连续系统

一维问题

• 杆、轴、弦
• 梁

杆纵向振动微分方程

• 平截面假设
• 纵向伸缩引起的横向变形是高阶小量

对杆的微元体列牛顿第二定律：

\[ f(x,t) \, \mathrm{d} x + N(x + \mathrm{d} x, t) - N(x,t) = \rho(x) A(x) \, \mathrm{d} x \frac{\partial^2 u}{\partial t^2} \]

略去高阶小量 \(\omicron(\mathrm{d} x)\)，得 \(N(x + \mathrm{d} x, t) \approx N(x,t) + \frac{\partial N}{\partial x} \mathrm{d} x\). 代入上式可得

\[ \frac{\partial N(x, t)}{\partial x} + f(x,t) = \rho(x) A(x) \frac{\partial^2 u}{\partial t^2} \qquad (\text{in } \Omega) \]

而

\[ N(x, t) = A(x) \sigma(x, t) = A(x) E(x) \varepsilon(x, t) = A(x) E(x) \frac{\partial u}{\partial x} \]

代入上式，得杆的纵向振动微分方程：

\[ \begin{equation} \label{eq:longitudinal-vibration} \tag{3.1} \underset{惯性力}{\underbrace{\rho(x) A(x) \frac{\partial^2 u}{\partial t^2}}} - \underset{恢复力}{\underbrace{\frac{\partial}{\partial x} \left[ E(x) A(x) \frac{\partial u}{\partial x} \right]}} = \underset{外力}{\underbrace{f(x,t)}} \end{equation} \]

注意到上式没有阻尼力，所以是无阻尼振动微分方程。实际的阻尼分为外阻尼和内阻尼，较为复杂。观察 \eqref{eq:longitudinal-vibration} 的系数，可知这是一个双曲型方程。

• 初始条件

\[ u(x, t)\bigg|_{t=t_0} = \varphi(x) , \quad \left.\frac{\partial u(x, t)}{\partial t} \right|_{t=t_0} = \psi(x). \]

• 边界条件（\(x = 0\)）
  • 固定端 \(u(0, t) = 0\)
  • 自由端 \(N(0, t) = 0 \iff \left.\frac{\partial u}{\partial x}\right|_{x=0} = 0\)
  • 弹性支撑 \(\left.\left(E A \frac{\partial u}{\partial x}\right)\right|_{x=0} - k_1 \left.u(x, t)\right|_{x=0} = 0\)

    实际上，令 \(k_1 \to \infty\) 可得固定端，\(k_1 = 0\) 可得自由端.

  • 集中质量 \(\left.\left(E A \frac{\partial u}{\partial x}\right)\right|_{x=0} - M \left.\frac{\partial^2 u}{\partial t^2}\right|_{x=0} = 0\)

回顾：定解问题的一般解法

1. 边界条件齐次化 找到只需满足边界条件的特解 \(\psi(x, t)\)，令 \(u(x, t) = \tilde{u}(x, t) + \psi(x, t)\)，则 \(\tilde{u}(x, t)\) 满足齐次边界条件.
2. 原方程 \(\mathcal{L} [\tilde{u}(x, t) + \psi(x, t)] = f(x, t)\) 转化为

\[ \mathcal{L} [\tilde{u}(x, t)] = f(x, t) - \mathcal{L} [\psi(x, t)] \]

• \(f(x, t)\) 为已知外力
• \(\mathcal{L} [\psi(x, t)]\) 为已知的，由边界条件齐次化引入的附加项

方程非线性 vs. 边界条件非线性

• 若考虑更高阶的应变：

\[ \varepsilon(x, t) = \frac{\partial u}{\partial x} + \frac{1}{2} \left(\frac{\partial u}{\partial x}\right)^2 + \cdots \]

则方程变为非线性微分方程.

• 若弹簧式非线性的：

\[ F = k_1 u + k_3 u^3 + \cdots \]

则边界条件变为非线性边界条件. 以上两种情况都属于非线性问题，不能使用分离变量法.

圆轴的扭转振动

对式 \eqref{eq:longitudinal-vibration} 中的变量进行替换：\(u(x, t) \mapsto \theta(x, t), \, A(x) \mapsto I(x), \, E(x) \mapsto G(x), \, f(x, t) \mapsto m(x, t)\)，即可得到圆轴的扭转振动微分方程：

\[ \begin{equation} \label{eq:torsional-vibration} \tag{3.2} \rho(x) I(x) \frac{\partial^2 \theta}{\partial t^2} - \frac{\partial}{\partial x} \left[ G(x) I(x) \frac{\partial \theta}{\partial x} \right] = m(x, t) \qquad (\text{in } \Omega) \end{equation} \]

1. 初始条件

\[ \theta(x, 0) = \varphi(x) , \quad \left.\frac{\partial \theta}{\partial t} \right|_{t=0} = \psi(x). \]

1. 边界条件（\(x = 0\)）
   • 固定端 \(\theta(0, t) = 0\)
   • 自由端 \(\left.\frac{\partial \theta}{\partial x}\right|_{x=0} = 0\)
   • 弹性支撑 \(\left.\left(G I \frac{\partial \theta}{\partial x}\right)\right|_{x=0} - k_1 \left.\theta(x, t)\right|_{x=0} = 0\)
   • 集中转动惯量 \(\left.\left(G I \frac{\partial \theta}{\partial x}\right)\right|_{x=0} - J \left.\frac{\partial^2 \theta}{\partial t^2}\right|_{x=0} = 0\)

弦振动方程

\[ \begin{aligned} \rho(x) A(x) \, \mathrm{d} x \frac{\partial^2 u}{\partial t^2} = \sum F &= f(x,t) \, \mathrm{d} x + T_0(x + \mathrm{d} x, t) \sin \theta(x + \mathrm{d} x, t) - T_0(x, t) \sin \theta(x, t) \\ &\approx f(x,t) \, \mathrm{d} x + T_0(x, t) \theta(x + \mathrm{d} x, t) - T_0(x, t) \theta(x, t) \\ &\xlongequal{\theta(x, t) \approx \frac{\partial u}{\partial x}} f(x,t) \, \mathrm{d} x + \frac{\partial}{\partial x} \left[ T_0(x, t) \frac{\partial u}{\partial x} \right] \mathrm{d} x \end{aligned} \]

得弦振动方程：

\[ \begin{equation} \label{eq:string-vibration} \tag{3.3} \rho(x) A(x) \frac{\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x} \left[ T_0(x, t) \frac{\partial u}{\partial x} \right] = f(x,t) \qquad (\text{in } \Omega) \end{equation} \]

对于弦振动的边界条件，一般只考虑两端固定（有自由端就没有张力了）。

杆自由振动方程求解

式 \eqref{eq:longitudinal-vibration} 的齐次方程为

\[ \rho(x) A(x) \frac{\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x} \left[ E(x) A(x) \frac{\partial u}{\partial x} \right] = 0 \]

设 \(\rho, E, A\) 均为常数，得到

\[ \begin{equation} \label{eq:wave-equation} \tag{3.4} \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} \end{equation} \]

其中 \(c = \sqrt{\frac{E}{\rho}}\) 为波速。下面采用分离变量法。设 \(u(x, t) = U(x) T(t)\)，代入 \eqref{eq:wave-equation} 得

\[ \frac{T''}{T} = c^2 \frac{U''}{U} \equiv -\omega^2 \]

得到两个 ODE：

\[ \left \{ \begin{aligned} &T'' + \omega^2 T = 0 \\ &U'' + \frac{\omega^2}{c^2} U = 0 \end{aligned} \right. \]