5.2高维波动方程习题参考解答

zhangzujin

# 高维波动方程习题参考解答 --- 1、 求解下列三维齐次波动方程的 Cauchy 问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=0, (x,y,z)\in\mathbb\{R\}^3, t > 0,\\\\ u|\_\{t=0\}=yz, u\_t|\_\{t=0\}=zx;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Poisson 公式, \begin\{aligned\} u(x,y,z,t)=&\frac\{\partial\}\{\partial t\}\left\[\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} \eta\zeta\mathrm\{ d\} S\right\] +\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} \zeta \xi\mathrm\{ d\} S\\\\ \equiv&I\_1+I\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \xi-x=at u, \eta-y=atv, \zeta-z=atw$, 则 \begin\{aligned\} I\_2&=\frac\{1\}\{4\pi a^2t\}\iint\_\{u^2+v^2+w^2=1\} (z+atw)(x+atu)\cdot (at)^2\mathrm\{ d\} S\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{4\pi a^2t\}\cdot zx (at)^2\cdot 4\pi =zxt. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同理, $\displaystyle I\_1=\frac\{\partial\}\{\partial t\}(yzt)=yz$. 故 \begin\{aligned\} u(x,y,z,t)=yz+zxt. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=0, (x,y,z)\in\mathbb\{R\}^3, t > 0,\\\\ u|\_\{t=0\}=f(x)+g(y), u\_t|\_\{t=0\}=\varphi(x)+\psi(z);\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Poisson 公式, \begin\{aligned\} u(x,y,z,t)=&\frac\{\partial\}\{\partial t\}\left\\{\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} \left\[f(\xi)+g(\eta)\right\]\mathrm\{ d\} S\right\\}\\\\ &\quad +\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} \left\[\varphi(\xi)+\psi(\zeta)\right\]\mathrm\{ d\} S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 把 $\displaystyle S\_\{at\}(M)$ 看成 $\displaystyle (\xi-x)^2+(\zeta-z)^2=(at)^2$ 绕着 $\displaystyle \zeta=z$ 旋转而得的旋转曲面, 则由 \begin\{aligned\} \sqrt\{(\mathrm\{ d\} \xi)^2+(\mathrm\{ d\} \zeta)^2\} =&\sqrt\{1+\left\[\frac\{\zeta-z\}\{\sqrt\{(at)^2-(\zeta-z)^2\}\}\right\]^2\}\mathrm\{ d\} \zeta\\\\ =&\frac\{at\}\{\sqrt\{(at)^2-(\zeta-z)^2\}\}\mathrm\{ d\} \zeta \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} \iint\_\{S\_\{at\}(M)\}\psi(\zeta)\mathrm\{ d\} S =&\int\_\{z-at\}^\{z+at\}\psi(\zeta)\cdot 2\pi \sqrt\{(at)^2-(\zeta-z)^2\}\cdot \frac\{at\mathrm\{ d\} \zeta\}\{\sqrt\{(at)^2-(\zeta-z)^2\}\}\\\\ =&2\pi at \int\_\{z-at\}^\{z+at\} \psi(\zeta)\mathrm\{ d\} \zeta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} u(x,y,z,t)=&\frac\{1\}\{2a\}\left\[\frac\{\partial\}\{\partial t\} \int\_\{x-at\}^\{x+at\}f(\xi)\mathrm\{ d\} \xi +\frac\{\partial\}\{\partial t\}\int\_\{y-at\}^\{y+at\}g(\eta)\mathrm\{ d\} \eta\right\]\\\\ &+\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\varphi(\xi)\mathrm\{ d\} \xi +\frac\{1\}\{2a\}\int\_\{z-at\}^\{z+at\}\psi(\zeta)\mathrm\{ d\} \zeta\\\\ =&\frac\{f(x+at)+f(x-at)+g(y+at)+g(y-at)\}\{2\}\\\\ &+\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\varphi(\xi)\mathrm\{ d\} \xi +\frac\{1\}\{2a\}\int\_\{z-at\}^\{z+at\}\psi(\zeta)\mathrm\{ d\} \zeta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=0, (x,y,z)\in\mathbb\{R\}^3, t > 0,\\\\ u|\_\{t=0\}=u\_t|\_\{t=0\}=\mathrm\{e\}^\{x^2+y^2+z^2\};\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 