﻿ 一类Liénard系统的零点个数最小上界研究

# 一类Liénard系统的零点个数最小上界研究A study on the Upper Number of Zeros for a Liénard System

Abstract: In this paper, we aim to use Chebyshev theory of Abel integral generator and polynomial symbolic computing technology to study and analyze the upper bound of the number of zeros of Abel integral corresponding to (4, 3) Liénard system. The minimum upper bound of the number of zeros for I(h,δ) is proved. The number of zeros of Abel integral of following Liénard system is considered . Through the in-depth study of Abel integral I(h,δ), it is proved that the generator of Abel integral can form Chebeyshev system, and a conclu-sion is shown.

1. 背景介绍

$\left\{\begin{array}{l}\frac{\text{d}x}{\text{d}t}={P}_{n}\left(x,y\right)\hfill \\ \frac{\text{d}y}{\text{d}t}={Q}_{n}\left(x,y\right)\hfill \end{array}$

$\stackrel{˙}{x}={H}_{y}\left(x,y\right),\text{ }\stackrel{˙}{y}=-{H}_{x}\left(x,y\right)$

2. 预备知识

1) 若任意的线性组合 ${k}_{0}{f}_{0}\left(x\right)+{k}_{1}{f}_{1}\left(x\right)+\cdots +{k}_{n-1}{f}_{n-1}\left(x\right)$ 在J上最多具有 $n-1$ 个孤立零点，则 $\left\{{f}_{0}\left(x\right),{f}_{1}\left(x\right),\cdots ,{f}_{n-1}\left(x\right)\right\}$ 称为Chebyshev系统(简称T系统)。

2) 若任意线性组合 ${k}_{0}{f}_{0}\left(x\right)+{k}_{1}{f}_{1}\left(x\right)+\cdots +{k}_{n-1}{f}_{n-1}\left(x\right)$，对于 $i=1,2,\cdots ,n$ 最多具有 $i-1$ 个零点(考虑重数)，则 $\left\{{f}_{0}\left(x\right),{f}_{1}\left(x\right),\cdots ,{f}_{n-1}\left(x\right)\right\}$ 称为扩充完全Chebyshev系统(简称ECT系统)。

3) 记 $W\left[{f}_{0}\left(x\right),{f}_{1}\left(x\right),\cdots ,{f}_{k-1}\left(x\right)\right]$ 为如下行列式：

$W\left[{f}_{0}\left(x\right),{f}_{1}\left(x\right),\cdots ,{f}_{k-1}\left(x\right)\right]=\mathrm{det}{\left({f}_{i}^{j}\right)}_{0\le ij\le i-1}=|\begin{array}{cccc}{f}_{0}\left(x\right)& {f}_{1}\left(x\right)& \cdots & {f}_{n-1}\\ {{f}^{\prime }}_{0}\left(x\right)& {{f}^{\prime }}_{1}\left(x\right)& \cdots & {{f}^{\prime }}_{n-1}\left(x\right)\\ ⋮& ⋮& \ddots & ⋮\\ {f}_{0}^{\left(n-1\right)}\left(x\right)& {f}_{1}^{\left(n-1\right)}\left(x\right)& \cdots & {f}_{n-1}^{\left(n-1\right)}\left(x\right)\end{array}|$

${l}_{i}\left(x\right):=\frac{{f}_{i}\left(x\right)}{{A}^{\prime }\left(x\right)}-\frac{{f}_{i}\left(z\left(x\right)\right)}{{A}^{\prime }\left( x \right)}$

1) 当 $i=0,1,\cdots ,n-2$ 时， $W\left[{l}_{0},{l}_{1},\cdots ,{l}_{i}\right]$$\left({x}_{l},{x}_{r}\right)$ 上存在，

2) $W\left[{l}_{0},{l}_{1},\cdots ,{l}_{i}\right]$$\left({x}_{l},{x}_{r}\right)$ 上有零点，用多重数计数，

3) $s>n+k-2$

${\oint }_{{\Gamma }_{h}}F\left(x\right){y}^{k-2}\text{d}x={\oint }_{{\Gamma }_{h}}G\left(x\right){y}^{k}\text{d}x,\text{}\forall k\in N$.

