}(h_1). \end{eqnarray} $

直接计算可的

$ I_{0, 0}(h_1) = \int_{L^+_{h_1, \hat{h}}}{\rm d}x_2 = 2\sqrt{2}\sqrt{h_1}, \ \ \ \ I_{1, 0}(h_1) = \int_{L^+_{h_1, \hat{h}}}x_1{\rm d}x_2 = \pi h_1, $

$ J_{0, 0}(h_1) = \int_{L^-_{h_1, \hat{h}}}{\rm d}x_2 = -2\sqrt{2}\sqrt{h_1}, \, \ J_{1, 0}(h_1) = \int_{L^-_{h_1, \hat{h}}}x_1{\rm d}x_2 = -\pi h_1. $

把它们代入(3.13)式,并注意到$ h_2 = \sqrt{h_1} $,即可得(3.12)式.证毕.

引理3.3  $ M_i(h) $, $ i = 2, 3, \cdots, n-1 $可表示为

(3.14) $ \begin{equation} \begin{array}{rl} M_2(h) = &{ }\sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^2_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}, \\ M_3(h) = &{ }\sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^3_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}, \\ &\vdots\\ M_{n-1}(h) = &{ }\sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^{n-1}_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}, \end{array} \end{equation} $

其中$ {\lambda}^i_{k_1k_2\cdots k_n} $, $ i = 2, 3, \cdots, n-1 $是常数.

证  不失一般性,我们只证明$ M_2(h) $, $ M_3(h) $, $ \cdots $, $ M_{n-1}(h) $可类似的进行证明.由定理2.1可得

$ \begin{eqnarray*} \bar{g}^+_3(x_1, x_2, \hat{h})& = &\int_0^{x_2}g_3^+(x_1, x_2, \hat{h}){\rm d}x_2\\ & = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^ma^3_{k_1k_2\cdots k_n}h_3^{k_3}\cdots h_n^{k_n}\int_0^{x_2}x_1^{k_1}x_2^{k_2}{\rm d}x_2\\ & = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^ma^3_{k_1k_2\cdots k_n}h_3^{k_3}\cdots h_n^{k_n}\frac{1}{k_2+1}x_1^{k_1}x_2^{k_2+1}. \end{eqnarray*} $

再注意到(3.2)式以及引理3.1可得

$ \begin{eqnarray*} \bar{M}_2^+(h)& = &\int_{L^+_{h_1, \hat{h}}}\bar{g}^+_3(x_1, x_2, \hat{h}){\rm d}x_1\\ & = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^ma^3_{k_1k_2\cdots k_n}h_3^{k_3}\cdots h_n^{k_n}\frac{1}{k_2+1}\int_{L^+_{h_1, \hat{h}}}x_1^{k_1}x_2^{k_2+1}{\rm d}x_1\\ & = &-\sum\limits_{k_1+k_2+\cdots+k_n = 0}^ma^3_{k_1k_2\cdots k_n}h_3^{k_3}\cdots h_n^{k_n}\frac{1}{k_1+1}\int_{L^+_{h_1, \hat{h}}}x_1^{k_1+1}x_2^{k_2}{\rm d}x_2\\ & = &-\sum\limits_{k_1+k_2+\cdots+k_n = 0}^ma^3_{k_1k_2\cdots k_n}h_3^{k_3}\cdots h_n^{k_n}\frac{1}{k_1+1}I_{k_1+1, k_2}(h_1)\\ & = &\sum\limits_{k_1+k_2+\cdots+k_n = 0}^m\bar{\lambda}^2_{k_1k_2\cdots k_n}h_2^{k_1+k_2+2}h_3^{k_3}\cdots h_n^{k_n}, \end{eqnarray*} $

其中$ h_2 = \sqrt{h_1} $.所以

$ \begin{eqnarray*} M^+_2(h) = \frac{\partial \bar{M}_2^+(h)}{\partial h_1} = \sum\limits_{k_1+k_2+\cdots+k_n = 0}^m\bar{\lambda}^2_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}. \end{eqnarray*} $

同理可得

$ \begin{eqnarray*} M^-_2(h) = \sum\limits_{k_1+k_2+\cdots+k_n = 0}^m\tilde{\lambda}^2_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}. \end{eqnarray*} $

因此

$ \begin{eqnarray*} M_2(h) = M^+_2(h)+M^-_2(h) = \sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^2_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}. \end{eqnarray*} $

引理3.3证毕.

定理1.1的证明  由引理3.2和引理3.3可得系统(1.1)的一阶Melnikov向量函数$ M(h) $为

$ \begin{eqnarray*} M(h) = \left(\begin{array}{c} M_1(h)\\ M_2(h)\\ M_3(h)\\ \vdots\\ M_{n-1}(h) \end{array}\right) = \left(\begin{array}{c} { } h_2\sum\limits_{k_1+k_2+\cdots+k_n = 0}^m\lambda^1_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}\\ { } \sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^2_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}\\ { } \sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^3_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n}\\ \vdots\\ { } \sum\limits_{k_1+k_2+\cdots+k_n = 0}^m{\lambda}^{n-1}_{k_1k_2\cdots k_n}h_2^{k_1+k_2}h_3^{k_3}\cdots h_n^{k_n} \end{array}\right). \end{eqnarray*} $

令$ M_1(h) = h_2\bar{M}_1(h) $,则$ \bar{M}_1(h) $是关于$ h = (h_2, h_3, \cdots, h_n)^T $的$ m $次多项式. $ M_i(h) $, $ i = 2, 3, $$ \cdots, n-1 $也是关于$ h = (h_2, h_3, \cdots, h_n)^T $的$ m $次多项式.由引理2.3可得, $ M(h) $最多有$ m^{n-1} $个零点.再由定理2.1即得定理1.1成立.