1.3定解问题习题参考解答

zhangzujin

# 定解问题习题参考解答 --- 1、 考虑 Poisson 方程的 Neumann 边值问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} \Delta u=f(x,y,z),&(x,y,z)\in\varOmega,\\\\ \left.\frac\{\partial u\}\{\partial n\}\right|\_\{\varGamma\}=0,&(x,y,z)\in\varGamma. \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\} (1)、 问上述边值问题的解是否唯一? (2)、 由散度定理证明上述边值问题有解的必要条件是 \begin\{aligned\} \iiint\_\varOmega f(x,y,z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z=0. \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) / (1)、 不唯一. 解之间可以相差任意一个常数. (2)、 \begin\{aligned\} \iiint\_\varOmega f(x,y,z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z =\iiint\_\varOmega \Delta u\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z =\iint\_\{\partial\varOmega\} \frac\{\partial u\}\{\partial n\}\mathrm\{ d\} S=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、 设物体表面的绝对温度为 $\displaystyle u$, 此时它向外界辐射出的热量按 Stefan-Bolzmann 定律正比于 $\displaystyle u^4$, 即 \begin\{aligned\} \mathrm\{ d\} Q=\sigma u^4\mathrm\{ d\} S\mathrm\{ d\} t, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} $\displaystyle \sigma > 0$ 为常数. 今假设物体与周围介质之间只有热辐射而没有热传导, 周围介质的温度为 $\displaystyle f(x, y, z, t)$, 试给出该热辐射问题的边界条件. [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\} -k\frac\{\partial u\}\{\partial n\}=-\sigma u^4+\sigma f^4. \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、 (J. Hadamard 的例子) 验证函数 $\displaystyle u\_m(x,y)=\frac\{1\}\{m^\{k+1\}\}\sin mx\sinh my$ 是定解问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{xx\}+u\_\{yy\}=0,&0 < x < \pi, y > 0,\\\\ u(x,0)=0, u\_y(x,0)=\frac\{1\}\{n^k\}\sin nx, &n\mbox\{是正常数\}, k > 0,\\\\ u(0,y)=0, u(\pi,y)=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\} 的解, 并说明该定解问题关于零解不是稳定的. [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) / 令 $\displaystyle u=u\_m$, 则 \begin\{aligned\} &u\_x=\frac\{1\}\{m^k\}\cos mx\sinh y, u\_y=\frac\{1\}\{m^k\}\sin mx \cosh my,\\\\ &u\_\{xx\}=-\frac\{1\}\{m^\{k-1\}\} \sin mx\sinh my, u\_\{yy\}=\frac\{1\}\{m^\{k-1\}\}\sin mx\sinh my \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\} &u\_\{xx\}+u\_\{yy\}=0, u(x,0)=0, u\_y(x,0)=\frac\{1\}\{m^k\}\sin mx, u(0,y)=0, u(\pi, y)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle u\_m(x,y)=\frac\{1\}\{m^\{k+1\}\}\sin mx\sinh my$ 是题中定解问题的解. 又由 \begin\{aligned\} \max\_\{x\in\mathbb\{R\}\}|(u\_m)\_y(0,0)|=\frac\{1\}\{m^k\}\xrightarrow\{m\to\infty \}0, \sup\_\{x\in\mathbb\{R\}, y\in\mathbb\{R\}\}|u\_m(x,y)|=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 即知该定解问题关于零解不是稳定的. 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、 验证函数 $\displaystyle u\_n(x,t)=1+\frac\{1\}\{n\}\mathrm\{e\}^\{n^2t\}\sin nx$ 是定解问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t=-u\_\{xx\},&(x,t)\in\mathbb\{R\}\times (0,\infty),\\\\ u(x,0)=1,&x\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\} 的解, 并说明该定解问题是不适定的. [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 $\displaystyle u=u\_n$, 则 \begin\{aligned\} &u\_t=\mathrm\{e\}^\{n^2t\}n\sin nx, u\_x=\mathrm\{e\}^\{n^2t\}\cos nx, u\_\{xx\}=-\mathrm\{e\}^\{n^2t\}n\sin nx \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle u\_t=-u\_\{xx\}$, $\displaystyle u(x,0)=1$. 这个 pde 有无穷多个解, 而唯一性不能保证, 是不适定的. 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)/