$8{h}^{3}{I}_{i}\left(h\right)={\oint }_{{\Gamma }_{h}}{f}_{i}\left(x\right){y}^{7}\text{d}x\equiv {\stackrel{˜}{I}}_{i}\left( h \right)$

$\left\{{I}_{0}\left(h\right),{I}_{1}\left(h\right),{I}_{2}\left(h\right),{I}_{3}\left(h\right)\right\}$ 是一个ECT系统，并且当且仅当 $\left\{{\stackrel{˜}{I}}_{0}\left(h\right),{\stackrel{˜}{I}}_{1}\left(h\right),{\stackrel{˜}{I}}_{3}\left(h\right)\right\}$ 也是一个ECT系统。

3. 证明过程

$\stackrel{˙}{x}=y,\text{}\stackrel{˙}{y}=x\left(x-1\right){\left(x-\frac{1}{2}\right)}^{2}+\epsilon \left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+{x}^{3}\right)y,\text{}0<\epsilon \ll 1.$

1977年，阿诺德(Arnold)提出弱化的希尔伯特(Hilbert)的第十六问题，目的是研究阿贝尔积分的零点数目

$I\left(h,\delta \right)={\oint }_{{\Gamma }_{h}}q\text{d}x-p\text{d}y,\text{ }h\in J$ (1)

$\stackrel{˙}{x}={H}_{y}\left(x,y\right)+\epsilon p\left(x,y\right),\text{ }\stackrel{˙}{y}=-{H}_{x}\left(x,y\right)+\epsilon q\left(x,y\right),$ (2)

$\stackrel{˙}{x}={H}_{y}\left(x,y\right),\text{ }\stackrel{˙}{y}=-{H}_{x}\left(x,y\right),$ (3)

$\stackrel{˙}{x}=y,\text{ }\stackrel{˙}{y}=\epsilon f\left(x\right)y+g\left(x\right),$ (4)

$\stackrel{˙}{x}=y,\text{}\stackrel{˙}{y}=x\left(x-1\right){\left(x-\frac{1}{2}\right)}^{2}+\epsilon \left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+{x}^{3}\right)y.$ (5)

$H\left(x,y\right)=\frac{1}{2}{y}^{2}-\frac{1}{5}{x}^{5}+\frac{1}{2}{x}^{4}-\frac{5}{12}{x}^{3}+\frac{1}{8}{x}^{2}=\frac{1}{2}{y}^{2}+A\left(x\right),$ (6)

$\left\{{\Gamma }_{h}\right\}\subset \left\{\left(x,y\right)|H\left(x,y\right)=h,\text{}{h}_{0}

${\Gamma }_{h}$ 上的Melnikov函数为：

$I\left(h,\delta \right)={\oint }_{{\Gamma }_{h}}\left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+{x}^{3}\right)\text{d}x\equiv {a}_{0}{I}_{0}\left(h\right)+{a}_{1}{I}_{1}\left(h\right)+{a}_{2}{I}_{2}\left(h\right)+{I}_{3}\left(h\right).$ (7)

Figure 1. The portrait of system (5) when ε = 0.

$I\left(h\right)=\frac{1}{2h}{\oint }_{{\Gamma }_{h}}\left(2A\left(x\right)+{y}^{2}\right){x}^{i}y\text{d}x=\frac{1}{2h}\left({\oint }_{{\Gamma }_{h}}2A\left(x\right)y\text{d}x+{\oint }_{{\Gamma }_{h}}{x}^{i}{y}^{3}\text{d}x\right),$ (8)

${\oint }_{{\Gamma }_{h}}2{x}^{i}A\left(x\right)y\text{d}x={\oint }_{{\Gamma }_{h}}{G}_{i}\left(x\right){y}^{3}\text{d}x,$ (9)

$\begin{array}{c}{G}_{i}\left(x\right)=\frac{1}{3}{\left(\frac{2{x}^{i}A\left(x\right)}{A\left(x\right)}\right)}^{\prime }\left(x\right)\\ =\frac{\left(48i{x}^{5}-192i{x}^{4}+48{x}^{5}+304i{x}^{3}-168{x}^{4}-240i{x}^{2}+236{x}^{3}+95ix-170{x}^{2}-15i+70x-15\right){x}^{i}}{45{\left(-1+2x\right)}^{3}{\left(x-1\right)}^{2}}\end{array}$

