数学物理方程 - 12.12 高维波动方程

三维波动方程

我们接下来研究三维波动方程 Cauchy 问题：

$$\{\begin{array}{ll}\frac{{\partial }^{2}u}{\partial t^2}-a^2\Delta u=f(x,t),& t>0,x\in {\mathbb{R}}^3\\ u{|}_{t=0}=g_0(x)& \\ \frac{\partial u}{\partial t}{|}_{t=0}=g_1(x)& \end{array}$$

其中 $f,g_0,g_1\in C^3$ . 这样的 Cauchy 问题方法和热传导方程类似：利用基本解进行直接求解.

设 $\tilde{u}(x,t)=H(t)u(x,t)$ ，

1️⃣ 先计算出

$$\left(\frac{{\partial }^2}{\partial t^2}-a^2\Delta \right)\tilde{u}(x,t)=F(x,t)$$

2️⃣ 然后利用计算结果 $F(x,t)$ 计算得到 $\tilde{u}(x,t)=E(x,t)\ast F(x,t)$ ，最后令

$$u(x,t)=\tilde{u}(x,t),t>0$$

那么我们现在进行 1️⃣ 步，依次有

$$\begin{array}{rl}\left(\frac{{\partial }^2}{\partial t^2}-a^2\Delta \right)\tilde{u}(x,t)& =\frac{{\partial }^2}{\partial t^2}[H(t)u(x,t)]-a^2\Delta [H(t)u(x,t)]\\ & ={\partial }_t\left[H^{\prime }(t)u(x,t)+H(t)\frac{\partial u}{\partial t}\right]-a^2\Delta [H(t)u(x,t)]\\ & =H(t)[u_{tt}-a^2\Delta u]+H^{\prime \prime }(t)u+2H^{\prime }(t)u_t\\ & =H(t)f(x,t)+{\delta }^{\prime }(t)u(x,t)+2\delta (t)u_t\end{array}$$

下面计算广义函数 ${\delta }^{\prime }(t)u(x,t)$ ，对任意 $\phi \in C_0^{\infty }$ 有

$$\begin{array}{rl}⟨{\delta }^{\prime }(t)u,\phi ⟩& =(-1)⟨\delta (t),{\partial }_t(u\phi )⟩\\ & =(-1)⟨\delta (t),u_t\phi +u{\phi }_t⟩\\ & =(-1)⟨\delta (t),u_t\phi ⟩+(-1)⟨\delta (t),u{\phi }_t⟩\\ & =(-1)⟨\delta (t)u_t,\phi ⟩+(-1)⟨\delta (t)u,{\phi }_t⟩\\ & =⟨-\delta (t)u_t+{\partial }_t(\delta (t)u),\phi ⟩\end{array}$$

因此

$${\delta }^{\prime }(t)u={\partial }_t(\delta (t)u)-\delta (t)u_t$$

同理我们可以计算 ${\partial }_t[\delta (t)u]$ 以及 $\delta (t)u_t$ ，有

$${\partial }_t(\delta (t)u)={\partial }_t(g_0(x)\delta (t)),\delta (t)u_t=g_1(x)\delta (t)$$

最后代入有

$$\left(\frac{{\partial }^2}{\partial t^2}-a^2\Delta \right)\tilde{u}(x,t)=H(t)f(x,t)+{\partial }_t[g_0(x)\delta (t)]+g_1(x)\delta (t)=F(x,t)$$

再进行 2️⃣ 步，考虑

$$\begin{array}{rl}& \tilde{u}(x,t)=E_{+}(x,t)\ast F(x,t)\\ & =E_{+}(x,t)\ast [H(t)f(x,t)+{\partial }_t(g_0(x)\delta (t))]\\ & =E_{+}(x,t)\ast (H(t)f(x,t))+E_{+}(x,t)\ast [{\partial }_t(\delta (t)g_0(x))]+E_{+}(x,t)\ast (\delta (t)g_1(x))\end{array}$$

然后又是广义函数的计算，分别计算可得

$$\begin{array}{rl}& E_{+}(x,t)\ast [{\partial }_t(\delta (t)g_0(x))]={\partial }_t(E_{+}(x,t)\ast g_0(x))\\ & E_{+}(x,t)\ast (\delta (t)g_1(x))=E_{+}(x,t)\ast g_1(x)\end{array}$$

代回有

$$\begin{array}{rl}& E_{+}(x,t)\ast F(x,t)\\ & =E_{+}(x,t)\ast [H(t)f(x,t)]+{\partial }_t(E_{+}(x,t)\ast g_0(x))+E_{+}(x,t)\ast g_1(x)\\ & =I_1+I_2+I_3\end{array}$$

接下来的计算更是又臭又长：