[偏微分方程教程习题参考解答]4.3高维波动方程

 

 

1. 求解下列齐次方程的 Cauchy 问题:

 

(1). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=0,&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=x^3+y^2z,\ u_t|_{t=0}=0. \ea}}$

 

解答: $$\beex \bea u(x,y,z,t)&=\frac{\p}{\p t}\sez{ \frac{1}{4\pi a^2t} \iint_{S_{at}(M)}(\xi^3+\eta^2\zeta)\rd S}\quad\sex{M:\ (x,y,z)}\\ &=\frac{\p}{\p t}\sez{ \frac{1}{4\pi a^2t} \iint_{S_1(0)} (x+at u)^3+(y+at v)^2(z+at w) (at)^2\rd S }\\ &=\frac{\p}{\p t}\sez{ \frac{t}{4\pi} \sex{ 3a^2t^2x\cdot \frac{4\pi}{3} +x^3\cdot 4\pi +a^2t^2z\cdot \frac{4\pi}{3} +y^2z\cdot 4\pi }}\\ &=x^3+y^2z+a^2t^2(3x+z). \eea \eeex$$

 

(2). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=0,&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=f(x)+g(y),\ u_t|_{t=0}=\varphi(x)+\psi(z). \ea}}$

 

解答: 直接代入 Poisson 公式即得.

 

(3). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=0,&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=u_t|_{t=0}=e^{x^2+y^2+z^2}. \ea}}$

 

解答: 先计算 $$\beex \bea U&=\frac{1}{4\pi a^2t}\iint_{S_{at}(M)} e^{\xi^2+\eta^2+\zeta^2}\rd S =\frac{t}{4\pi}\iint_{S_1(0)} e^{(x+atu)^2+(y+at v)^2+(z+at w)^2}\rd S\\ &=\frac{t}{4\pi} e^{x^2+y^2+z^2+a^2t^2} \iint_{S_1(0)} e^{2at xu+2at yv+2at zw}\rd S\\ &=\frac{t}{4\pi}e^{x^2+y^2+z^2+a^2t^2} \iint_{S_1(0)}e^{2at \sqrt{x^2+y^2+z^2}Z}\rd S\\ &\quad\sex{\mbox{正交变换 }X=\cdots,\ Y=\cdots,\ Z=\frac{2at(xu+yv+zw)}{2at\sqrt{x^2+y^2+z^2}}}\\ &=\frac{t}{4\pi}e^{x^2+y^2+z^2+a^2t^2} \int_0^{2\pi}\rd \varphi \int_0^\pi e^{2at\sqrt{x^2+y^2+z^2}\cos \tt}\sin\tt \rd \tt\\ &\quad\sex{X=\sin\tt\cos \varphi,\ Y=\sin\tt\sin \varphi,\ Z=\cos\tt,\ 0\leq \tt\leq \pi,\ 0\leq \varphi\leq 2\pi}\\ &=\frac{t}{2}e^{x^2+y^2+z^2+a^2t^2}\cdot \frac{1}{2at \sqrt{x^2+y^2+z^2}} \sex{e^{2at\sqrt{x^2+y^2+z^2}}-e^{-2at\sqrt{x^2+y^2+z^2}}}\\ &=\frac{e^{x^2+y^2+z^2+a^2t^2}}{2a\sqrt{x^2+y^2+z^2}}\sinh(2at\sqrt{x^2+y^2+z^2}), \eea \eeex$$ 然后由 Poisson 公式, $$\beex \bea u(x,t)&=\frac{\p U}{\p t}+U\\ &=\frac{e^{x^2+y^2+z^2+a^2t^2}}{2a\sqrt{x^2+y^2+z^2}} \sez{ 2a\sqrt{x^2+y^2+z^2} \cosh \sex{2at \sqrt{x^2+y^2+z^2}} \atop+(1+2a^2t)\sinh\sex{\sqrt{x^2+y^2+z^2}}}. \eea \eeex$$

 

(4). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy})=0,&(x,y)\in\bbR^2,\ t>0,\\ u|_{t=0}=0,\ u_t|_{t=0}= x+y. \ea}}$

 

