1.2几个经典方程习题参考解答

zhangzujin

# 几个经典方程习题参考解答 --- 1、 验证函数 \begin\{aligned\} u(x,y,t)=\frac\{1\}\{\sqrt\{a^2t^2-(x-\xi)^2-(y-\eta)^2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在区域 $\displaystyle \varOmega=\left\\{(x,y,t);(x-\xi)^2+(y-\eta)^2 < a^2t^2\right\\}$ 内满足方程 \begin\{aligned\} u\_\{tt\}=a^2(u\_\{xx\}+u\_\{yy\}), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle a$ 为正常数, $\displaystyle \xi,\eta$ 为任意实数. [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 $\displaystyle u=\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{1\}\{2\}\}$ 知 \begin\{aligned\} u\_t=&-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\}a^2\cdot 2t,\\\\ u\_\{tt\}=&\frac\{3\}\{4\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{5\}\{2\}\}a^4\cdot 4t^2\\\\ &-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\}a^2\cdot 2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 及 \begin\{aligned\} u\_x=&-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\} \left\[-2(x-\xi)\right\],\\\\ u\_\{xx\}=&-\frac\{3\}\{4\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{5\}\{2\}\} 4(x-\xi)^2\\\\ &-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\} (-2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由对称性知 \begin\{aligned\} u\_\{yy\}=&-\frac\{3\}\{4\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{5\}\{2\}\} 4(y-\eta)^2\\\\ &-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\}(-2). \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\_\{xx\}+u\_\{yy\} =&-\frac\{3\}\{4\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{5\}\{2\}\} 4\left\[(x-\xi)^2+(y-\eta)^2\right\]\\\\ &-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\} (-4). \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\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\})\\\\ =&\frac\{3\}\{4\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{5\}\{2\}\} 4a^2 \left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]\\\\ &-\frac\{1\}\{2\}\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\} \cdot a^2[2-(-4)]\\\\ =&\left\[a^2t^2-(x-\xi)^2-(y-\eta)^2\right\]^\{-\frac\{3\}\{2\}\} \left\[3a^2-\frac\{1\}\{2\}a^2\cdot 6\right\]=0. \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、 设三维热传导方程具有球对称形式 $\displaystyle u(x,y,z,t)=u(r,t)$ ($r=\sqrt\{x^2+y^2+z^2\}$) 的解, 试证: \begin\{aligned\} u\_t=a^2\left(u\_\{rr\}+\frac\{2u\_r\}\{r\}\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 \begin\{aligned\} u\_x=u\_r\frac\{x\}\{r\},\quad u\_\{xx\}=u\_\{rr\}\frac\{x^2\}\{r^2\} +u\_r\frac\{r-x\frac\{x\}\{r\}\}\{r^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 及对称性知 \begin\{aligned\} \Delta u=a^2\left(u\_\{rr\}+u\_r\frac\{3r-r\}\{r^2\}\right) =a^2\left(u\_\{rr\}+\frac\{2u\_r\}\{r\}\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)/ --- 3、 若 $\displaystyle n$ ($n\geq 2$) 维 Laplace 方程 \begin\{aligned\} \frac\{\partial^2u\}\{\partial x\_1^2\}+\cdots+\frac\{\partial^2u\}\{\partial x\_n^2\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 具有球对称形式的解 $\displaystyle u(x\_1,\cdots,x\_n)=f(r)$, 其中 $\displaystyle r=\sqrt\{x\_1^2+\cdots+x\_n^2\}$, 则 \begin\{aligned\} f(r)=\left\\{\begin\{array\}\{ll\} C\_1+C\_2\dfrac\{1\}\{r^\{n-2\}\},&n\neq 2,\\\\ C\_1+C\_2\ln \dfrac\{1\}\{r\},&n=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\} 其中 $\displaystyle C\_1,C\_2$ 为任意常数. [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 \begin\{aligned\} u\_\{x\_i\}=f'\frac\{x\_i\}\{r\}, u\_\{x\_ix\_i\}=f''\frac\{x\_i^2\}\{r^2\} +f'\frac\{r-x\_i\frac\{x\_i\}\{r\}\}\{r^2\} \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\} \Delta u=f''+(n-1)\frac\{f'\}\{r\}=0. \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\} 0=r^\{n-1\}f''+(n-1)r^\{n-2\} f' =(r^\{n-1\}f')'\Rightarrow C=r^\{n-1\} f'\Rightarrow f'=\frac\{C\}\{r^\{n-1\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle n=2$, 则 \begin\{aligned\} f'=\frac\{C\}\{r\}\Rightarrow f=C\_1+C\_2\ln \frac\{1\}\{r\},\quad C\_2=-C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle n=3$, 则 \begin\{aligned\} f=C\_1+C\_2\frac\{1\}\{r^\{n-2\}\},\quad C\_2=\frac\{C\}\{-n+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)/ --- 4、 有一柔软的均匀细线, 在阻尼介质中作微小横振动, 单位长度弦受的阻力 $\displaystyle F=-Ru\_t$ (其中负号表示阻力方向与振动方向相反). 试推导其振动方程. [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\} \rho u\_\{tt\}=Tu\_\{xx\}-Ru\_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)/ --- 5、 设包围在圆柱形管道中的理想气体发生微小的纵向振动, 并设在振动过程中气体总是沿着圆柱体的轴向运动, 且气体的速度、密度、压力在同一横截面上的各点都是相同的, 试求管中气体的纵向运动方程. [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) / 设圆柱体的轴向为 $\displaystyle x$, 设在 $\displaystyle t$ 时刻 $\displaystyle x$ 处气体的速度为 $\displaystyle u(x,t)$, 密度为 $\displaystyle \rho(x,t)$. 考虑 $\displaystyle [x,x+\Delta x]$ 段在时间 $\displaystyle [t,t+\Delta t]$ 内质量的变化. 流入的净质量为 \begin\{aligned\} -\rho(x+\Delta x,t)u(x+\Delta x)\Delta t +\rho(x,t)u(x,t)\Delta t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 因密度变化而增加的质量为 \begin\{aligned\} \left\[\rho(x,t+\Delta t)-\rho(x,t)\right\]\Delta x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由质量守恒即知 \begin\{aligned\} &\left\[\rho(x,t+\Delta t)-\rho(x,t)\right\]\Delta x\\\\ &=-\rho(x+\Delta x,t)u(x+\Delta x)\Delta t +\rho(x,t)u(x,t)\Delta t\\\\ \Rightarrow&\rho\_t+(\rho u)\_x=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这就是管中气体的纵向运动方程. 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、 一薄膜紧绷在容器口, 考查容器内膜振动所引起的压力变化. 假设容器中的空气在膜振动过程中永远保持为均匀的 (如果小扰动在空气中的传播速度比波在膜上的传播速度大得多的话, 即可作此假定), 试导出膜的微小横振动所满足的方程. [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) / 设容器口为区域 $\displaystyle \varOmega\subset \mathbb\{R\}^2$, 在 $\displaystyle x\in \varOmega$ 处 $\displaystyle t$ 时刻薄膜的位置为 $\displaystyle u(x,y,t)$, 则容器内的体积变化量为 \begin\{aligned\} \iint\_\varOmega u(x,y,t)\mathrm\{ d\} x\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 一般来说, 体积越大, 压力越大 (比如理想气体). 又由于模作微小横振动, 而可设外力为 $\displaystyle \frac\{k\}\{\iint\_\varOmega u(x,y,t)\mathrm\{ d\} x\mathrm\{ d\} y\}$. 故膜的微小横振动所满足的方程为 \begin\{aligned\} u\_\{tt\}-a^2u\_\{xx\}=\frac\{k\}\{\iint\_\varOmega u(x,y,t)\mathrm\{ d\} x\mathrm\{ d\} y\}. \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)/