三维波动方程的泊松公式
首先我们直接给出泊松公式：
$$ u(x,y,z,t)=\frac{\partial}{\partial{t}}(\frac{1}{4\pi{a}^2t})\iint_{S_{at}(M)}\varphi{dS}+\frac{1}{4\pi{a^2}t}\iint_{S_{at}(M)}\psi{d}S $$
我们直接解习题好吧：
$$ u_{tt}=a^2(u_{xx}+u_{yy}+u_{zz}) $$
的柯西问题：
$$ \begin{cases} u|_{t=0}=x^3+y^2z\\ u_t|_{t=0}=0 \end{cases} $$
解：由泊松公式
$$ u=\frac{\partial}{\partial{t}}\frac{1}{4\pi{a}}\iint_{S_{at}(M)}\frac{\phi}{r}dS+\frac{1}{4{\pi}a}\iint_{S_{at}(M)}\frac{\psi}{r}dS $$
现在$\psi=0,\phi=x^3+y^2z$。且
$$ \iint_{S_{at}(M)}\frac{\phi}{r}dS=\int^{\pi}_0\int^{2\pi}_0\phi(r,\theta,\varphi)r\sin{\theta}d{\theta}d{\varphi}|_{r=at} $$
其中
$$ \begin{aligned} \phi(r,\theta,\varphi)=&\phi(x+r\sin{\theta}\cos{\varphi},y+r\sin{\theta}\sin{\varphi},z+r\cos{\theta})\\ =&(x+r\sin{\theta}\cos{\varphi})^3+(y+\sin{\theta}\sin{\varphi})^2(z+r\cos{\theta})\\ =&x^3+y^2z+3x^2r\sin{\theta}\cos{\varphi}+3xr^2\sin^2{\theta}\cos^2{\varphi}+r^2r\sin^3{\theta}\cos^3{\varphi}\\ +&2yzr\sin{\theta}\sin{\varphi}+rz\sin^2{\theta}\sin^2{\varphi}+y^2r\cos{\theta}+2yr^2\sin{\theta}\cos{\theta}\sin{\varphi}+r^3r\sin{\theta}\sin^2{\varphi}\cos{\theta} \end{aligned} $$
计算：
$$ \int^{\pi}_0\int^{2\pi}_0\phi(r,\theta,\varphi)r\sin{\theta}d{\theta}d{\varphi} $$
$$ \begin{aligned} \int^{\pi}_0\int^{2\pi}_0(x^3+y^2z)r\sin{\theta}d{\theta}d{\varphi}=&r(x^3+y^2z){\cdot}2\pi(-\cos{\theta})^{\pi}_0\\ =&4\pi{r}(x^3+y^2z) \end{aligned} $$
$$ \int^{\pi}_0\int^{2\pi}_03x^2r\sin{\theta}\cos{\varphi}{\cdot}rsin{\theta}d{\theta}d{\varphi}=3x^2r^2\int^{\pi}_0\sin^2{\theta}d{\theta}\int^{2\pi}_0\cos{\varphi}d{\varphi}=0 $$
$$ \int^{\pi}_0\int^{2\pi}_03xr^2\sin^2{\theta}\cos^2{\varphi}{\cdot}r\sin{\theta}d{\theta}d{\varphi}=3xr^3\int^{\pi}_0\sin^3{\theta}d{\theta}\int^{2\pi}_0\cos^2{\varphi}d{\varphi}\\ =3xr^3[\frac{1}{3}\cos^2{\theta}-\cos{\theta}]^{\pi}_0{\cdot}[\frac{\phi}{2}+\frac{1}{4}\sin{2\phi}]^{2\pi}_0\\ =4xr^3\pi\int^{\pi}_0\int^{2\pi}_0r^3\sin{\theta}\cos^3{\varphi}{\cdot}rsin{\theta}d{\theta}d{\varphi}\\ =r^4\int^{\pi}_0\sin^4{\theta}d{\theta}\int^{2\pi}_0\cos^3{\varphi}d{\varphi}=4\pi{r}^3\\ $$
