多自由微振动

处于平衡位形的体系受到微扰时, 体系可能在平衡位形附近以简正模式做微振动.

如何求解简正模式, 判断振动的稳定性.

力学体系满足:

1. 位形空间广义坐标不显含时间 \(q_k = q_k(s)\).
2. 体系中各质点受到力均为保守力.
3. 体系无约束或只受到完整约束 \(G_{\alpha}(q) = 0\).

对约束体系,拉格朗日量已消去受限坐标.

位形空间的平衡点(一个平衡位形)由力学体系的一套广义坐标值描述: \(q_1^{(e)}, q_2^{(e)}, \dots,q_D^{(e)}\). 该广义坐标值满足:

• 初始时刻有 \(q_k(0) = q_k^{(e)}\), 且 \(\dot{q}_k(0) = 0\).
• 体系将永远静止于各广义坐标值 \(q_k(0) = q_k^{(e)}\).

几种不同的平衡位形:

稳定(stable)

微扰使体系在平衡位形附近做微振动.

非稳定(unstable)

微扰使得远离平衡位形.

随遇(conditional)

改变广义坐标,重置广义速度为零,体系在新坐标系中达到平衡.

虚功原理: 平衡条件满足 \(\dot{q}_k(0) = 0, \ddot{q}_k(0)\). 可得 \[\frac{\partial U(q)}{\partial q_k}\Big{|}_{q = q^{(e)}} = 0\] 其中 \(k = 1, 2, \dots, D\). 可解得平衡点.

设已求出位形空间平衡点. 当各广义坐标微微偏离平衡点时, 求系统运动状态. 引入小坐标 \(r_i\) 为小量, 进行 Taylor 展开. \[r_i = q_i - q_i^{(e)},\; q_i = q_i^{(e)} + r_i \Rightarrow \dot{r}_i = \dot{q}_i.\] 设 \(\dot{r}_i\) 为与 \(r_i\) 同阶小量: \(\dot{r}_i = O(r_i)\). \[L(s, \dot{s}) = \frac{1}{2}M^k\dot{s}_k^2 - U(s), \;L(q, \dot{q}) = L \left[ s(q), \dot{s}(q, \dot{q}) \right].\] 又因有变换:

\begin{equation*} \begin{cases} q = q(s),\\ \dot{q} = \dot{q}(s, \dot{s}) \end{cases} \Rightarrow\begin{cases} s = s(q), \\ \dot{s} = \dot{s}(q, \dot{q}) \end{cases} \end{equation*}

可得:

\begin{aligned} L(q, \dot{q}) &= \frac{1}{2}\left[ M^k \left( \frac{\partial s_k}{\partial q_j}\dot{q}^j \right) \left( \frac{\partial s_k}{\partial q_i}\dot{q}^i \right) \right] - U(q)\\ &= \frac{1}{2} M^k \frac{\partial s_k}{\partial q_i}\frac{\partial s_k}{\partial q_j} \dot{q}^i \dot{q}^j - U(q)\\ &= \frac{1}{2} m_{ij}\dot{q}^i\dot{q}^j - U(q). \end{aligned}

其中 \[m_{ij} = M^k \frac{\partial s_k}{\partial q_i} \frac{\partial s_k}{\partial q_j} = m_{ij}(q).\] 最终得到 \(L(r, \dot{r})\): \[L(r, \dot{r}) = L \left[ q(r), \dot{q}(r, \dot{r}) \right] = \frac{1}{2}m_{ij}\dot{r}^i \dot{r}^j - U \left( q^{(e)} + r\right).\] 其中 \(m_{ij} = m_{ij}\left( q^{(e)}+r \right)\). 作 Taylor 展开,保留到二阶:

