简正模小记

普物小笔记。

考察一个有着 \(n\) 个自由度的系统 \((\psi_1, \psi_2, \psi_3, \cdots, \psi_n)\),若其受以下运动方程约束:

\[\ddot \psi_i + \sum_{j = 1}^n A_{ij} \psi_j = 0,\qquad i = 1, 2, 3, \cdots, n \]

我们令:

\[A = \begin{pmatrix} A_{11} & A_{12} & \cdots & A_{1n}\\ A_{21} & A_{22} & \cdots & A_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ A_{n1} & A_{n2} & \cdots & A_{nn}\\ \end{pmatrix}, \qquad B = \begin{pmatrix} B_{11} & B_{12} & \cdots & B_{1n}\\ B_{21} & B_{22} & \cdots & B_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ B_{n1} & B_{n2} & \cdots & B_{nn}\\ \end{pmatrix} \]

即可将运动方程写成矩阵形式,即:

\[\begin{pmatrix} \ddot \psi_1 \\ \ddot \psi_2 \\ \vdots \\ \ddot \psi_n \end{pmatrix} + A \begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_n \end{pmatrix} = 0 \]

则有:

\[\begin{align*} \psi_i & = \sum_{j = 1}^n B_{ij} \cos(\omega_j t + \phi_j)\\ \ddot \psi_i & = - \sum_{j = 1}^n \omega_j^2 B_{ij} \cos(\omega_j t + \phi_j)\\ \end{align*} \]

将上式带入运动方程中可得:

\[\begin{align*} \ddot \psi_i + \sum_{j = 1}^n A_{ij} \psi_j & = \sum_{k = 1}^n A_{ik} \sum_{j = 1}^n B_{kj} \cos(\omega_j t + \phi_j) - \sum_{j = 1}^n \omega_j^2 B_{ij} \cos(\omega_j t + \phi_j)\\ & = \sum_{j = 1}^n \cos(\omega_j t + \phi_j) \sum_{k = 1}^n A_{ik} B_{kj} - \sum_{j = 1}^n \omega_j^2 B_{ij} \cos(\omega_j t + \phi_j)\\ & = \sum_{j = 1}^n \cos(\omega_j t + \phi_j) \left(\sum_{k = 1}^n A_{ik} B_{kj} - \omega_j^2 B_{ij}\right)\\ & = 0,\qquad i = 1, 2, 3, \cdots, n \end{align*} \]

又由于 \(\cos(\omega_j t + \phi_j)\) 间线性独立,则:

\[\sum_{k = 1}^n A_{ik} B_{kj} = \omega_j^2 B_{ij},\qquad i = 1, 2, 3, \cdots, n \]

将其写成矩阵形式,即得:

\[\begin{pmatrix} A_{i1} & A_{i2} & A_{i3} & \cdots & A_{in} \end{pmatrix} \begin{pmatrix} B_{1j} \\ B_{2j} \\ B_{3j} \\ \vdots \\ B_{nj} \end{pmatrix} = \omega_j^2 B_{ij} \]

所以有:

\[\begin{pmatrix} A_{i1} & A_{i2} & A_{i3} & \cdots & A_{in} \end{pmatrix} B = \begin{pmatrix} \omega_1^2 B_{i1} & \omega_2^2 B_{i2} & \omega_3^2 B_{i3} & \cdots & \omega_n^2 B_{in} \end{pmatrix} \]

因此可得:

\[AB = \begin{pmatrix} \omega_1^2 B_{11} & \omega_2^2 B_{12} & \cdots & \omega_n^2 B_{1n}\\ \omega_1^2 B_{21} & \omega_2^2 B_{22} & \cdots & \omega_n^2 B_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ \omega_1^2 B_{n1} & \omega_2^2 B_{n2} & \cdots & \omega_n^2 B_{nn}\\ \end{pmatrix} = B \begin{pmatrix} \omega_1^2\\ & \omega_2^2\\ & & \omega_3^2\\ & & & \ddots\\ & & & & \omega_n^2 \end{pmatrix} \]

又因为 \((\psi_1, \psi_2, \cdots, \psi_n)\) 线性无关,所以 \(\mathrm{rank}(B) = n\)\(B^{-1}\) 存在,则:

\[B^{-1}AB = \mathrm{diag}(\omega_1^2 \quad \omega_2^2 \quad \omega_3^2 \quad \cdots \quad \omega_n^2) \]

抽出 \(B\) 的列向量 \(v_1, v_2, \cdots, v_n\)\((v_1~v_2~\cdots~v_n) = B\),则:

\[\begin{align*} AB & = B~\mathrm{diag}(\omega_1^2 \quad \omega_2^2 \quad \omega_3^2 \quad \cdots \quad \omega_n^2)\\ & = (\omega_1^2 v_1 \quad \omega_2^2 v_2 \quad \omega_3^2 v_3 \quad \cdots \quad \omega_n^2 v_n)\\ & = (Av_1 \quad Av_2 \quad Av_3 \quad \cdots \quad Av_n) \end{align*} \]

因此 \((A - \omega_i^2 I_n) v_i = 0\),此时 \(\omega_i^2\)\(A\) 的特征值,\(v_i\) 为其对应的特征向量。又有:

\[\psi_i = \sum_{j = 1}^n B_{ij} \cos(\omega_j t + \phi_j) \]

将其写成矩阵形式,即得:

\[\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_n \end{pmatrix} = \begin{pmatrix} B_{11} & B_{12} & \cdots & B_{1n}\\ B_{21} & B_{22} & \cdots & B_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ B_{n1} & B_{n2} & \cdots & B_{nn}\\ \end{pmatrix} \begin{pmatrix} \cos(\omega_1 t + \phi_1)\\ \cos(\omega_2 t + \phi_2)\\ \vdots\\ \cos(\omega_n t + \phi_n)\\ \end{pmatrix} = B \begin{pmatrix} \cos(\omega_1 t + \phi_1)\\ \cos(\omega_2 t + \phi_2)\\ \vdots\\ \cos(\omega_n t + \phi_n)\\ \end{pmatrix} \]

又因为 \((v_1~v_2~\cdots~v_n) = B\),所以可将上式重写为:

\[\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_n \end{pmatrix} = B \begin{pmatrix} \cos(\omega_1 t + \phi_1)\\ \cos(\omega_2 t + \phi_2)\\ \vdots\\ \cos(\omega_n t + \phi_n)\\ \end{pmatrix} = \begin{pmatrix} v_1 & v_2 & \cdots & v_n \end{pmatrix} \begin{pmatrix} \cos(\omega_1 t + \phi_1)\\ \cos(\omega_2 t + \phi_2)\\ \vdots\\ \cos(\omega_n t + \phi_n)\\ \end{pmatrix} = \sum_{i = 1}^n v_i \cos(\omega_i t + \phi_i) \]

其中 \(v_i \cos(\omega_i t + \phi_i)\) 即为假设所有自由度均以圆频率 \(\omega_i\) 和初相位 \(\phi_i\) 振动时所求出的特解,也即圆频率为 \(\omega_i\) 的简正模式,\(v_i\) 即为该模式下各个自由度上振动的振幅所组成的列向量。

posted @ 2025-11-14 08:49  Neuro-Reimu  阅读(9)  评论(0)    收藏  举报