解答: $$\beex \bea u(x,t)&=\frac{1}{4\pi a^2t}\cdot 2\iint_{\vSa_{at}(M)} (\xi+\eta)\cdot \frac{at}{\sqrt{a^2t^2-(\xi-x)^2-(\eta-y)^2}}\rd \xi\rd \eta\\ &=\frac{1}{2\pi a}\int_0^1 \int_0^{2\pi} (x+at\rho\cos\tt +y+at\rho\sin\tt)\frac{1}{\sqrt{a^2t^2-a^2t^2\rho^2}}a^2t^2\rho\rd \tt\rd \rho\\ &=\frac{1}{2\pi a}\cdot 2\pi (x+y)at\int_0^1 \frac{\rho}{\sqrt{1-\rho^2}}\rd \rho\\ &=(x+y)t. \eea \eeex$$

 

(5). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy})=0,&(x,y)\in\bbR^2,\ t>0,\\ u|_{t=0}=x^2(x+y),\ u_t|_{t=0}=0. \ea}}$

 

解答: $$\beex \bea &\quad u(x,t)\\ &=\frac{\p}{\p t}\sez{ \frac{1}{4\pi a^2t}\cdot 2\iint_{\vSa_{at}(M)}(\xi^3+\xi^2\eta)\frac{at}{\sqrt{a^2t^2-(\xi-x)^2-(\eta-y)^2}}\rd \xi\rd \eta }\\ &=\frac{\p}{\p t}\sed{ \frac{1}{2\pi a} \int_0^1 \int_0^{2\pi} \sez{(x+at\rho \cos\tt)^3\atop +(x+at\rho\cos\tt)^2(y+at\rho\sin\tt)} \frac{1}{\sqrt{a^2t^2-a^2t^2\rho^2}}\cdot a^2t^2\rho\rd \tt\rd \rho }\\ &=\frac{\p}{\p t}\sez{ \frac{t}{2\pi} \int_0^1 (x^3\cdot 2\pi +3xa^2t^2\rho^2\cdot\pi +x^2y\cdot 2\pi +ya^2t^2\rho^2\cdot \pi )\frac{\rho}{\sqrt{1-\rho^2}}\rd \rho }\\ &=\frac{\p}{\p t} \sez{ t(x^3+x^2y) +\frac{t}{2}a^2t^2(3x+y)\int_0^1 \frac{\rho^3}{\sqrt{1-\rho^2}} \rd\rho}\\ &=x^2(x+y)+a^2t^2(3x+y),\quad\sex{\int_0^1 \frac{\rho^3}{\sqrt{1-\rho^2}}\rd \rho=\int_0^\frac{\pi}{2}\sin^3\tt\rd \tt=\frac{2}{3}}. \eea \eeex$$

 

(6). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy})=0,&(x,y)\in\bbR^2,\ t>0,\\ u|_{t=0}=\varphi(r),\ u_t|_{t=0}=\psi(r),\ r=\sqrt{x^2+y^2}. \ea}}$

 

解答: 代入 Poisson 公式即得结果, 不算啦.

 

(7). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy})=0,&(x,y)\in\bbR^2,\ t>0,\\ u|_{t=0}=\varphi(x)+f(y),\ u_t|_{t=0}=\psi(x)+g(y). \ea}}$

 

解答: 代入 Poisson 公式即得结果, 不算啦.

 

2. 试用降维法导出弦振动方程的 D'Alembert 公式.

 

解答: 由定理 4.1, 只需求出 $$\bex \sedd{\ba{ll} u_{tt}-a^2u_{xx}=0,\\ u|_{t=0}=0,\ u_t|_{t=0}=\psi(x) \ea} \eex$$ 的解 $u$. 将其看成二维弦振动方程, 由 Poisson 公式, $$\beex \bea u&=\frac{1}{4\pi a^2t}\cdot 2\iint_{\vSa_{at}(M)} \psi(\xi)\frac{at}{\sqrt{a^2t^2-(\xi-x)^2-(\eta-y)^2}}\rd \xi\rd \eta\\ &=\frac{1}{2\pi a}\int_{x-at}^{x+at} \psi(\xi)\rd \xi \int_{y-\sqrt{a^2t^2-(\xi-x)^2}} ^{y+\sqrt{a^2t^2-(\xi-x)^2}} \frac{1}{\sqrt{a^2t^2-(\xi-x)^2-(\eta-y)^2}}\rd \eta\\ &=\frac{1}{2\pi a}\int_{x-at}^{x+at} \psi(\xi)\cdot \pi \rd \xi\quad\sex{\eta=y+\sqrt{a^2t^2-(\xi-x)^2}\sin\tt}\\ &=\frac{1}{2a}\int_{x-at}^{x+at}\psi(\xi)\rd \xi. \eea \eeex$$