$\begin{array}{c}{J}_{i}\left(h\right)=\frac{1}{2h}{\oint }_{{\Gamma }_{h}}\left({G}_{i}\left(x\right)+{x}^{i}\right){y}^{3}\text{d}x=\frac{1}{4{h}^{2}}{\oint }_{{\Gamma }_{h}}\left(2A\left(x\right)+{y}^{2}\right)\left({G}_{i}\left(x\right)+{x}^{i}\right){y}^{5}\text{d}\varphi \\ =\frac{1}{4{h}^{2}}{\oint }_{{\Gamma }_{h}}2A\left(x\right)\left({x}^{i}+{G}_{i}\left(x\right)\right){y}^{3}\text{d}x+\frac{1}{4{h}^{2}}{\oint }_{{\Gamma }_{h}}\left({x}^{i}+{G}_{i}\left(x\right)\right){y}^{5}\text{d}x.\end{array}$ (10)

${\oint }_{{\Gamma }_{h}}2A\left(x\right)\left({x}^{i}+{G}_{i}\left(x\right)\right){y}^{3}\text{d}x\text{=}{\oint }_{{\Gamma }_{h}}{E}_{i}\left(x\right){y}^{5}\text{d}x.$ (11)

$\begin{array}{c}{r}_{i}\left(x\right)=2304{i}^{2}{x}^{10}-18432{i}^{2}{x}^{9}+66408{i}^{2}{x}^{8}-163008i{x}^{9}+19584{x}^{10}-139776{i}^{2}{x}^{2}+541920i{x}^{3}\\ -137088{x}^{4}+193696{i}^{2}{x}^{5}-1060176i{x}^{7}+428832{x}^{8}-183840{i}^{2}{x}^{5}+1353896i{x}^{6}-790848{x}^{7}\\ +121120{i}^{2}{x}^{4}-1182420i{x}^{5}+955848{x}^{6}-54720{i}^{2}{x}^{3}+718010i{x}^{4}-796200{x}^{5}+16225{i}^{2}{x}^{2}\\ -301035i{x}^{3}+467270{x}^{4}-2850{i}^{2}x+84050i{x}^{2}-193220{x}^{3}+225i{\text{}}^{2}-14250ix+54825{x}^{2}\\ +1125i-9900x+900.\end{array}$

$\begin{array}{c}{J}_{i}\left(h\right)=\frac{1}{4{h}^{2}}{\oint }_{{\Gamma }_{h}}\left({E}_{i}\left(x\right)+{G}_{i}\left(x\right)+{x}^{i}\right){y}^{5}\text{d}x=\frac{1}{8{h}^{3}}{\oint }_{{\Gamma }_{h}}\left(2A\left(x\right)+{y}^{2}\right)\left({E}_{i}\left(x\right)+{G}_{i}\left(x\right)+{x}^{i}\right){y}^{5}\text{d}\\ =\frac{1}{8{h}^{3}}{\oint }_{{\Gamma }_{h}}2A\left(x\right)\left({E}_{i}\left(x\right)+{G}_{i}\left(x\right)+{x}^{i}\right){y}^{5}\text{d}x+\frac{1}{8{h}^{3}}{\oint }_{{\Gamma }_{h}}\left({E}_{i}\left(x\right)+{G}_{i}\left(x\right)+{x}^{i}\right){y}^{7}\text{d}x.\end{array}$ (12)

${\oint }_{{\Gamma }_{h}}2A\left(x\right)\left({E}_{i}\left(x\right)+{G}_{i}\left(x\right)+{x}^{i}\right){y}^{5}\text{d}x={\oint }_{{\Gamma }_{h}}{D}_{i}\left(x\right){y}^{7}\text{d}x.$ (13)

${D}_{i}\left(x\right)=\frac{1}{7}{\left(\frac{2A\left(x\right)\left({x}^{i}+{G}_{i}\left(x\right)+{E}_{i}\left(x\right)}{A\left(x\right)}\right)}^{\prime }\left(x\right)=\frac{1}{354375{\left(-1+2x\right)}^{9}{\left(x-1\right)}^{6}}{x}^{i}{\stackrel{¯}{g}}_{i}\left(x\right)$，其中

