## Periodic Solutions of a Class of Nonlinear Hill's Type Equations with Bounded Restoring Force

Wang Chao,

 基金资助: 国家自然科学基金.  11571249江苏省自然科学基金.  BK20171275

 Fund supported: the NSFC.  11571249the Natural Science Fundation of Jiangsu Province.  BK20171275

Abstract

In this pater, we study the existence and multiplicity of the periodic solutions of a class of Hill's type equations with bounded restoring force. We prove the existence of infinite of subharmonic solutions when the weight is positive. We also consider the existence, multiplicity and dense distribution of symmetric periodic solutions in case of even and periodic weight functions.

Keywords： Hill's type equation ; Weight functions ; Bounded restoring force ; Sub-harnomic solutions ; Symmetric periodic solutions

Wang Chao. Periodic Solutions of a Class of Nonlinear Hill's Type Equations with Bounded Restoring Force. Acta Mathematica Scientia[J], 2019, 39(4): 761-772 doi:

## 1 引言

$x''+a(t)g(x)= 0$

$2\pi$ -周期解的存在性和多解性问题,其中$a(t):\mathbb{R} \rightarrow \mathbb{R} ^+_0:=\{x:x\in\mathbb{R}, x>0\}$是关于$t$的$2\pi$-周期的连续函数;

$g:\mathbb{R} \rightarrow \mathbb{R}$是局部Lipschitz连续函数, $g(0)=0,$满足

$(g_1)\quad \exists d_0>0$和$G_0>0$使得对$\forall t\in[0, 2\pi]$和$|x|\geq d_0$都有

$(g_2)\quad \exists M_0>0, \forall s\in\mathbb{R}, |g(s)|\leq M_0.$

$x''+a(t)|x|^{\gamma-1}x=0, \ \gamma>0,$

Waltman[1]首先研究了方程

## 2 引理

$\left\{ \begin{array}{lll}x'=y, \\y'=-a(t)g(x). \end{array}\right.$

当$(x(t), y(t))\in {\rm I}$时,易见$-\sin^2\theta(t)\leq -\sin^2\delta.$由方程(2.2)知存在$M_1>0,$当$r>M_1$时有$\frac{a_0M_0}{r} < \frac{1}{3}\sin^2\delta.$从而有

对充分大的$R>A_0,$我们首先计算解在$\{(x, y):|(x, y)|\geq R\}$内通过区域$\Omega:=\{(x, y): -x < y < x, x>0\}$所需时间.

(1)任给$s>\tilde{R},$有$\gamma(s)>s, \ \gamma_{(-1)}(s) < s$;

(2)任给$A>\tilde{R}$以及方程(2.2)的定义在$[t_0, t_1](t_1>t_0)$上的解$(r(t), \theta(t))$,若$r(t_0) < A,$ $r(t_1)>\gamma(A)$;或者$r(t_0)>A, $$r(t_1) < \gamma_{(-1)}(A)则有 首先,我们在XOY平面上定义一些区域.记 其中d_0在条件(g_1)中给出. 设(x(t), y(t))是方程(2.1)的定义在[\tau, \tau+L]上的解,满足 其中\tau\in\mathbb{R} 是任意的一个实数, L>0是一个正实数, M_1:= a_0M_0, M_0在条件(g_2)中定义.下面我们将在各个区域讨论r(t)的变化情况. (ⅰ)设\forall t\in I_1\subset [\tau, \tau+L],有(x(t), y(t))\in \mbox{I}.因为 所以\forall t, s\in I_1, 有|y(t)-y(s)|\leq 2M_1.从而, |r(t)-r(s)|\leq 2(d_0^2+M_1^2)^{\frac{1}{2}}.所以 易见F_1(r)\hbox{、}F_{-1}(r):\mathbb{R} \rightarrow\mathbb{R} 都是单调递增的函数.在区域\mbox{V}中有类似的讨论,并且存在单调递增的函数 (ⅱ)设\forall t\in I_2\subset [\tau, \tau+L],有(x(t), y(t))\in \mbox{II}.由系统(2.1),得 设t_0\hbox{、}t_1\in I_2, t_0 < t_1且(x_0, y_0)\hbox{、} (x_1, \ y_1)是轨线上分别对应于t_0和t_1的两个点.对上面的方程积分,有 \int_{y_0}^{y_1}y{\rm d}y = \int_{x_0}^{x_1}(-a(t(x))g(x)){\rm d}x, 注意到x(t)在I_2上单调递增,从而有 \frac{y_1^2-y_0^2}{2} = -a(t(\xi))g(\xi)(x_1-x_0), 其中 t = t(x)$$ x = x(t)$的反函数且$\xi\in[x_0, x_1]$.这样,我们有

$$$x_1-x_0 = \frac{y_1+y_0}{-2a(t(\xi))g(\xi)}(y_1-y_0)$$$

$$$y_1-y_0 = \frac{-2a(t(\xi))g(\xi)}{y_1+y_0}(x_1-x_0).$$$

$$$F_{-2}(r_0)<r_1< F_2(r_0).$$$

(ⅲ)设$\forall t\in I_3\subset [\tau, \tau+L]$,有$(x(t), y(t))\in \mbox{III}. $$\forall t_0, t_1\in I_3, 因为 $$|r(t_1)-r(t_0)|\leq\sqrt{(x(t_1)-x(t_0))^2+(y(t_1)-y(t_0))^2}\leq d_0\sqrt{L+4},$$ 从而 $$F_{-3}(r_0)\equiv r_0-d_0\sqrt{L+4}\leq r_1\leq r_0+ d_0\sqrt{L+4}: = F_3(r_0).$$ 易见, F_{-3}$$ F_3$都是单调递增函数.在区域$\mbox{VII}$中有类似的讨论,并且存在单调递增的函数

(ⅳ)类似于(ⅱ)中的讨论,若$\forall t\in I_4\subset [\tau, \tau+L]$,有$(x(t), y(t))\in \mbox{IV},$则可以找到两个单调递增的函数$F_{\pm 4}(r)$使得对任意的$t_0\hbox{、}t_1\in I_4, t_0<t_1,$都有

$$$F_{-4}(r(t_0))\leq r(t_1)\leq F_4(r(t_0)).$$$

$$$F_{-8}(r(t_0))\leq r(t_1)\leq F_8(r(t_0)).$$$

$m_0\in \Bbb{Z}^+$使得$2m_0\pi\geq T_0.$则由(3.1)式知,对任给的$\theta_0\in {\Bbb R}$及正整数$m\geq m_0$

## 4 对称方程的周期运动

如定理3.1的证明中那样定义${\cal A}(r_2) $$T_0,$$ m_1$为充分大的正整数且满足

第一步.设$p_1 $$p_2 分别是 m_1 阶和 m_2 阶的 {\cal E} -点,且设 \hat{m} = m_1m_2. 类似于定理3.1的证明,令 r_1>\max\{A_0, \tilde{R}\} 充分大使得以原点为圆心 r_1 为半径的圆包含 p_1, p_2 作为内点,像在定理3.1的证明中那样定义 r_2$$ T_0. $$m = l\hat{m} 使得 l\hat{m}\pi>T_0, 其中 l 是某一个正整数. 定义方程(2.1)的Poincare映射 l_0 是连接 p_1$$ p_2$的开的直线段,即

$L $$x -轴,即 下面,我们用反证法.假设 l_0 不包含 {\cal E} 点,即对所有 k\geq1$$ l^{k}\cap L = \phi$.

$q_3 = (a', b')\in {\Bbb R}^2$,则存在$p_3\in l_0$使得

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