 

3. 求解下列非齐次方程的 Cauchy 问题:

 

(1). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=2(y-t),\\ u|_{t=0}=0,\ u_t|_{t=0}=x^2+y^2. \ea}}$

 

解答: $$\beex \bea u&=\frac{1}{4\pi a^2t}\iint_{S_{at}(M)} (\xi^2+\eta^2)\rd S +\frac{1}{4\pi a^2}\iiint_{K_{at}(M)} \frac{2\sez{\eta-\sex{t-\frac{\sqrt{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2}}{a}}}} {\sqrt{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2}}\rd \Omega\\ &=\frac{t}{4\pi}\iint_{S_1(0)} (x+at u)^2+(y+at v)^2\rd S +\frac{1}{4\pi a^2}\iiint_{K_1(0)} \frac{2\sez{(y+at v)-\sex{t-\frac{at\sqrt{u^2+v^2+w^2}}{a}}}}{at\sqrt{u^2+v^2+w^2}}a^3t^3\rd \Omega\\ &=\frac{t}{4\pi}\int_0^\pi \sin\tt\rd \tt \int_0^{2\pi} (x^2+y^2+a^2t^2\sin^2\tt)\rd \varphi +\frac{at^3}{4\pi}\int_0^1 \frac{2\sex{y-t+\frac{at\tau}{a}}}{at\tau}\cdot 4\pi \tau^2\rd \tau\\ &=\frac{2a^2t^3}{3}+t(x^2+y^2)+t^2y-\frac{t^3}{3}. \eea \eeex$$

 

(2). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=x^2t^2,\\ u|_{t=0}=y^2,\ u_t|_{t=0}=z^2. \ea}}$

 

解答: $$\beex \bea u&=\frac{\p}{\p t}\sez{\frac{1}{4\pi \cdot 8t}\iint_{S_{2\sqrt{2}t}(M)} \eta^2\rd S} +\frac{1}{4\pi \cdot 8t} \iint_{S_{2\sqrt{2}t}(M)}\zeta^2\rd S\\ &\quad +\frac{1}{4\pi\cdot 8}\iiint_{K_{2\sqrt{2}t}(M)}\frac{\xi^2\sex{t-\frac{r}{2\sqrt{2}}}^2}{r}\rd \Omega\\ &=\frac{\p}{\p t}\sez{\frac{t}{4\pi} \iint_{S_1(0)} (y+2\sqrt{2}t v)^2\rd S} +\frac{t}{4\pi}\iint_{S_1(0)} (z+2\sqrt{2}t w)^2\rd S\\ &\quad +\frac{1}{4\pi \cdot 8}\iiint_{K_1(0)} \frac{(x+2\sqrt{2}t u)^2\sex{t-\sqrt{u^2+v^2+w^2}t}^2}{2\sqrt{2}t\sqrt{u^2+v^2+w^2}}\cdot (2\sqrt{2}t)^3\rd \Omega\\ &=\frac{\p}{\p t}\sez{\frac{t}{4\pi}\sex{y^2\cdot 4\pi+8t^2\cdot \frac{4\pi}{3}}} +\frac{t}{4\pi}\sex{z^2\cdot4\pi +8t^2\cdot \frac{4\pi}{3}}\\ &\quad +\frac{t^4}{4\pi}\int_0^1 \sex{x^2+\frac{8t^2}{3}\rho^2} \frac{(1-\rho)^2}{\rho} 4\pi \rho^2\rd \rho\\ &=8t^2+\frac{8t^3}{3}+\frac{2t^6}{45}+\frac{x^2t^4}{12} +y^2+z^2t. \eea \eeex$$

 

(3). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=\cos x\cdot \sin y e^z,\\ u|_{t=0}=x^2e^{y+z},\ u_t|_{t=0}=\sin x\cdot e^{y+z}. \ea}}$

 

解答: 懒得做了...