6 知取 \begin\{aligned\} \psi(r)=\mathrm\{e\}^\{r^2\}, r=\sqrt\{x^2+y^2+z^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 后, \begin\{aligned\} u(x,y,z,t)=\frac\{1\}\{2ar\}\int\_\{r-at\}^\{r+at\}\rho \mathrm\{e\}^\{\rho^2\}\mathrm\{ d\} \rho =\frac\{\mathrm\{e\}^\{(r+at)^2\}-\mathrm\{e\}^\{(r-at)^2\}\}\{4ar\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 2、 设 $\displaystyle \varphi(x,y,z)\in C^2(\mathbb\{R\}^3)$, 且 \begin\{aligned\} \lim\_\{r\to\infty\}\frac\{\varphi(x,y,z)\}\{r^\{\alpha-1\}\}=A, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle r=\sqrt\{x^2+y^2+z^2\}$, $\displaystyle \alpha$ 为一常数. 又 $\displaystyle u(x,y,z,t)$ 为 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=0,&(x,y,z)\in\mathbb\{R\}^3,\ t > 0,\\\\ u|\_\{t=0\}=0,\quad u\_t|\_\{t=0\}=\varphi(x,y,z) \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解. 试证: \begin\{aligned\} \lim\_\{t\to\infty\}\frac\{u(x,y,z,t)\}\{t^\alpha\}=C, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 并计算常数 $\displaystyle C$ 之值. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 对任意固定的 $\displaystyle (x,y,z)$, 记 \begin\{aligned\} \tilde\{r\}=\sqrt\{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 \begin\{aligned\} &\lim\_\{r\to\infty\}\frac\{\varphi(\xi,\eta,\zeta)\}\{\tilde\{r\}^\{\alpha-1\}\}\\\\ =&\lim\_\{r\to\infty\} \frac\{\varphi(\xi,\eta,\zeta)\}\{(\xi^2+\eta^2+\zeta^2)^\frac\{\alpha-1\}\{2\}\} \cdot\left\[ \frac\{\xi^2+\eta^2+\zeta^2\}\{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2\}\right\]^\frac\{\alpha-1\}\{2\}=A, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中最后一步是因为 \begin\{aligned\} P\ (\xi,\eta,\zeta),\ Q\ (x,y,z)\Rightarrow \left\\{\begin\{array\}\{ll\} |OP|\leq |OQ|+\tilde\{r\}\Rightarrow \frac\{|OP|\}\{\tilde\{r\}\}\leq \frac\{|OQ|\}\{\tilde\{r\}\}+1,\\\\ |OP|\geq \tilde\{r\}-|OQ|\Rightarrow \frac\{|OP|\}\{\tilde\{r\}\}\geq 1-\frac\{|OQ|\}\{\tilde\{r\}\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} \exists\ T > 0,\mathrm\{ s.t.\} t > T,\ \tilde\{r\}=at\Rightarrow \left|\frac\{\varphi(\xi,\eta,\zeta)\}\{\tilde\{r\}^\{\alpha-1\}\}-A\right| < \varepsilon, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} \frac\{u(x,y,z,t)\}\{t^\alpha\} =&\frac\{1\}\{4\pi a^2t^\{\alpha+1\}\}\iint\_\{S\_\{at\}(M)\}\varphi(\xi,\eta,\zeta)\mathrm\{ d\} S\\\\ =&a^\{\alpha-1\} \frac\{1\}\{4\pi a^2t^2\}\iint\_\{S\_\{at\}(M)\}\frac\{\varphi(\xi,\eta,\zeta)\}\{\tilde\{r\}^\{\alpha-1\}\}\mathrm\{ d\} S, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} t > T\Rightarrow (A-\varepsilon)a^\{\alpha-1\}\leq \frac\{u(x,y,z,t)\}\{t^\alpha\}\leq (A+\varepsilon)a^\{\alpha-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle t\to\infty$ 有 \begin\{aligned\} (A-\varepsilon)a^\{\alpha-1\}\leq \varliminf\_\{t\to\infty\}\frac\{u(x,y,z,t)\}\{t^\alpha\} \leq \varlimsup\_\{t\to\infty\}\frac\{u(x,y,z,t)\}\{t^\alpha\}\leq (A+\varepsilon)a^\{\alpha-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再令 $\displaystyle \varepsilon\to 0^+$ 有 \begin\{aligned\} \lim\_\{t\to\infty\}\frac\{u(x,y,z,t)\}\{t^\alpha\}=Aa^\{\alpha-1\}\equiv C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 3、 证明球面波问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}=a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\}),&(x,y,z)\in\mathbb\{R\}^3,\ t > 0,\\\\ u|\_\{t=0\}=\varphi(r),\ u\_t|\_\{t=0\}=\psi(r) \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解为 \begin\{aligned\} 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)\mathrm\{ d\} \rho, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle r=\sqrt\{x^2+y^2+z^2\}$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Duhamel 原理知仅需证明 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}=a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\}),&(x,y,z)\in\mathbb\{R\}^3,\ t > 0,\\\\ u|\_\{t=0\}=0,\ u\_t|\_\{t=0\}=\psi(r) \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解为 \begin\{aligned\} u(r,t)=\frac\{1\}\{2ar\}\int\_\{r-at\}^\{r+at\}\rho\psi(\rho)\mathrm\{ d\} \rho. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这可直接由 Poisson 公式得到: \begin\{aligned\} u&=\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} \psi(\sqrt\{\xi^2+\eta^2+\zeta^2\})\mathrm\{ d\} S\\\\ &=\frac\{t\}\{4\pi\}\iiint\_\{S\_1(0)\} \psi(\sqrt\{(x+at u)^2+(y+at v)^2+(z+at w)^2\})\mathrm\{ d\} S\\\\ &=\frac\{t\}\{4\pi\}\iint\_\{S\_1(0)\} \psi(\sqrt\{r^2+a^2t^2+2at(xu+yv+zw)\})\mathrm\{ d\} S\\\\ &=\frac\{t\}\{4\pi\}\iint\_\{S\_1(0)\} \psi(\sqrt\{r^2+a^2t^2+2atr W\})\mathrm\{ d\} S\ \left(\begin\{array\}\{c\}\mbox\{正交变换 \}U=\cdots,\ V=\cdots,\\\\ W=\frac\{xu+yv+zw\}\{r\}\end\{array\}\right)\\\\ &=\frac\{t\}\{4\pi\}\int\_\{-1\}^1 \psi(\sqrt\{r^2+a^2t^2+2atr W\})\cdot 2\pi \sqrt\{1-W^2\}\\\\ &\quad \cdot \sqrt\{1+\left(\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} W\}\sqrt\{1-W^2\}\right)^2\}\mathrm\{ d\} W\\\\ &=\frac\{t\}\{2\}\int\_\{-1\}^1 \psi(\sqrt\{r^2+a^2t^2+2atrW\}) \mathrm\{ d\} W =\frac\{t\}\{2\}\int\_\{r-at\}^\{r+at\} \psi(\rho)\frac\{\rho\}\{atr\}\mathrm\{ d\} \rho\\\\ &\quad \left(\rho=\sqrt\{r^2+a^2t^2+2atr W\}\Rightarrow \frac\{\mathrm\{ d\} \rho\}\{\mathrm\{ d\} W\}=\frac\{atr\}\{\rho\}\right)\\\\ &=\frac\{1\}\{2ar \}\int\_\{r-at\}^\{r+at\}\rho\psi(\rho)\mathrm\{ d\} \rho. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 4、 求解 Cauchy 问题: \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}=a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\}),&(x,y,z)\in\mathbb\{R\}^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\mathbb\{R\}^3. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由第 3 题, \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}=a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\}),&(x,y,z)\in\mathbb\{R\}^3,\ t > 0,\\\\ u|\_\{t=0\}=\varphi(\sqrt\{x^2+y^2+z^2\}),\ u\_t|\_\{t=0\}=0,&(x,y,z)\in\mathbb\{R\}^3 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解为 \begin\{aligned\} u\_1=\frac\{(r-at)\varphi(r-at)+(r+at)\varphi(r+at)\}\{2r\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle r=\sqrt\{x^2+y^2+z^2\}$. 往求 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}=a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\}),&(x,y,z)\in\mathbb\{R\}^3,\ t > 0,\\\\ u|\_\{t=0\}=0,\ u\_t|\_\{t=0\}=\psi(x+y+z),&(x,y,z)\in\mathbb\{R\}^3. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解 \begin\{aligned\} u\_2&=\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} \psi(\xi+\eta+\zeta)\mathrm\{ d\} S\\\\ &=\frac\{t\}\{4\pi\}\iint\_\{S\_1(0)\} \psi(x+at u+y+atv +z+at w)\mathrm\{ d\} S\\\\ &=\frac\{t\}\{4\pi\}\iint\_\{S\_1(0)\}\psi(x+y+z+a+\sqrt\{3\}W)\mathrm\{ d\} S\\\\ &\quad \left(\mbox\{正交变换 \}U=\cdots,\ V=\cdots,\ W=\frac\{u+v+w\}\{\sqrt\{3\}\}\right)\\\\ &=\frac\{t\}\{4\pi\}\int\_\{-1\}^1 \psi(x+y+z+\sqrt\{3\}at W) \cdot 2\pi \sqrt\{1-W^2\} \\\\ &\quad \cdot \sqrt\{1+\left(\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} W\}\sqrt\{1-W^2\}\right)^2\} \mathrm\{ d\} W\\\\ &=\frac\{t\}\{2\}\int\_\{-1\}^1 \psi(x+y+z+\sqrt\{3\}at W)\mathrm\{ d\} W =\frac\{t\}\{2\}\int\_\{x+y+z-\sqrt\{3\}at\}^\{x+y+z+\sqrt\{3\}at\}\psi(\rho)\frac\{1\}\{\sqrt\{3\}at\}\mathrm\{ d\} \rho\\\\ &=\frac\{1\}\{2\sqrt\{3\}a\}\int\_\{x+y+z-\sqrt\{3\}at\}^\{x+y+z+\sqrt\{3\}at\}\psi(\rho)\mathrm\{ d\}\rho. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由叠加原理, 原 Cauchy 问题的解为 \begin\{aligned\} 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)\mathrm\{ d\}\rho, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle r=\sqrt\{x^2+y^2+z^2\}$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 5、 设 $\displaystyle u(r,t)$ 是三维波动方程初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=0, (x,y,z)\in\mathbb\{R\}^3, t > 0,\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=(1+4r^2)^\{-\frac\{1\}\{2\}\}, (x,y,z)\in\mathbb\{R\}^3\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解, 其中 $\displaystyle r=\sqrt\{x^2+y^2+z^2\}$, 求 $\displaystyle \lim\_\{t\to\infty\}u(0,t)$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 3 知 \begin\{aligned\} u(r,t)=&\frac\{1\}\{2r\}\int\_\{r-t\}^\{r+t\}\rho(1+4\rho^2)^\{-\frac\{1\}\{2\}\}\mathrm\{ d\} \rho\\\\ =&\frac\{1\}\{2r\}\frac\{\sqrt\{1+4(r+t)^2\}-\sqrt\{1+4(r-t)^2\}\}\{4\}\\\\ =&\frac\{16rt\}\{8r\left\[\sqrt\{1+4(r+t)^2\}+\sqrt\{1+4(r-t)^2\}\right\]\}\\\\ =&\frac\{2t\}\{\sqrt\{1+4(r+t)^2\}+\sqrt\{1+4(r-t)^2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} \lim\_\{t\to\infty\}u(0,t)=\lim\_\{t\to\infty\}\frac\{t\}\{\sqrt\{1+4t^2\}\}=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 6、 求解下列二维齐次波动方程的 Cauchy 问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\})=0, (x,y)\in\mathbb\{R\}^2, t > 0,\\\\ u|\_\{t=0\}=2x^2-y^2, u\_t|\_\{t=0\}=2x^2+y^2;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Poisson 公式知 \begin\{aligned\} u(x,t)=&\frac\{\partial\}\{\partial t\}\left\[\frac\{1\}\{2\pi a\}\iint\_\{\varSigma\_\{at\}(M)\} \frac\{2\xi^2-\eta^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi \mathrm\{ d\} \eta\right\]\\\\ &+\frac\{1\}\{2\pi a\}\iint\_\{\varSigma\_\{at\}(M)\} \frac\{2\xi^2+\eta^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi \mathrm\{ d\} \eta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 \begin\{aligned\} I&=\iint\_\{\varSigma\_\{at\}(M)\} \frac\{(\xi-x)^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \iint\_\{\varSigma\_\{at\}(M)\} \frac\{(\eta-y)^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ &=\frac\{1\}\{2\}\iint\_\{\varSigma\_\{at\}(M)\} \frac\{(\xi-x)^2+(\eta-y)^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ &=\frac\{1\}\{2\}\int\_0^1 \frac\{(at)^2r^2\}\{\sqrt\{(at)^2-(atr)^2\}\}\cdot (at)^22\pi r\mathrm\{ d\} r =\pi(at)^3\int\_0^1 \frac\{r^3\}\{\sqrt\{1-r^2\}\}\mathrm\{ d\} r\\\\ &\stackrel\{r=\sin\theta\}\{=\}\cdots=\frac\{2\pi\}\{3\}(at)^3, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} J&=\iint\_\{\varSigma\_\{at\}(M)\} \frac\{1\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ &=\int\_0^1 \frac\{1\}\{at\sqrt\{1-r^2\}\}\cdot (at)^22\pi r\mathrm\{ d\} r =2\pi at, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 \begin\{aligned\} &\iint\_\{\varSigma\_\{at\}(M)\} \frac\{\xi^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ =&\iint\_\{\varSigma\_\{at\}(M)\} \frac\{(\xi-x)^2+2(\xi-x)x+x^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&I+x^2J=\frac\{2\pi\}\{3\}(at)^3+2\pi x^2at,\\\\ &\iint\_\{\varSigma\_\{at\}(M)\} \frac\{\eta^2\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta =\frac\{2\pi\}\{3\}(at)^3+2\pi y^2at. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} u(x,t)=&\frac\{\partial\}\{\partial t\}\left\\{\frac\{1\}\{2\pi a\} \left\[\frac\{2\pi\}\{3\}(at)^3+2\pi(2x^2-y^2)at\right\]\right\\}\\\\ &+\frac\{1\}\{2\pi a\}\left\[2\pi(at)^3+2\pi(2x^2+y^2)at\right\]\\\\ =&a^2t^2+2x^2-y^2+a^2t^3+(2x^2+y^2)t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\})=0, (x,y)\in\mathbb\{R\}^2, t > 0,\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=2xy.\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} u(x,y,t)=&\frac\{1\}\{2\pi a\}\iint\_\{\varSigma\_\{at\}(M)\} \frac\{2\xi\eta\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ =&\frac\{1\}\{\pi a\}\iint\_\{\varSigma\_\{at\}(M)\} \frac\{(\xi-x)(\eta-y)+(\xi-x)y+x(\eta-y)+xy\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\frac\{1\}\{\pi a\}\iint\_\{\varSigma\_\{at\}(M)\} \frac\{xy\}\{\sqrt\{(at)^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ =&\frac\{xy\}\{\pi a\}\int\_0^1 \frac\{1\}\{at\sqrt\{1-r^2\}\}\cdot (at)^2\cdot 2\pi r\mathrm\{ d\} r=2xyt. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 7、 试用降维法导出弦振动方程的 D'Alembert 公式. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由定理 Duhamel 原理, 只需求出 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}-a^2u\_\{xx\}=0,\\\\ u|\_\{t=0\}=0,\ u\_t|\_\{t=0\}=\psi(x) \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解 $\displaystyle u$. 