$$ \int^{\pi}_0\int^{2\pi}_02yzr\sin{\theta}\sin{\varphi}{\cdot}r\sin{\theta}d{\theta}d{\varphi}=2yzr^2\int^{\pi}_0\sin^2{\theta}d{\theta}\int^{2\pi}_0\sin{\varphi}d{\varphi}=0 $$
$$ \int^{\pi}_0\int^{2\pi}_0r^2z\sin^2{\theta}\sin^2{\varphi}{\cdot}r\sin{\theta}d{\theta}d{\varphi}=rz\int^{\pi}_0\sin^3{\theta}d{\theta}\int^{2\pi}_0\sin^2{\varphi}d{\varphi}\\=r^3z[\frac{1}{3}\cos^3{\theta}-\cos{\theta}]^{\pi}_0{\cdot}[\frac{\varphi}{2}-\frac{1}{4}\sin{2\varphi}]^{2\pi}_0=\frac{4}{3}\pi{r}^3z $$
$$ \int^{\pi}_0\int^{2\pi}y^2r\cos{\theta}{\cdot}r\sin{\theta}d{\theta}d{\varphi}=y^2r^2\int^{\pi}_0\cos{\theta}\sin{\theta}d{\theta}\int^{2\pi}_0d{\varphi}=0 $$
$$ \int^{\pi}_0\int^{2\pi}_02yr^2\sin{\theta}\cos{\theta}\sin{\varphi}{\cdot}r\sin{\theta}d{\theta}d{\varphi}=2yr^3\int^{\pi}_0\sin^2{\theta}\cos{\theta}d{\theta}\int^{2\pi}_{0}\sin\varphi{d}{\varphi}=0 $$
$$ \int^{\pi}_0\int^{2\pi}_0r^3\sin^2{\theta}\sin^2{\varphi}\cos{\theta}{\cdot}r\sin{\theta}d{\theta}d{\varphi}=r^4\int^{\pi}_0\sin^3{\theta}\cos{\theta}d{\theta}\int^{2\pi}_0\sin^2{\varphi}d{\varphi}=0 $$
$$ \iint_{S_{at}(M)}\frac{\phi}{r}dS=[4\pi{r}(x^2+y^2z)+4\pi{r}^3+\frac{4}{3}\pi{r}^3z]_{r=at}\\=4\pi{at}[x^2+y^2z+xa^2t^2+\frac{1}{3}a^2t^2z] $$
$$ u(x,y,z)=\frac{\partial}{\partial{t}}\frac{1}{4\pi{a}}\iint_{S_{at}(M)}\frac{\phi}{r}dS=\frac{\partial}{\partial{t}}[tx^3+ty^2z+xa^2t^3+\frac{1}{3}a^2t^2z]\\ =x^3+y^2z+3a^2t^2x+a^2t^2z $$
二维波动方程的泊松公式
$$ \begin{aligned} u(x,y,z,t)=&u(M,t)\\ =&\frac{1}{2{\pi}a}\left[\frac{\partial}{\partial{t}}\iint_{\sum^M_{at}}\frac{\varphi(\xi,\eta)d{\xi}d{\eta}}{\sqrt{(at)^2-(\xi-x)^2-(\eta-y)^2}}\right]+\iint_{\sum^M_{at}}\frac{\psi(\xi,\eta)d{\xi}d{\eta}}{\sqrt{(at)^2-(\xi-x)^2-(\eta-y)^2}}\\ =&\frac{1}{2\pi{a}}[\frac{\partial}{\partial{t}}\int^{at}_0\int^{2\pi}_0\frac{\varphi(x+r\cos{\theta},y+r\sin{\theta})}{\sqrt{(at)^2-r^2}}rd{\theta}dr+\int^{at}_{0}\int^{2\pi}_0\frac{\psi(x+r\cos{\theta},y+r\sin{\theta})}{\sqrt{(at)^2-r^2}}rd{\theta}dr] \end{aligned} $$
