积分变换
$\quad\\$
傅里叶变换
$\quad\\$
无界自由振动
$$u_{tt}-a^2\Delta u=0\\
\quad\\u|_{t=0}=\varphi\qquad u_t|_{t=0}=\psi$$
  进行傅里叶变换$F(\boldsymbol{k},t)=\displaystyle\frac{1}{(2\pi)^n}\int_{\text{total}} f(\boldsymbol{r},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{r}}\dd V$
$$
\quad\\
U’’+k^2a^2U=0\\
\quad\\U|_{t=0}=\Phi(\boldsymbol{k})\qquad U’|_{t=0}=\Psi(\boldsymbol{k})$$
  $U$解具有形式
$$U(\boldsymbol{k},t)=\Phi(\boldsymbol{k})\cos(kat)+\frac{1}{ka}\Psi(\boldsymbol{k})\sin(kat)$$
  进行傅里叶逆变换
  一维情形，得到达朗贝尔公式
$$u(x,t)=\frac{1}{2}[\varphi(x+at)+\varphi(x-at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(\xi)\dd\xi$$
  三维情形，得到泊松公式
$$u(\boldsymbol{r},t)=\iiint_{\boldsymbol{k}}\left[\frac{1}{8\pi^3}\iiint_{\boldsymbol{r’}}\varphi(\boldsymbol{r’})e^{-i\boldsymbol{k}\cdot\boldsymbol{r’}}\dd V_{r’}\right]\cos(kat)e^{i\boldsymbol{k}\cdot\boldsymbol{r}}\dd V_k
\\+\iiint_{\boldsymbol{k}}\left[\frac{1}{8\pi^3}\iiint_{\boldsymbol{r’}}\psi(\boldsymbol{r’})e^{-i\boldsymbol{k}\cdot\boldsymbol{r’}}\dd V_{r’}\right]\frac{1}{ka}\sin(kat)e^{i\boldsymbol{k}\cdot\boldsymbol{r}}\dd V_k\\
\quad\\
=\frac{1}{8\pi^3}\iiint_{\boldsymbol{r’}}\varphi(\boldsymbol{r’})\left[\iiint_{\boldsymbol{k}}\cos(kat)e^{i\boldsymbol{k}\cdot\boldsymbol{(r-r’)}}\dd V_{k}\right]\dd V_{r’}
\\+\frac{1}{8\pi^3 a}\iiint_{\boldsymbol{r’}}\psi(\boldsymbol{r’})\left[\iiint_{\boldsymbol{k}}\frac{1}{k}\sin(kat)e^{i\boldsymbol{k}\cdot\boldsymbol{(r-r’)}}\dd V_{k}\right]\dd V_{r’}\\
\quad\\
=\frac{1}{4\pi a}\frac{\partial }{\partial t}\iiint_{\boldsymbol{r’}}\varphi(\boldsymbol{r’})\left[\iiint_{\boldsymbol{k}}\frac{1}{2\pi^2 k}\sin(kat)e^{i\boldsymbol{k}\cdot\boldsymbol{(r-r’)}}\dd V_{k}\right]\dd V_{r’}
\\+\frac{1}{4\pi a}\iiint_{\boldsymbol{r’}}\psi(\boldsymbol{r’})\left[\iiint_{\boldsymbol{k}}\frac{1}{2 \pi^2 k}\sin(kat)e^{i\boldsymbol{k}\cdot\boldsymbol{(r-r’)}}\dd V_{k}\right]\dd V_{r’}\\
\quad\\
\frac{1}{4\pi a}\frac{\partial }{\partial t}\iiint_{\boldsymbol{r’}}\varphi(\boldsymbol{r’})\left(\mathscr{F^{-1}}\left\{\mathscr{F}\left[\frac{1}{r}\delta(r-at)\right]e^{-i\boldsymbol{k}\cdot\boldsymbol{r’}}\right\}\right)\dd V_{r’}
\\+\frac{1}{4\pi a}\iiint_{\boldsymbol{r’}}\psi(\boldsymbol{r’})\left(\mathscr{F^{-1}}\left\{\mathscr{F}\left[\frac{1}{r}\delta(r-at)\right]e^{-i\boldsymbol{k}\cdot\boldsymbol{r’}}\right\}\right)\dd V_{r’}\\$$
$$
u(\boldsymbol{r},t)=\frac{1}{4\pi a}\frac{\partial }{\partial t}\iint_{|\boldsymbol{r}-\boldsymbol{r’}|=at}\frac{\varphi(\boldsymbol{r’})}{at}\dd S’+\frac{1}{4\pi a}\iint_{|\boldsymbol{r}-\boldsymbol{r’}|=at}\frac{\psi(\boldsymbol{r’})}{at}\dd S’$$