 

(4). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=xe^t\cos (3y+4z),\\ u|_{t=0}=xy \cos z,\ u_t|_{t=0}=e^x yz. \ea}}$

 

解答: 懒得做了...

 

(5). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=(x^2+y^2+z^2)^2e^t,\\ u|_{t=0}=u_t|_{t=0}=0. \ea}}$

 

解答: 懒得做了...

 

(6). $\dps{\sedd{\ba{ll} u_{tt}-(u_{xx}+u_{yy})=t\sin y,\\ u|_{t=0}=x^2,\ u_t|_{t=0}=\sin y. \ea}}$

 

解答: 懒得做了...

 

(7). $\dps{\sedd{\ba{ll} u_{tt}-3(u_{xx}+u_{yy})=x^3+y^3,\\ u|_{t=0}=u_t|_{t=0}=x^2. \ea}}$

 

解答: 懒得做了...

 

(8). $\dps{\sedd{\ba{ll} u_{tt}-(u_{xx}+u_{yy})=e^{x+2y},\\ u|_{t=0}=u_t|_{t=0}=e^{x+2y}. \ea}}$

 

解答: 懒得做了...

 

(9). $\dps{\sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy})=(x^2+y^2)^2e^t,\\ u|_{t=0}= u_t|_{t=0}=0. \ea}}$

 

解答: 懒得做了...

 

4. 求解定解问题 (3.26) 及 (3.27).

 

解答: 就是利用波的反射原理求解, 进行奇延拓或偶延拓啦.

 

5. 设 $\varphi(x,y,z)\in C^2(\bbR^3)$, 且 $$\bex \vlm{r}\frac{\varphi(x,y,z)}{r^{\al-1}}=A, \eex$$ 其中 $r=\sqrt{x^2+y^2+z^2}$, $\al$ 为一常数. 又 $u(x,y,z,t)$ 为 Cauchy 问题: $$\bex \sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=0,&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=0,\quad u_t|_{t=0}=\varphi(x,y,z) \ea} \eex$$ 的解. 试证: $$\bex \vlm{t}\frac{u(x,y,z,t)}{t^\al}=C, \eex$$ 并计算常数 $C$ 之值.

 

证明: 对任意固定的 $(x,y,z)$, 记 $$\bex \tilde r=\sqrt{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2}, \eex$$ 则 $$\bex \vlm{r}\frac{\varphi(\xi,\eta,\zeta)}{\tilde r^{\al-1}} =\vlm{r} \frac{\varphi(\xi,\eta,\zeta)}{(\xi^2+\eta^2+\zeta^2)^\frac{\al-1}{2}} \cdot\sez{ \frac{\xi^2+\eta^2+\zeta^2}{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2}}^\frac{\al-1}{2}=A, \eex$$ 其中最后一步是因为 $$\bex P\ (\xi,\eta,\zeta),\ Q\ (x,y,z)\ra \sedd{\ba{ll} |OP|\leq |OQ|+\tilde r\ra \frac{|OP|}{\tilde r}\leq \frac{|OQ|}{\tilde r}+1,\\ |OP|\geq \tilde r-|OQ|\ra \frac{|OP|}{\tilde r}\geq 1-\frac{|OQ|}{\tilde r}. \ea} \eex$$ 于是 $$\bex \exists\ T>0,\st t>T,\ \tilde r=at\ra \sev{\frac{\varphi(\xi,\eta,\zeta)}{\tilde r^{\al-1}}-A}<\ve, \eex$$ $$\bex \frac{u(x,y,z,t)}{t^\al} =\frac{1}{4\pi a^2t^{\al+1}}\iint_{S_{at}(M)}\varphi(\xi,\eta,\zeta)\rd S =a^{\al-1} \frac{1}{4\pi a^2t^2}\iint_{S_{at}(M)}\frac{\varphi(\xi,\eta,\zeta)}{\tilde r^{\al-1}}\rd S, \eex$$ $$\bex t>T\ra (A-\ve)a^{\al-1}\leq \frac{u(x,y,z,t)}{t^\al}\leq (A+\ve)a^{\al-1}. \eex$$ 令 $t\to\infty$ 有 $$\bex (A-\ve)a^{\al-1}\leq \vli{t}\frac{u(x,y,z,t)}{t^\al} \leq \vls{t}\frac{u(x,y,z,t)}{t^\al}\leq (A+\ve)a^{\al-1}. \eex$$ 再令 $\ve\to0$ 有 $$\bex \vlm{t}\frac{u(x,y,z,t)}{t^\al}=Aa^{\al-1}. \eex$$