\begin{aligned} L(r, \dot{r}) = L(r, \dot{r})\big{|}_{r, \dot{r} = 0} &+ \underbrace{\left( \frac{\partial L}{\partial r_i}\Big{|}_{r,\dot{r}=0} r^i + \frac{\partial L}{\partial \dot{r}_i}\Big{|}_{r,\dot{r}=0} \dot{r}^i \right)}_0\\ &+\frac{1}{2}\left( \frac{\partial^2 L(r, \dot{r})}{\partial r_i\partial r_j} \Big{|}_{r,\dot{r}=0} r^ir^j + \underbrace{\frac{\partial^2 L(r, \dot{r})}{\partial r_i\partial \dot{r}_j} \Big{|}_{r,\dot{r}=0} r^i\dot{r}^j }_0 + \frac{\partial^2 L(r, \dot{r})}{\partial \dot{r}_i\partial \dot{r}_j} \Big{|}_{r,\dot{r}=0} \dot{r}^i\dot{r}^j \right)\\ &+ o(h^2). \end{aligned}

其中 \(h = \max \left\{ r_i, \dot{r}_i \right\}\). 上式标注项总和为零的证明如下:

显然 \(L(r, \dot{r})|_{r, \dot{r} = 0} = L(q, \dot{q})|_{q=q^{(e)}, \dot{q}^{(e)} = 0} = -U(q^{(e)}) \). 由平衡条件可知: \[\frac{\partial U(q)}{\partial q_k}\Big{|}_{q=q^{(e)}} = 0.\] 可得标注项为零. \(\blacksquare\)

忽略高阶项, 有:

\begin{aligned} L(r, \dot{r}) &= -U(q^{(e)}) + \frac{1}{2}\left( \frac{\partial^2 L(r, \dot{r})}{\partial r_i\partial r_j}\Big{|}_{r,\dot{r}=0} r^ir^j + \frac{\partial^2 L(r, \dot{r})}{\partial \dot{r}_i\partial \dot{r}_j} \Big{|}_{r,\dot{r}=0} \dot{r}^i\dot{r}^j \right)\\ &=-U(q^{(e)}) + \frac{1}{2}\left( T_{ij}\dot{r}^i\dot{r}^j - V_{ij}r^ir^j \right). \end{aligned}

其中 \(\mathbf{T}, \mathbf{V}\) 分别有: \[\mathbf{T} = (T_{ij})_{D\times D}, \;T_{ij} = \frac{\partial^2 L(r, \dot{r})}{\partial \dot{r}_i\partial \dot{r}_j} \Big{|}_{r, \dot{r} = 0} = m_{ij}\left( q^{(e)} \right).\] \[\mathbf{V} = (V_{ij})_{D\times D}, \;V_{ij} = -\frac{\partial^2 L(r, \dot{r})}{\partial r_i\partial r_j} \Big{|}_{r, \dot{r} = 0} = \frac{\partial^2 U(q)}{\partial q_i\partial q_j} \Big{|}_{q = q^{(e)}}.\] 易证 \(\mathbf{T}\) 实对称正定, \(\mathbf{V}\) 实对称.

仍有拉格朗日方程: \[\frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L(r, \dot{r})}{\partial \dot{r}_k} \right) - \frac{\partial L(r, \dot{r})}{\partial r_k} = 0.\] 代入可得: \[\frac{\mathrm{d}}{\mathrm{d}t} \left( T_{ik}\dot{r}^i + T_{kk}\dot{r}_k + V_{ik}r^i + V_{kk}r_k\right) = 0.\] 近似为: \[\frac{\mathrm{d}}{\mathrm{d}t}\left( T_{ik}\dot{r}^i \right) + V_{ik}r^i + o(h) = 0.\] 即: \[T_{ik}\ddot{r}^i + V_{ik}r^i + o(h) = 0.\]