将其看成二维弦振动方程, 由 Poisson 公式, \begin\{aligned\} u&=\frac\{1\}\{4\pi a^2t\}\cdot 2\iint\_\{\varSigma\_\{at\}(M)\} \psi(\xi)\frac\{at\}\{\sqrt\{a^2t^2-(\xi-x)^2-(\eta-y)^2\}\}\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ &=\frac\{1\}\{2\pi a\}\int\_\{x-at\}^\{x+at\} \psi(\xi)\mathrm\{ d\} \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\}\}\mathrm\{ d\} \eta\\\\ &=\frac\{1\}\{2\pi a\}\int\_\{x-at\}^\{x+at\} \psi(\xi)\cdot \pi \mathrm\{ d\} \xi\quad \left(\eta=y+\sqrt\{a^2t^2-(\xi-x)^2\}\sin\theta\right)\\\\ &=\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\xi)\mathrm\{ d\} \xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 8、 写出 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-4(u\_\{xx\}+u\_\{yy\})=0,&(x,y)\in\mathbb\{R\}^2, t > 0,\\\\ u|\_\{t=0\}=x+y, u\_t|\_\{t=0\}=2xy,&(x,y)\in\mathbb\{R\}^2\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 关于点 $\displaystyle (x,y,t)=(0,0,2)$ 的依赖区域. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \left\\{(x,y)\in\mathbb\{R\}^2; x^2+y^2\leq 2^2\cdot 2^2=16\right\\}$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 9、 求解下列非齐次方程的 Cauchy 问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=2(y-t),\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=x^2+y^2;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} u(x,y,z,t)=&\frac\{1\}\{4\pi a^2t\}\iint\_\{S\_\{at\}(M)\} (\xi^2+\eta^2)\mathrm\{ d\} S\\\\ &+\int\_0^t \frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{ d\} \tau \iint\_\{S\_\{a(t-\tau)\}(M)\} 2(\eta-\tau)\mathrm\{ d\} S\\\\ \equiv&I\_1+I\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 \begin\{aligned\} I\_1&=\frac\{1\}\{4\pi a^2t\}\iint\_\{\varSigma\_\{at\}(M)\} \left\[\begin\{array\}\{c\} (\xi-x)^2+2(\xi-x)x+x^2\\\\ (\eta-y)^2+2(\eta-y)y+y^2\end\{array\}\right\]\mathrm\{ d\} S\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{4\pi a^2t\} \left\\{\frac\{2\}\{3\} \iint\_\{\varSigma\_\{at\}(M)\} \left\[(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2\right\]\mathrm\{ d\} S\right\\}\\\\ &\quad +\frac\{1\}\{4\pi a^2t\}(x^2+y^2)\iint\_\{\varSigma\_\{at\}(M)\}\mathrm\{ d\} S\\\\ &=\frac\{1\}\{4\pi a^2t\}\cdot\frac\{2\}\{3\}(at)^2\cdot 4\pi(at)^2 +(x^2+y^2)t\\\\ &=\frac\{2\}\{3\}a^2t^3+(x^2+y^2)t, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} I\_2&=\int\_0^t \frac\{1\}\{4\pi a^2(t-\tau)\} \iint\_\{S\_\{a(t-\tau)\}(M)\} \left\[2(\eta-y)+2(y-\tau)\right\]\mathrm\{ d\} \tau\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \int\_0^t \frac\{2(y-\tau)\}\{4\pi a^2(t-\tau)\} \cdot 4\pi a^2(t-\tau)^2\mathrm\{ d\} \tau\\\\ &=\int\_0^t 2(y-\tau)(t-\tau)\mathrm\{ d\} \tau =yt^2-\frac\{t^3\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} u(x,y,z,t)=I\_1+I\_2=\frac\{2\}\{3\}a^2t^3+(x^2+y^2)t+yt^2-\frac\{t^3\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=\cos x\cdot\sin y\mathrm\{e\}^z,\\\\ u|\_\{t=0\}=x^2\mathrm\{e\}^\{y+z\}, u\_t|\_\{t=0\}= \sin x\cdot \mathrm\{e\}^\{y+z\};\end\{array\}\right.$ [能力有限, 算到后面会有超越函数...呜呜, 没有给出解答.] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [能力有限, 算到后面会有超越函数...呜呜, 没有给出解答.] 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\}+u\_\{zz\})=(x^2+y^2+z^2)^2\mathrm\{e\}^t,\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=0;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 为方便记, 将 $\displaystyle (x,y,z)$ 写成 $\displaystyle (x\_1,x\_2,x\_3)$, 则 \begin\{aligned\} u(x\_1,x\_2,x\_3,t)=\int\_0^t \frac\{1\}\{4\pi a^2(t-\tau)\}\iint\_\{S\_\{a(t-\tau)\}(M)\} (\xi\_1^2+\xi\_2^2+\xi\_3^2)^2\mathrm\{e\}^\tau \mathrm\{ d\} S\mathrm\{ d\} \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 \begin\{aligned\} &\iint\_\{S\_\{at\}(M)\}\left\[\left(\sum \xi\_i^2\right)^2-\left(\sum x\_i^2\right)^2\right\]\mathrm\{ d\} S\\\\ =&\iint\_\{S\_1(0)\}\left\\{ \left\[\sum (x\_i+atu\_i)^2\right\]^2 -\left\[\sum x\_i^2\right\]^2\right\\}(at)^2\mathrm\{ d\} S\\\\ =&(at)^2\iint\_\{S\_1(0)\}\sum \left\[(x\_i+atu\_i)^2+x\_i^2\right\] \sum \left\[(x\_i+atu\_i)^2-x\_i^2\right\]\mathrm\{ d\} S\\\\ =&(at)^2\iint\_\{S\_1(0)\}\left\[\sum (2x\_i^2+2atx\_iu\_i+a^2t^2u\_i^2) \sum\_i (2x\_i+atu\_i)atu\_i\right\]\mathrm\{ d\} S\\\\ =&(at)^2\iint\_\{S\_1(0)\} \left(2\sum x\_i^2+a^2t^2 +2at\sum x\_iu\_i\right) \left(2at \sum x\_iu\_i+a^2t^2\right)\mathrm\{ d\} S\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&(at)^2\left\[\left(2\sum x\_i^2+a^2t^2\right)\cdot a^2t^2\cdot 4\pi +4a^2t^2\sum x\_i^2 \frac\{1\}\{3\}\iint\_\{S\_1(0)\}\sum\_j u\_j^2\mathrm\{ d\} S\right\]\\\\ =&\frac\{4\}\{3\}\pi (at)^4\left(3a^2t^2+10\sum x\_i^2\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} &u(x,y,z,t)=\int\_0^t \frac\{1\}\{3\}(t-\tau)\left\\{\begin\{array\}\{c\} \left\[3a^2(t-\tau)^2+x^2+y^2+z^2\right\]\\\\ \left\[a^2(t-\tau)^2+3(x^2+y^2)+z^2\right\] \end\{array\}\right\\}\mathrm\{e\}^\tau \mathrm\{ d\} \tau\\\\ =&(\mathrm\{e\}^t-1-t)(x^2+y^2+z^2)^2\\\\ &+\frac\{10a^2\}\{3\}(6\mathrm\{e\}^t-6-6t-3t^2-t^3)(x^2+y^2+z^2)\\\\ &-a^4\left(t^5+5t^4+20t^3+60t^2 +120t-120\mathrm\{e\}^t+120\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 10、 求解下列二维非齐次波动方程的 Cauchy 问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-(u\_\{xx\}+u\_\{yy\})=t\sin y,&(x,y)\in\mathbb\{R\}^2, t > 0, \\\\ u|\_\{t=0\}=x^2, u\_t|\_\{t=0\}=\sin y,&(x,y)\in\mathbb\{R\}^2;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_\{1,tt\}-u\_\{1,xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ u\_1(x,0)=x^2,u\_\{1,t\}(x,0)=0, &x\in\mathbb\{R\};\end\{array\}\right.\\\\ \left\\{\begin\{array\}\{llllllllllll\}u\_\{2,tt\}-u\_\{2,yy\}=0,&y\in\mathbb\{R\}, t > 0,\\\\ u\_2(y,0)=0, u\_\{2,y\}(y,0)=\sin y,&y\in\mathbb\{R\}.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 D'Alembert 公式知 \begin\{aligned\} u\_1(x,t)=&\frac\{1\}\{2\}[(x+t)^2+(x-t)^2]=x^2+t^2,\\\\ u\_2(y,t)=&\frac\{1\}\{2\}\int\_\{x-t\}^\{x+t\} \sin \xi \mathrm\{ d\} \xi +\frac\{1\}\{2\}\int\_0^t \mathrm\{ d\} \tau \int\_\{y-(t-\tau)\}^\{y+(t-\tau)\} \tau \sin \xi\mathrm\{ d\} \xi=t\sin y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再由叠加原理知 $\displaystyle u(x,y,t)=u\_1(x,t)+u\_2(y,t)=x^2+t^2+t\sin y$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-(u\_\{xx\}+u\_\{yy\})=\mathrm\{e\}^\{x+2y\},&(x,y)\in\mathbb\{R\}^2, t > 0, \\\\ u|\_\{t=0\}=u\_t|\_\{t=0\}=\mathrm\{e\}^\{x+2y\},&(x,y)\in\mathbb\{R\}^2.\end\{array\}\right.$ [能力有限, 算到后面会有超越函数...呜呜, 没有给出解答.] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [能力有限, 算到后面会有超越函数...呜呜, 没有给出解答.] 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/