$\begin{array}{c}{\stackrel{¯}{g}}_{i}\left(x\right)=110592{i}^{3}{x}^{15}-1327104{i}^{3}{x}^{14}+4478976{i}^{2}{x}^{15}+7409664{i}^{3}{x}^{13}-51342336{i}^{2}{x}^{14}+57701376i{x}^{15}\\ -25546752{i}^{3}{x}^{12}+273341952{i}^{2}{x}^{13}-632530944i{x}^{14}+234772992{x}^{15}+60855552i{x}^{11}\\ -896894208{i}^{2}{x}^{12}+3214916352i{x}^{13}-2465116416{x}^{14}-106142976{i}^{3}{x}^{10}+2029183488{i}^{2}{x}^{11}\\ -10052225280i{x}^{12}+11981620224{x}^{13}+140091904{i}^{3}{x}^{9}-3354500736{i}^{2}{x}^{10}+21630292992i{x}^{11}\end{array}$

$\begin{array}{c}-35763372288{x}^{12}-142525440{i}^{3}{x}^{8}+4187744064{i}^{2}{x}^{9}-33942149568i{x}^{10}+73328718528{x}^{11}\\ +112724880{i}^{3}{x}^{7}-4022191200{i}^{2}{x}^{8}+40144961792i{x}^{9}-109441242720{x}^{10}-69325920{i}^{3}{x}^{6}\\ +2998330200{i}^{2}{x}^{7}-36465079440i{x}^{8}+122889085920{x}^{9}+32887200{i}^{3}{x}^{5}-1735866300{i}^{2}{x}^{6}\\ +25667156060i{x}^{7}-105792127920{x}^{8}-11818800{i}^{3}{x}^{4}+774744750{i}^{2}{x}^{5}-14013994940i{x}^{6}\end{array}$

$\begin{array}{c}+70464218410{x}^{7}+3114575{i}^{3}{x}^{3}-262037325{i}^{2}{x}^{4}+5894048125i{x}^{5}-36356166065{x}^{6}\\ -568125{i}^{3}{x}^{2}+65093175{i}^{2}{x}^{3}-1878268450i{x}^{4}+14433723450{x}^{5}+64125{i}^{3}x-11228625{i}^{2}{x}^{2}\\ +439913425i{x}^{3}-4338507050{x}^{4}-3375{i}^{3}+1204875{i}^{2}x-71683875i{x}^{2}+958177950{x}^{3}\\ -60750{i}^{2}+7293375ix-147305250{x}^{2}-351000i+14168250x-648000.\end{array}$

$\frac{\text{d}}{\text{d}x}{l}_{i}\left(x\right)=\frac{\text{d}}{\text{d}x}\left(\frac{{f}_{i}}{{A}^{\prime }}\right)\left(x\right)-\left[\frac{\text{d}}{\text{d}x}\left(\frac{{f}_{i}}{{A}^{\prime }}\right)\left(z\left(x\right)\right)\right]\frac{\text{d}z}{\text{d}x},$ (14)

${x}_{r} (15)

$\begin{array}{l}W\left[{l}_{1}\left(x\right)\right]={l}_{1}\left(x\right)=\frac{4\left(x-z\right){w}_{1}\left(x,z\right)}{3375{\left(-1+2x\right)}^{8}{\left(x-1\right)}^{5}{\left(-1+2z\right)}^{8}{\left(z-1\right)}^{5}},\hfill \\ W\left[{l}_{1}\left(x\right),{l}_{2}\left(x\right)\right]=\frac{64{\left(x-z\right)}^{3}{w}_{2}\left(x,z\right)}{\text{11390625}{\left(z-1\right)}^{9}{\left(-1+2z\right)}^{16}{\left(x-1\right)}^{9}{\left(-1+2x\right)}^{16}p\left(x,z\right)},\hfill \\ W\left[{l}_{1}\left(x\right),{l}_{2}\left(x\right),{l}_{0}\left(x\right)\right]=-\frac{512{\left(x-z\right)}^{6}{w}_{3}\left(x,z\right)}{\text{38443359375}{x}^{3}{\left(x-1\right)}^{12}{\left(-1+2x\right)}^{22}{z}^{3}{\left(-1+2z\right)}^{22}{\left(z-1\right)}^{12}{p}^{3}\left(x,z\right)},\hfill \\ W\left[{l}_{1}\left(x\right),{l}_{2}\left(x\right),{l}_{0}\left(x\right),{l}_{3}\left(x\right)\right]=\frac{4096{\left(x-z\right)}^{10}{w}_{4}\left(x,z\right)}{43248779296875{x}^{4}{\left(x-1\right)}^{15}{\left(-1+2x\right)}^{30}{z}^{4}{\left(-1+2z\right)}^{30}{\left(z-1\right)}^{15}{p}^{6}\left(x,z\right)}.\hfill \end{array}$

$\begin{array}{c}R\left(q,p,z\right)=573308928000{x}^{12}-3439853568000{x}^{11}+87907368960000{x}^{10}-12421693440000{x}^{9}\\ +10439000064000{x}^{8}-5064228864000{x}^{7}+1076281344000{x}^{6}+163233792000{x}^{5}-135336960000{x}^{4}\\ +14542848000{x}^{3}+4700160000{x}^{2}-691200000x-103680000.\end{array}$

1) 计算 $q\left(x,z\right)$${w}_{1}\left(x,z\right)$ 关于z的结式得：

$R\left(q,{w}_{1},z\right)=1099511627776{\left(x-1\right)}^{4}{\left(-1+2x\right)}^{14}{\phi }_{1}\left(x\right)$，其中 ${\phi }_{1}\left(x\right)$ 为关于x的70次多项式。由Sturm定理，当 $x\in \left(0,1\right)$ 时， ${\phi }_{1}\left(x\right)\ne 0$ 因此 ${w}_{1}\left(x,z\right)$$q\left(x,z\right)$ 没有满足条件(15)的公共根。得出结论：在(0, 1)上， ${W}_{1}\left(x,z\right)\ne 0$

2) 类似地，检查 $q\left(x,z\right)$${w}_{2}\left(x,z\right)$ 之间是否存在公共根，用 ${w}_{2}\left(x,z\right)$ 代替程序中的 ${w}_{1}\left(x,z\right)$，得到 $R\left(q,{w}_{2},z\right)=3518437208883200{\left(x-1\right)}^{6}{\left(-1+2x\right)}^{28}{\phi }_{2}\left(x\right)$，其中 ${\phi }_{2}\left(x\right)$ 为关于x的142次多项式。将Sturm定理应用于 ${\phi }_{2}\left(x\right)$，有两个点，用 ${x}_{1}$${x}_{2}$ 表示，使得 ${\phi }_{2}\left(x\right)=0\text{}$${x}_{1}\approx 0.3194871852,\text{}{x}_{2}\approx 0.8780598147$。因此，将检查 $q\left(x,z\right)$${w}_{2}\left(x,z\right)$ 在(0, 1)上是否有公共根，使用带有Maple 19的程序来查找所有可能的区间。

> with(RegularChains);

> with(ChainTools);

> with(SemiAlgebraicSetTools);

> sys := [w2(x,z), q(x,z)];

> R2 := PolynomialRing([x,z]);

> dec := Triangularize(sys, R2);

[regular chain]

> L := map(Equations, dec, R2);

$\left[\left[2x-1,2z-1\right],\left[x-1,z-1\right],\left[{\phi }_{3}\left(x,z\right),{\phi }_{4}\left(z\right)\right]\right]$

$\left[x=\left[\frac{147953}{131072},\frac{73977}{65536}\right],\text{}z=\left[-\frac{23003894633164393}{144115188075855872},-\frac{2875486829145549}{18014398509481984}\right]\right],$

$\left[x=\left[\frac{115089}{131072},\frac{57545}{65536}\right],\text{}\left[z=\frac{3181281276395}{1759218644416},-\frac{12725125105579}{70368744177664}\right]\right],$