从 \(r\)-系线性变换到 \(\rho\)-系 (\(r_i = C_{ik}\rho^k,\;\dot{r}_i = C_{ik}\dot{\rho}^k\)) \(\rho\)-系称为简正坐标系. \[\begin{bmatrix}r_1\\ r_2\\ \vdots \\ r_D\end{bmatrix} = \mathbf{C} \begin{bmatrix}\rho_1\\ \rho_2\\ \vdots \\ \rho_D\end{bmatrix},\; \mathbf{C} = \begin{bmatrix} C_{11}&C_{12}&\cdots&C_{1D}\\ C_{21}&C_{22}&\cdots&C_{2D}\\ \vdots&\vdots&\ddots&\vdots\\ C_{D1}&C_{D2}&\cdots&C_{DD} \end{bmatrix}. \] 由前述拉格朗日量: \[L(r, \dot{r}) = -U \left( q^{(e)} \right) + \frac{1}{2} \left( T_{ij}\dot{r}^i\dot{r}^j - V_{ij}r^ir^j \right).\] 忽略常量 \(U\). 可得: \[L(\rho, \dot{\rho}) = \frac{1}{2}\left( \dot{\rho}^T\mathbf{C}^T \mathbf{T} \mathbf{C}^T\dot{\rho} - \rho^T\mathbf{C}^T \mathbf{V} \mathbf{C}^T\rho \right)\] 其中 \(\rho = \left[ \rho_1, \rho_2,\dots, \rho_D\right]^T\). 能证明: \(\mathbf{C}^T\mathbf{T}\mathbf{C} = \mathbf{I}_n\), \(\mathbf{C}^T \mathbf{V}\mathbf{C} = \Theta = \text{diag}(\theta_k)\). 可得 \[L(\rho, \dot{\rho}) = \frac{1}{2}\left( \dot{\rho}_k\dot{\rho}^k - \theta^k\rho_k^2\right).\]

此时拉格朗日方程为: \(\ddot{\rho} + \theta_k\rho^k = 0\), \(k = 1, 2, \dots, D\). 可解得:

\begin{equation*} \rho_k = \begin{cases} A_k\cos\omega_k t + B_k\sin\omega_k t,& \theta_k > 0,\\ A_k + B_k t,& \theta_k = 0,\\ A_k\cosh \gamma_k t + B_k\sinh \gamma_k t,& \theta_k < 0. \end{cases} \end{equation*}

其中 \(\omega_k = \sqrt{\theta_k}\), \(\gamma_k = \sqrt{-\theta_k}\).

唯一的问题在于常数矩阵 \(\mathbf{C}\) 的存在性⇒ 广义本征值问题.

\[\mathbf{A}\vec{z}^{(k)} = \theta_k \mathbf{g}\vec{z}^{(k)}.\] 转化为: \[\left( \mathbf{A} - \theta_k\mathbf{g} \right)\vec{z}^{(k)} = 0 \Rightarrow \left| \mathbf{A}-\theta_k\mathbf{g} \right| = 0.\] 其中 \(\mathbf{A}\) 为实对称矩阵, \(\mathbf{g}\) 为实对称正定矩阵.

定理一

\(\mathbf{A}\) 为\(N\)阶实对称矩阵, \(\mathbf{g}\)为实对称正定矩阵. 则广义本征值方程\(\mathbf{A}\vec{z}^{(k)}=\theta_k\mathbf{g}\vec{z}^{(k)}\)有:

• 存在 \(N\) 个本征值 \(\theta\in \mathbb{R}\)
• 存在 \(N\) 个对应的归一化实本征矢量 \(\vec{z}^{(k)}\).
• \(\vec{z}^{(k)}\) 满足如下正交特性: \[\left[ \vec{z}^{(k)} \right]^T\mathbf{g}\vec{z}^{(l)}=\delta_{kl}.\] 记为 \(\left[ \vec{z}^{(k)} \right]^T\odot\vec{z}^{(l)}=\delta_{kl}\).

