解答

1. 原问题的求解过程

\left\{ \begin{align} \cfrac{\partial^2u}{\partial t^2} = a^2 \cfrac{\partial^2u}{\partial x^2} \; (0<x<l, t>0) \tag{3}\label{eq3} \\ u(t,0) = 0, \, u(t,l) =\sin\omega t \tag{4}\label{eq4} \\ u(0,x)=0, \, u_t(0,x) = 0 \end{align} \right.

$v(t,x)=X(x)\sin\omega t$

$X''+\cfrac{\omega^2}{a^2}X=0$

$X(x)=C_1\cos\cfrac{\omega x}a+C_2\sin\cfrac{\omega x}a$

$C_2=\cfrac 1{\color{red}{\sin\cfrac{\omega l}a}}$

$X(x) = \cfrac 1{\sin \cfrac{\omega l}a}\sin\cfrac{\omega x}a \\ v(t,x) = \cfrac {\sin\cfrac{\omega x}a}{\sin \cfrac{\omega l}a}\sin\omega t$

$\left\{ \begin{array}l \cfrac{\partial^2w}{\partial t^2} = a^2 \cfrac{\partial^2w}{\partial x^2} \; (0<x<l, t>0) \\ w(t,0) = w(t,l) = 0 \\ w(0,x)=0, \, w_t(0,x) = -\omega\cfrac{\sin\cfrac{\omega x}a}{\sin\cfrac{\omega l}a} \end{array} \right.$

$w(t,x) = 2\omega al\sum_{n=1}^{+\infty}\cfrac{(-1)^{n+1}}{\color{red}{(\omega l)^2-(n\pi a)^2}}\sin\cfrac{n\pi at}l\sin\cfrac{n\pi x}l$

$u(t,x) = \cfrac {\sin\cfrac{\omega x}a}{\color{red}{\sin \cfrac{\omega l}a}}\sin\omega t + \\ 2\omega al\sum_{n=1}^{+\infty}\cfrac{(-1)^{n+1}}{\color{red}{(\omega l)^2-(n\pi a)^2}}\sin\cfrac{n\pi at}l\sin\cfrac{n\pi x}l$

2. 解的分析

$\lim_{\omega \to \frac{n\pi l}a}u(t,x)=\infty ,\, n=1,2,\ldots$

3. 对所求得的解进行物理解释

3.1 问题的提出

$\cfrac{\partial^2u}{\partial t^2} = a^2 \cfrac{\partial^2u}{\partial x^2} + f(t,x) \\ \left( a = \sqrt{\cfrac T\rho} ,\, f(t,x)=\cfrac{g(t,x)}\rho \right)$

$\cfrac{\partial^2u}{\partial t^2} = a^2 \cfrac{\partial^2u}{\partial x^2} ,\; \left( a = \sqrt{\cfrac T\rho} \right)$

3.2 一维波动方程中参数 $a$ 的物理意义

$a_\bot = -\cfrac{v^2}{R}$

$F_\bot = -2T\sin\theta \approx -2T\theta$

$l=2R\theta$

$F_\bot = -2T\theta = \rho \cdot 2R\theta\left(-\cfrac {v^2}{R}\right) \\ \therefore v = \sqrt{\cfrac T\rho}$

3.3 驻波与固有频率

$2l=n\lambda ,\, n=1,2,\ldots$

$f=\cfrac v\lambda=\cfrac{nv}{2l} ,\, n=1,2,\ldots$

$\omega = 2\pi f = \cfrac{n\pi v}l$

$\color{red}{\omega = \cfrac{n\pi a}l} \color{black}{,\, n=1,2,\cdots}$

4. 计算机模拟

4.1 准备工作

u[t_, x_] :=
2*a*l*$Omega]* Sum[ ((-1)^(n + 1)*Sin[(Pi*n*x)/l]*Sin[(Pi*a*n*t)/l])/((l*\[Omega])^2 - (Pi*a*n)^2), {n, 1, 20} ] + (Sin[t*\[Omega]]*Sin[(x*\[Omega])/a])/Sin[(l*\[Omega])/a]  这里由于计算机计算方式的限制，无法将 \sum_{n=1}^{+\infty} 导入计算。经过测试，当累加上限为 20 时，该函数已经能够提供很好的近似结果了，因此出于各种考虑（尤其是计算资源的利用），这里就取累加上限为 20 了，因此实际绘图的函数是： \[u(t,x) = \cfrac {\sin\cfrac{\omega x}a}{\sin \cfrac{\omega l}a}\sin\omega t + 2\omega al\sum_{n=1}^{20}\cfrac{(-1)^{n+1}}{(\omega l)^2-(n\pi a)^2}\sin\cfrac{n\pi at}l\sin\cfrac{n\pi x}l$

4.2 静态 3D 绘图

l = 10; a = 1; \[Omega] = 0.5;
Plot3D[u[t, x], {x, 0, 10}, {t, 0, 50}]


l = 10; a = 1; \[Omega] = 0.31;
Plot3D[u[t, x], {x, 0, 10}, {t, 0, 200}]


l = 10; a = 1; \[Omega] = 3*Pi/10 + 1/1000;
Plot3D[u[t, x], {x, 0, 10}, {t, 0, 60}]


4.3 动态图像

l = 10; a = 1; \[Omega] = 0.15;
Animate[
Plot[u[t, x], {x, 0, 10}, PlotRange -> {-5, 5}], {t, 0, 300},
AnimationRate -> 30]


l = 10; a = 1; \[Omega] = 0.31;
Animate[
Plot[u[t, x], {x, 0, 10}, PlotRange -> {-5, 5}], {t, 0, 120, 0.4},
AnimationRate -> 30]


l = 10; a = 1; \[Omega] = 0.63;
Animate[
Plot[u[t, x], {x, 0, 10}, PlotRange -> {-5, 5}], {t, 0, 120, 0.4},
AnimationRate -> 30]


l = 10; a = 1; \[Omega] = 1.257;
Animate[
Plot[u[t, x], {x, 0, 10}, PlotRange -> {-5, 5}], {t, 0, 120, 0.4},
AnimationRate -> 30]


Tags:

Updated: