In this paper we consider the following nonlinear quasi-periodic system:$$\begin{eqnarray}{\dot{x}}=(A+\unicode[STIX]{x1D716}P(t,\unicode[STIX]{x1D716}))x+\unicode[STIX]{x1D716}g(t,\unicode[STIX]{x1D716})+h(x,t,\unicode[STIX]{x1D716}),\quad x\in \mathbb{R}^{d},\end{eqnarray}$$ where $A$ is a $d\times d$ constant matrix of elliptic type, $\unicode[STIX]{x1D716}g(t,\unicode[STIX]{x1D716})$ is a small perturbation with $\unicode[STIX]{x1D716}$ as a small parameter, $h(x,t,\unicode[STIX]{x1D716})=O(x^{2})$ as $x\rightarrow 0$, and $P,g$ and $h$ are all analytic quasi-periodic in $t$ with basic frequencies $\unicode[STIX]{x1D714}=(1,\unicode[STIX]{x1D6FC})$, where $\unicode[STIX]{x1D6FC}$ is irrational. It is proved that for most sufficiently small $\unicode[STIX]{x1D716}$, the system is reducible to the following form: $$\begin{eqnarray}{\dot{x}}=(A+B_{\ast }(t))x+h_{\ast }(x,t,\unicode[STIX]{x1D716}),\quad x\in \mathbb{R}^{d},\end{eqnarray}$$ where $h_{\ast }(x,t,\unicode[STIX]{x1D716})=O(x^{2})~(x\rightarrow 0)$ is a high-order term. Therefore, the system has a quasi-periodic solution with basic frequencies $\unicode[STIX]{x1D714}=(1,\unicode[STIX]{x1D6FC})$, such that it goes to zero when $\unicode[STIX]{x1D716}$ does.

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一类具有刘维尔基本频率的非线性准周期系统的可约性

在本文中，我们考虑以下非线性准周期系统：$$\begin{eqnarray}{\dot{x}}=(A+\unicode[STIX]{x1D716}P(t,\unicode[STIX]{x1D716}))x+\unicode[STIX]{x1D716}g( t,\unicode[STIX]{x1D716})+h(x,t,\unicode[STIX]{x1D716}),\quad x\in \mathbb{R}^{d},\end{eqnarray}$$在哪里$澳元是一个$d\倍 d$椭圆型常数矩阵，$\unicode[STIX]{x1D716}g(t,\unicode[STIX]{x1D716})$是一个小扰动$\unicode[STIX]{x1D716}$作为一个小参数，$h(x,t,\unicode[STIX]{x1D716})=O(x^{2})$作为$x\右箭头 0$， 和$P,g$和$h$都是解析准周期的$t$基本频率$\unicode[STIX]{x1D714}=(1,\unicode[STIX]{x1D6FC})$， 在哪里$\unicode[STIX]{x1D6FC}$是不合理的。证明对于大多数足够小的$\unicode[STIX]{x1D716}$，系统可简化为以下形式：$$\begin{eqnarray}{\dot{x}}=(A+B_{\ast }(t))x+h_{\ast }(x,t,\unicode[STIX]{x1D716}),\四元 x\in \mathbb{R}^{d},\end{eqnarray}$$在哪里$h_{\ast }(x,t,\unicode[STIX]{x1D716})=O(x^{2})~(x\rightarrow 0)$是高阶项。因此，系统有一个具有基本频率的准周期解$\unicode[STIX]{x1D714}=(1,\unicode[STIX]{x1D6FC})$, 这样当它变为零时$\unicode[STIX]{x1D716}$做。