 

6. 设 $\varphi(x,y,z)\in C^3(\bbR^3),\ \psi(x,y,z)\in C^2(\bbR^3)$ 且具有紧支集 (即在一个有界区域外为零), $u(x,y,z,t)$ 为 Cauchy 问题: $$\bex \sedd{\ba{ll} u_{tt}-a^2(u_{xx}+u_{yy}+u_{zz})=0,&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=\varphi(x,y,z),\ u_t|_{t=0}=\psi(x,y,z) \ea} \eex$$ 的解, 试证明存在常数 $C$, 使得 $$\bex |u(x,y,z,t)|\leq \frac{C}{t},\quad \forall\ (x,y,z)\in\bbR^3. \eex$$

 

证明: 由 $$\beex \bea u(x,y,z,t)&=\frac{\p}{\p t}\sez{\frac{1}{4\pi a^2t}\iint_{S_{at}(M)} \varphi\rd S} +\frac{1}{4\pi a^2t} \iint_{S_{at}(M)} \psi\rd S\\ &=\frac{\p}{\p t} \sez{\frac{t}{4\pi} \iint_{S_1(0)} \varphi(x+atu, y+atv,z+at w)\rd S}\\ &\quad +\frac{t}{4\pi}\iint_{S_1(0)} \psi(x+atu, y+atv, z+at w)\rd S\\ &=\frac{1}{4\pi} \iint_{S_1(0)}\varphi(\cdot )\rd S +\frac{t}{4\pi}\iint_{S_1(0)}a\frac{\p \varphi}{\p n}(\cdot)\rd S +\frac{t}{4\pi}\iint_{S_1(0)}\psi(\cdot)\rd S\\ &=\frac{1}{4\pi a^2t^2}\iint_{S_{at}(M)} \varphi+at\frac{\p \varphi}{\p n} +t \psi\rd S\\ &=\frac{1}{4\pi a^2t^2}\iint_{S_{at}(M)\cap K} \varphi+at\frac{\p \varphi}{\p n} +t \psi\rd S\quad\sex{K=\supp(\varphi,\psi)} \eea \eeex$$ 知 $$\bex |u(x,y,z,t)|\leq \frac{1}{4\pi a^2t^2}(C_1t+C_2)\cdot |K|\leq \frac{C}{t}. \eex$$

 

7. 证明球面波问题 $$\bex \sedd{\ba{ll} u_{tt}=a^2(u_{xx}+u_{yy}+u_{zz}),&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=\varphi(r),\ u_t|_{t=0}=\psi(r) \ea} \eex$$ 的解为 $$\bex u(r,t)=\frac{(r-at)\varphi(r-at)+(r+at)\varphi(r+at)}{2r} +\frac{1}{2ar}\int_{r-at}^{r+at}\rho\psi(\rho)\rd \rho, \eex$$ 其中 $r=\sqrt{x^2+y^2+z^2}$.

 

证明: 由定理 4.1, 仅需证明 $$\bex \sedd{\ba{ll} u_{tt}=a^2(u_{xx}+u_{yy}+u_{zz}),&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=0,\ u_t|_{t=0}=\psi(r) \ea} \eex$$ 的解为 $$\bex u(r,t)=\frac{1}{2ar}\int_{r-at}^{r+at}\rho\psi(\rho)\rd \rho. \eex$$ 这可直接由 Poisson 公式得到: $$\beex \bea u&=\frac{1}{4\pi a^2t}\iint_{S_{at}(M)} \psi(\sqrt{\xi^2+\eta^2+\zeta^2})\rd S\\ &=\frac{t}{4\pi}\iiint_{S_1(0)} \psi(\sqrt{(x+at u)^2+(y+at v)^2+(z+at w)^2})\rd S\\ &=\frac{t}{4\pi}\iint_{S_1(0)} \psi(\sqrt{r^2+a^2t^2+2at(xu+yv+zw)})\rd S\\ &=\frac{t}{4\pi}\iint_{S_1(0)} \psi(\sqrt{r^2+a^2t^2+2at W})\rd S\\ &\quad\sex{\mbox{正交变换 }U=\cdots,\ V=\cdots,\ W=\frac{xu+yv+zw}{r}}\\ &=\frac{t}{4\pi}\int_{-1}^1 \psi(\sqrt{r^2+a^2t^2+2atr W})\cdot 2\pi \sqrt{1-W^2}\cdot \sqrt{1+\sex{\frac{\rd }{\rd W}\sqrt{1-W^2}}^2}\rd W\\ &=\frac{t}{2}\int_{-1}^1 \psi(\sqrt{r^2+a^2t^2+2atrW}) \rd W\\ &=\frac{t}{2}\int_{r-at}^{r+at} \psi(\rho)\frac{\rho}{atr}\rd \rho\\ &\quad\sex{\rho=\sqrt{r^2+a^2t^2+2atr W}\ra \frac{\rd \rho}{\rd W}=\frac{atr}{\rho}}\\ &=\frac{1}{2ar }\int_{r-at}^{r+at}\rho\psi(\rho)\rd \rho. \eea \eeex$$

