7.2Green 函数习题参考解答

zhangzujin

# Green 函数习题参考解答 --- 1、 对于二维情形的调和函数, 建立类似于 (2.1) 的基本积分公式. [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) / 由书第 161 页 (1.8) 知如下结论. 设 $\displaystyle \varOmega\subset \mathbb\{R\}^2$ 是有界区域, $\displaystyle u\in C^2(\varOmega)\cap C^1(\bar\{\\}varOmega)$ 是调和函数, 则对 $\displaystyle \forall\ P\_0(x\_1^0,x\_2^0)\in\varOmega$, 有 \begin\{aligned\} u(P\_0)=\frac\{1\}\{2\pi\}\int\_\{\partial\varOmega\} \left\[\ln \frac\{1\}\{r\} \frac\{\partial u\}\{\partial \nu\} -u\frac\{\partial\}\{\partial \nu\}\ln \frac\{1\}\{r\}\right\]\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\} 其中 $\displaystyle \mathrm\{ d\} s$ 是 $\displaystyle \partial\varOmega$ 的线元, $\displaystyle \mathrm\{ d\} \sigma$ 是 $\displaystyle \varOmega$ 上的面元. 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、 试给出二维 Laplace 方程的 Dirichlet 问题的 Green 函数的定义. [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 $\displaystyle \varOmega\subset \mathbb\{R\}^2$ 是有界区域, $\displaystyle \varGamma$ 是它的边界, 则对 $\displaystyle \forall\ P\_0\in\varOmega$, 相应的 Laplace 方程的基本解为 $\displaystyle \frac\{1\}\{2\pi\}\ln\frac\{1\}\{r\_\{PP\_0\}\}$. 作调和函数 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta g=0, x\in\varOmega,\\\\ g(P)=\frac\{1\}\{2\pi\}\ln \frac\{1\}\{r\_\{PP\_0\}\}, x\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\} 则 $\displaystyle G(x,y)=\frac\{1\}\{2\pi\}\ln \frac\{1\}\{r\_\{PP\_0\}\}-g(P,P\_0)$ 就是二维 Laplace 方程的 Dirichlet 问题的 Green 函数. 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、 试利用第二 Green 公式及方程 $\displaystyle \Delta\_3v-v=0$ 的球对称解 $\displaystyle v=\frac\{1\}\{r\mathrm\{e\}^r\}$, 求出方程 \begin\{aligned\} \Delta\_3u-u=f \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解的积分表达式. [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) / 在 $\displaystyle \varOmega\_\varepsilon=\varOmega\backslash K\_\varepsilon(P\_0)$ ($K\_\varepsilon(P\_0)$ 的边界为 $\displaystyle \varGamma\_\varepsilon$) 对 $\displaystyle v=\frac\{1\}\{r\mathrm\{e\}^r\}$, $\displaystyle u$ 应用第二 Green 公式知 \begin\{aligned\} &\int\_\{\varOmega\_\varepsilon\} \left(u\frac\{\partial v\}\{\partial \nu\}-v\frac\{\partial u\}\{\partial \nu\}\right)\mathrm\{ d\} S =\int\_\{\varOmega\_\varepsilon\} (u\Delta v-v\Delta u)\mathrm\{ d\} x\\\\ =&\int\_\{\varOmega\_\varepsilon\} \left\[u(\Delta v-v)-v(\Delta u-u)\right\]\mathrm\{ d\} x =-\int\_\{\varOmega\_\varepsilon\} vf\mathrm\{ d\} x =-\int\_\{\varOmega\_\varepsilon\} \frac\{f\}\{r\mathrm\{e\}^r\}\mathrm\{ d\} x. \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\} -\int\_\{\varOmega\_\varepsilon\}\frac\{f\}\{r\mathrm\{e\}^r\}\mathrm\{ d\} x =\int\_\{\partial\varOmega\} \left(u\frac\{\partial v\}\{\partial \nu\}-v\frac\{\partial u\}\{\partial \nu\}\right)\mathrm\{ d\} S -\int\_\{\varGamma\_\varepsilon\}\left(u\frac\{\partial v\}\{\partial \nu\}-v\frac\{\partial u\}\{\partial \nu\}\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\} 注意到 \begin\{aligned\} &-\int\_\{\varGamma\_\varepsilon\}u\frac\{\partial v\}\{\partial \nu\}\mathrm\{ d\} S =-\int\_\{\varGamma\_\varepsilon\} u\frac\{\partial\}\{\partial r\}(r^\{-1\}\mathrm\{e\}^\{-r\})\mathrm\{ d\} S\\\\ =&\int\_\{\varGamma\_\varepsilon\} u\left\[r^\{-2\}\mathrm\{e\}^\{-r\}+r^\{-1\}\mathrm\{e\}^\{-r\}\right\]\mathrm\{ d\} S =\frac\{\mathrm\{e\}^\{-\varepsilon\}\}\{\varepsilon^2\}\int\_\{\varGamma\_\varepsilon\} u\mathrm\{ d\} S +\frac\{\mathrm\{e\}^\{-\varepsilon\}\}\{\varepsilon\}\int\_\{\varGamma\_\varepsilon\}u\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\} 我们令 $\displaystyle \varepsilon\to 0^+$ 得 \begin\{aligned\} 4\pi u(P\_0)=\int\_\{\partial\varOmega\} \left\[\frac\{1\}\{r\mathrm\{e\}^r\}\frac\{\partial u\}\{\partial \nu\} -u\frac\{\partial\}\{\partial \nu\}\left(\frac\{1\}\{r\mathrm\{e\}^r\}\right)\right\]\mathrm\{ d\} S -\int\_\varOmega \frac\{f\}\{r\mathrm\{e\}^r\}\mathrm\{ d\} 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 $\displaystyle r=r\_\{PP\_0\}$. 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、 设二元函数 $\displaystyle u(x,y)$ 是单位圆上的调和函数, 且在单位圆周上取值为 $\displaystyle u=\sin\theta$, 其中 $\displaystyle \theta$ 表示极角, 试求函数 $\displaystyle u$ 在原点的值. [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) / 由平均值公式知 \begin\{aligned\} u(0)=\frac\{1\}\{2\pi\}\int\_\{x^2+y^2=1\}u\mathrm\{ d\} s =\frac\{1\}\{2\pi\} \int\_\{-\pi\}^\pi \sin\theta \mathrm\{ d\} \theta\xlongequal\{\tiny\mbox\{对称性\}\} 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)/