定理二

\(\mathbf{C} = \left[\vec{z}^{(1)},\dots,\vec{z}^{(N)}\right]\), 满足 \[\mathbf{C}^T \mathbf{g}C = \mathbf{I}_n,\; \mathbf{C}^T\mathbf{A}\mathbf{C} = \Theta = \text{diag}(\theta_k).\]

由前述: \[L(\rho, \dot{\rho}) = \frac{1}{2}\left( \dot{\rho}^T\mathbf{C}^T \mathbf{T} \mathbf{C}^T\dot{\rho} - \rho^T\mathbf{C}^T \mathbf{V} \mathbf{C}^T\rho \right)\] 其中 \(\mathbf{T}\) 实对称正定, \(\mathbf{V}\) 实对称. 可取 \(\mathbf{V} = \mathbf{A}\), \(\mathbf{T}=\mathbf{g}\), 可得 \(\mathbf{V}\vec{z}^{(k)} = \theta_k \mathbf{T}\vec{z}^{(k)}\). 根据定理二, \(\mathbf{C}^T\mathbf{T}\mathbf{C} = \mathbf{I}_n\), \(\mathbf{C}^T\mathbf{V}\mathbf{C} = \text{diag}(\theta_k)\). 可得拉格朗日量: \[L(\rho, \dot{\rho}) = \frac{1}{2} \left( \dot{\rho}_k\dot{\rho}^k - \theta^k\rho_k^2\right).\]

由前述\(\rho_k\) 解, 对应三种平衡情况:

1. \(\theta_k > 0\) 时, 稳定平衡,
2. \(\theta = 0 \wedge B_k = 0\) 时, 随遇平衡,
3. \(\theta_k < 0\) 时, 不稳定平衡.

关于体系的平衡模式, 有:

• 仅当体系所有模式都是稳定平衡模式,体系才处于稳定平衡位形.
• 只要有一个模式是不稳定模式,体系平衡不再是稳定的.
• 求解 \(\left| \mathbf{V}-\theta_k\mathbf{T} \right|= 0\) 即可判断平衡位形是否为稳定平衡.

有: \(\mathbf{C} = \left[\vec{z}^{(1)},\dots,\vec{z}^{(N)}\right]\), \(\vec{z}^{(k)} = \left[ z_1^{(k)}, \dots ,z_D^{(k)},\right]\). 可得:

\begin{aligned} \vec{r} &= \mathbf{C}\vec{\rho} = \sum_{k=1}^D\vec{z}^{(k)}\rho_k\\ &=\sum_{\theta_k>0}\vec{z}^{(k)} \left( A_k\cos\omega_k t + B_k\sin\omega_k t \right) + \sum_{\theta_k=0}\vec{z}^{(k)}\left( A_k + B_k t \right)\\ &+\sum_{\theta_k=0}\vec{z}^{(k)} \left( A_k\cosh \gamma_k t + B_k\sinh \gamma_k t \right). \end{aligned}

广义变换不显含时间等价于广义能量函数表示能量. 引回常量 \(-U \left(q^{(e)}\right)\). 得到: \[L(\rho, \dot{\rho}) = -U \left( q^{(e)} \right) + \frac{1}{2} \left( \dot{\rho}_k\dot{\rho}^k - \theta^k\rho_k^2\right).\] 广义能量函数有:

\begin{aligned} H(\rho, \dot{\rho}) &= \frac{\partial L}{\partial \dot{\rho}_k}\dot{\rho}^k- L(\rho, \dot{\rho})\\ &= \frac{1}{2}\left( \dot{\rho}_k\dot{\rho}^k - \theta^k\rho_k^2\right) + U \left( q^{(e)} \right)\\ &=U \left( q^{(e)} \right) + \frac{1}{2}\sum_{\theta_k>0}\omega_k^2 \left( A_k^2 + B_k^2 \right)\\ &+\frac{1}{2}\sum_{\theta_k=0}B_k^2 + \frac{1}{2}\sum_{\theta_k < 0}\gamma_k^2 \left( -A_k^2 + B_k^2 \right) \end{aligned}

注意非稳定模式 (\(\theta_k < 0\)) 中初始位移 \(A_k\) 总是减小能量.

只激发 \(D\) 个简正模中第\(k\)个, 即 \(A_l=B_l = 0,\;l\ne k\). 可得 \[\vec{r} = \vec{z}^{(k)}\left( A_k\cos\omega_k t + B_k\sin\omega_k t \right).\] 即: \[\frac{r_i(t)}{r_j(t)} = \frac{z_i^{(k)}}{z_j^{(k)}}.\]