 

8. 求解下面 Cauchy 问题: $$\bex \sedd{\ba{ll} u_{tt}=a^2(u_{xx}+u_{yy}+u_{zz}),&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=\varphi(\sqrt{x^2+y^2+z^2}),\ u_t|_{t=0}=\psi(x+y+z),&(x,y,z)\in\bbR^3. \ea} \eex$$

 

解答: 由第 7 题, $$\bex \sedd{\ba{ll} u_{tt}=a^2(u_{xx}+u_{yy}+u_{zz}),&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=\varphi(\sqrt{x^2+y^2+z^2}),\ u_t|_{t=0}=0,&(x,y,z)\in\bbR^3 \ea} \eex$$ 的解为 $$\bex u_1=\frac{(r-at)\varphi(r-at)+(r+at)\varphi(r+at)}{2r}, \eex$$ 其中 $r=\sqrt{x^2+y^2+z^2}$. 往求 $$\bex \sedd{\ba{ll} u_{tt}=a^2(u_{xx}+u_{yy}+u_{zz}),&(x,y,z)\in\bbR^3,\ t>0,\\ u|_{t=0}=0,\ u_t|_{t=0}=\psi(x+y+z),&(x,y,z)\in\bbR^3. \ea} \eex$$ 的解 $$\beex \bea u_2&=\frac{1}{4\pi a^2t}\iint_{S_{at}(M)} \psi(\xi+\eta+\zeta)\rd S =\frac{t}{4\pi}\iint_{S_1(0)} \psi(x+at u+y+atv +z+at w)\rd S\\ &=\frac{t}{4\pi}\iint_{S_1(0)}\psi(x+y+z+a+\sqrt{3}W)\rd S\\ &\quad\sex{\mbox{正交变换 }U=\cdots,\ V=\cdots,\ W=\frac{u+v+w}{\sqrt{3}}}\\ &=\frac{t}{4\pi}\int_{-1}^1 \psi(x+y+z+\sqrt{3}at W) \cdot 2\pi \sqrt{1-W^2} \cdot \sqrt{1+\sex{\frac{\rd }{\rd W}\sqrt{1-W^2}}^2} \rd W\\ &=\frac{t}{2}\int_{-1}^1 \psi(x+y+z+\sqrt{3}at W)\rd W =\frac{t}{2}\int_{x+y+z-\sqrt{3}at}^{x+y+z+\sqrt{3}at}\psi(\rho)\frac{1}{\sqrt{3}at}\rd \rho\\ &=\frac{1}{2\sqrt{3}a}\int_{x+y+z-\sqrt{3}at}^{x+y+z+\sqrt{3}at}\psi(\rho)\rd\rho. \eea \eeex$$ 由叠加原理, 原 Cauchy 问题的解为 $$\beex \bea u&=u_1+u_2\\ &=\frac{(r-at)\varphi(r-at)+(r+at)\varphi(r+at)}{2r}+\frac{1}{2\sqrt{3}a}\int_{x+y+z-\sqrt{3}at}^{x+y+z+\sqrt{3}at}\psi(\rho)\rd\rho, \eea \eeex$$ 其中 $r=\sqrt{x^2+y^2+z^2}$.