$\begin{array}{l}\left[x=\left[\frac{41875}{131072},\frac{10469}{32768}\right],\\ z=\left[\frac{27313710324978218727081661251716190518835493283418984010717824408507818285254767}{27371421987580235682274733772977928352834969285952318751528091320482060895025888928},\\ \text{}\frac{17071073145311138670442603828232261907427218330213686500669864025531738642828423}{14821387422376473014270860811112052205218558037201992197050570753012880593911808}\right]\right],\end{array}$

$\begin{array}{l}\left[x=\left[-\frac{10461}{65536},-\frac{20921}{131072}\right],\\ z=\left[\frac{17964076768096284321771248998958445677497741050123422581583682197739874395984749117530235}{15914343565113172548972231940698266883214596825515126958094847260581103904401068017057792},\\ \text{}\frac{143712614144770274574169991991667565419981928400987380652669457581918995167877992940241881}{127314748520905380391777855525586135065716774604121015664758778084648831235208544136462336}\right]\right],\end{array}$

$\begin{array}{l}\left[x=\left[\frac{56447}{65536},\frac{451579}{524288}\right],\\ z=\left[\frac{63300383292229564562517090328922307792455814963777798670531323163388007775889}{57896044618658097711785492504343953926634992332820282019728792003956564819968},\\ \text{}\frac{31650191646114782281258545164461153896227907481888899335265661581694003887945}{28948022309329048855892746252171976963317496166410141009864396001978282409984}\right]\right],\end{array}$

$\begin{array}{l}\left[x=\left[-\frac{23703}{131072},-\frac{11851}{65536}\right],\\ z=\left[\frac{325351617328729556618428350174928032828431538540974018963326935493166395828803}{3705346855594118253554271520278013051304639509300498049262642688253220148477952},\\ \text{}\frac{13014064693149182264737134006998112131313726154163896075853307741972665583315213}{14821387422376473014217086081112052205218558037207992197050570753012880593911808}\right]\right],\end{array}$

$\begin{array}{l}\left[x=\left[\frac{71653}{65536},\frac{143307}{131072}\right],\\ z=\left[\frac{12174518124912944273199013256495173628060618648103873508325294960539283985}{14134776518227074636666380005943348126619871175004951664972849610340958208},\\ \text{}\frac{24349036249825888546398026512990347256121237296027747016650589921078567971}{28269553036454149273332760011886696253239742350009903329945699220681916416}\right]\right],\end{array}$

$\begin{array}{l}\left[x=\left[\frac{75483}{65536},\frac{150967}{131072}\right],z=\left[\frac{3478907379019106586553154394578987}{10889035741470030830827987437816582766592},\\ \text{}\frac{13915629516076426342612639720617578315949}{43556142965880123323311949751266331066331066368}\right]\right],\end{array}$

3) 接下来，要检查 $q\left(x,z\right)$${w}_{3}\left(x,z\right)$ 之间是否存在公共根，我们用 ${w}_{3}\left(x,z\right)$ 代替程序中的 ${w}_{1}\left(x,z\right)$，我们得到：

$R\left(q,{w}_{3},z\right)=65841835161401738294227131414080654928331879219200000{\left(x-1\right)}^{6}{\left(-1+2x\right)}^{38}{\phi }_{5}\left(x\right),$

> with(RegularChains);

> with(ChainTools);

> with(SemiAlgebraicSetTools);

> sys := [w3(x,z), q(x,z)];

> R2 := PolyomialRing([x,z]);

> dec := Triangularize(sys, R2);

[regular chain]

> L := map(Equations, dec, R2);

$\left[\left[2x-1,2z-1\right],\left[x-1,z-1\right],\left[{\phi }_{6}\left(x,z\right),{\phi }_{7}\left(z\right)\right]\right]$

4) 类似地，我们使用相同的程序来查找所有可能存在公共根的区间，这些区间可能存在 ${w}_{4}\left(x,z\right)$$q\left(x,z\right)$ 的公共根，然后得到以下正则链：

$\left[\left[2x-1,2z-1\right],\left[x-1,z-1\right],\left[{\phi }_{8}\left(x,z\right),{\phi }_{9}\left(z\right)\right]\right]$

$W\left[{l}_{1}\left(x\right),{l}_{2}\left(x\right),{l}_{0}\left(x\right),{l}_{3}\left(x\right)\right]\ne 0$.

NOTES

*通讯作者。

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