4.3应用习题参考解答

zhangzujin

# 应用习题参考解答 --- 1、 求解混合问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-16u\_\{xx\}=\frac\{1\}\{2\}(\pi-x)t,&0 < x < \pi, t > 0,\\\\ u|\_\{t=0\}=\pi-x, u\_t|\_\{t=0\}=1,&0\leq x\leq \pi,\\\\ u|\_\{x=0\}=\pi+t, u|\_\{x=\pi\}=2\pi+t,&t\geq 0;\end\{array\}\right.$ [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) / 作辅助函数 \begin\{aligned\} U(x,t)=&(\pi+t)+\frac\{x\}\{\pi\}[(2\pi+t)-(\pi+t)]=\pi+x+t,\\\\ v(x,t)=&u(x,t)-U(x,t), \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\} \left\\{\begin\{array\}\{llllllllllll\} v\_\{tt\}-16v\_\{xx\}=\frac\{(\pi-x)t\}\{2\},&0 < x < \pi, t > 0,\\\\ v|\_\{t=0\}=-2x, v\_t|\_\{t=0\}=0,&0\leq x\leq \pi,\\\\ v|\_\{x=0\}=0, v|\_\{x=\pi\}=0.\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 Duhamel 原理知 \begin\{aligned\} v(x,t)=\int\_0^t w(x,t;\tau)\mathrm\{ d\} \tau+z(x,t), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle w,z$ 分别满足 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}w\_\{tt\}-16w\_\{xx\}=0,&0 < x < \pi, t > \tau,\\\\ w|\_\{t=\tau\}=0, w\_t|\_\{t=\tau\}=\frac\{(\pi-x)\tau\}\{2\},&0\leq x\leq \pi,\\\\ w|\_\{x=0\}=0, w|\_\{x=\pi\}=0,&t\geq \tau;\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\} \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}z\_\{tt\}-16z\_\{xx\}=0,&0 < x < \pi, t\geq 0,\\\\ z|\_\{t=0\}=-2x, z\_t|\_\{t=0\}=0, &0\leq x\leq \pi,\\\\ z|\_\{x=0\}=0, z|\_\{x=\pi\}=0,&t\geq 0.\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\} 用分离变量法, 可设 \begin\{aligned\} w(x,t;\tau)=\sum\_\{k=1\}^\infty \left\[a\_k\cos 4k(t-\tau)+b\_k\sin 4k(t-\tau)\right\]\sin kx. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将初值条件代入知 $\displaystyle a\_k=0$, \begin\{aligned\} &\sum\_\{k=1\}^\infty b\_k4k\sin kx=\frac\{(\pi-x)\tau\}\{2\} \Rightarrow b\_k4k\cdot\frac\{\pi\}\{2\}=\int\_0^\pi \frac\{(\pi-x)\tau\}\{2\}\sin kx\mathrm\{ d\} x\\\\ \Rightarrow&b\_k=\frac\{\tau\}\{4k^2\} \Rightarrow w(x,t;\tau)=\frac\{\tau\}\{4\}\sum\_\{k=1\}^\infty \frac\{\sin 4k(t-\tau)\}\{k^2\}\sin kx. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再者, 容易知道 \begin\{aligned\} z(x,t)=\sum\_\{k=1\}^\infty (a\_k\cos 4kt+b\_k\sin 4kt)\sin kx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 \begin\{aligned\} b\_k=0, -2x=\sum\_\{k=1\}^\infty a\_k\sin kx\Rightarrow a\_k=\frac\{4(-1)^k\}\{k\}. \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\} z(x,t)=&4\sum\_\{k=1\}^\infty \frac\{(-1)^k\}\{k\}\cos 4kt \sin kx,\\\\ u(x,t)=&\pi+x+t+\sum\_\{k=1\}^\infty \frac\{4kt-\sin 4kt\}\{64k^4\}\sin kx +4\sum\_\{k=1\}^\infty \frac\{(-1)^k\}\{k\}\cos 4kt \sin kx. \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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-u\_\{xx\}=\boxed\{\begin\{array\}\{c\}(5+2x+4\cos 2\pi x+3\cos 5\pi x)t\\\\ -x+1\end\{array\}\},&0 < x < 1, t > 0,\\\\ u|\_\{t=0\}=x-1, &0\leq x\leq 1,\\\\ u|\_\{x=0\}=t-1, u|\_\{x=1\}=t^2,&t\geq 0.\end\{array\}\right.$ [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) / 作辅助函数 \begin\{aligned\} U(x,t)=t-1+x(t^2-t+1), v(x,t)=u(x,t)-U(x,t), \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\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-v\_\{xx\}=(5+4\cos2\pi x+3\cos 5\pi x)t, &0 < x < 1, t > 0,\\\\ v|\_\{t=0\}=0,&0\leq x\leq 1,\\\\ v|\_\{x=0\}=0, v|\_\{x=1\}=0, &t\geq 0.\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 Duhamel 原理, $\displaystyle v(x,t)=\int\_0^t w(x,t;\tau)\mathrm\{ d\} \tau$, 其中 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}w\_t-w\_\{xx\}=0,&0 < x < 1, t > 0,\\\\ w|\_\{t=\tau\}=(5+4\cos2\pi x+3\cos 5\pi x)\tau, &0\leq x\leq 1,\\\\ w|\_\{x=0\}=0, w|\_\{x=1\}=0,&t\geq 0.\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\} 由分离变量法知 \begin\{aligned\} w(x,t;\tau)=\sum\_\{k=1\}^\infty a\_k\mathrm\{e\}^\{-(k\pi)^2(t-\tau)\}\sin k\pi x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 \begin\{aligned\} &\sum\_\{k=1\}^\infty a\_k\sin k\pi x =(5+4\cos 2\pi x+3\cos 5\pi x)\tau\\\\ \Rightarrow&a\_k=\left\\{\begin\{array\}\{llllllllllll\}-\frac\{24i\}\{25\pi-4i^2\pi\}\tau,&k=2i,\\\\ \frac\{4(36i^2-36i-11)\}\{(8i^3-12i^2-2i+3)\pi\}\tau,&k=2i-1.\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 $\displaystyle u(x,t)=\sum\_\{k=1\}^\infty \frac\{a\_k\}\{\tau\}\frac\{\mathrm\{e\}^\{-(k\pi)^2t\}+(k\pi)^2t-1\}\{(k\pi)^4\} \sin k\pi x$. 这里我就不展开来了, 注意 $\displaystyle \frac\{a\_k\}\{\tau\}$ 与 $\displaystyle \tau$ 已经无关了. 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)/