7.3球和半空间上的 Dirichlet 问题习题参考解答
<div id="marked"># 球和半空间上的 Dirichlet 问题习题参考解答
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1、 证明定理 7.4.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 定理 7.4 若 $\displaystyle \varphi(x)$ 在 $\displaystyle \mathbb\{R\}^2$ 上有界连续, 则函数
\begin\{aligned\} u(x,y,z)=\frac\{z\}\{2\pi\} \iint\_\{\mathbb\{R\}^2\} \frac\{\varphi(\xi,\eta)\mathrm\{ d\} \xi\mathrm\{ d\}\eta\} \{[(x-\xi)^2+(y-\eta)^2+z^2]^\frac\{3\}\{2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
属于 $\displaystyle C^\infty(\mathbb\{R\}^3\_+)$ 是 Dirichlet 问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_3u=0, (x,y,z)\in\left\\{(x,y,z); (x,y)\in\mathbb\{R\}^2, z > 0\right\\},\\\\ u|\_\{z=0\}=\varphi(x,y), (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\}
的解. 事实上, 对 $\displaystyle P\_0\in\mathbb\{R\}^3\_+$, 设它关于 $\displaystyle \partial \mathbb\{R\}^3\_+$ 的对称点为 $\displaystyle P\_1$, 则
\begin\{aligned\} G(P,P\_0)=&\frac\{1\}\{4\pi r\_\{PP\_0\}\}-\frac\{1\}\{4\pi r\_\{PP\_1\}\},\\\\ u(P\_0)=&=-\int\_\{\mathbb\{R\}^2\}\frac\{\partial G(P,P\_0)\}\{\partial\nu\} \varphi(P)\mathrm\{ d\} S\_P. \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\} \Delta\_0u(P\_0)=&-\iint\_\{\mathbb\{R\}^2\}\frac\{\partial\}\{\partial \nu\} \left\[\Delta\_0G(P,P\_0)\right\]\varphi(P)\mathrm\{ d\} S\_P\\\\ =&-\iint\_\{\mathbb\{R\}^2\}\frac\{\partial\}\{\partial \nu\} \left\[\Delta\_0G(P\_0,P)\right\]\varphi(P)\mathrm\{ d\} S\_P =-\iint\_\{\mathbb\{R\}^2\}\frac\{\partial\}\{\partial \nu\} \left\\varphi(P)\mathrm\{ d\} S\_P=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle M\_0\in\partial\mathbb\{R\}^3\_+$,
\begin\{aligned\} &|u(P\_0)-\varphi(M\_0)| =\left|-\iint\_\{\mathbb\{R\}^2\}\frac\{\partial G(P,P\_0)\}\{\partial \nu\} [\varphi(P)-\varphi(P\_0)]\mathrm\{ d\} S\_P\right|\\\\ \leq&\left|-\iint\_\{|PM\_0|<\delta\}\frac\{\partial G(P,P\_0)\}\{\partial \nu\} [\varphi(P)-\varphi(P\_0)]\mathrm\{ d\} S\_P\right|\\\\ &+\left|-\iint\_\{|PM\_0|\geq \delta\}\frac\{\partial G(P,P\_0)\}\{\partial \nu\} [\varphi(P)-\varphi(P\_0)]\mathrm\{ d\} S\_P\right|\\\\ \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 $\displaystyle \varphi$ 在 $\displaystyle M\_0$ 处连续知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta\_1 > 0,\mathrm\{ s.t.\} \forall\ P\in\mathbb\{R\}^2:|PM\_0|<\delta\_1, |\varphi(P)-\varphi(M\_0)|<\frac\{\varepsilon\}\{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\} I\_1\leq&\sup\_\{|PP\_0|<\delta\_1\}|\varphi(P)-\varphi(M\_0)| \cdot \left|-\iint\_\{|PM\_0|<\delta\} \frac\{\partial G(P,P\_0)\}\{\partial \nu\} \mathrm\{ d\} S\_P\right|\\\\ <&\frac\{\varepsilon\}\{2\} \left\[-\iint\_\{|PM\_0|<\delta\} \frac\{\partial G(P,P\_0)\}\{\partial \nu\} \mathrm\{ d\} S\_P\right\] \leq\frac\{\varepsilon\}\{2\}\left\[-\iint\_\{\mathbb\{R\}^2\} \frac\{\partial G(P,P\_0)\}\{\partial \nu\} \mathrm\{ d\} S\_P\right\]=\frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对上述 $\displaystyle \delta\_1 > 0$, 若 $\displaystyle |P\_0M\_0|<\frac\{\delta\_1\}\{2\}$, 则
\begin\{aligned\} |PM\_0|\geq \delta\_1\Rightarrow |P\_0P|&\geq |PM\_0|-|P\_0M\_0| \geq |PM\_0|-|P\_0M\_0|\\\\ &\geq |PM\_0|-\frac\{\delta\_1\}\{2\}> |PM\_0|-\frac\{|PM\_0|\}\{2\}=\frac\{|PM\_0|\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} I\_2\leq&\sup\_\{\mathbb\{R\}^2\}|\varphi| \cdot \iint\_\{|PM\_0|\geq \delta\_1\}\frac\{z\mathrm\{ d\} S\_P\}\{2\pi |P\_0P|^3\} \leq \sup\_\{\mathbb\{R\}^2\}|\varphi| \cdot \iint\_\{|PM\_0|\geq \delta\_1\}\frac\{z\mathrm\{ d\} S\_P\}\{2\pi\left\[\frac\{|PM\_0|\}\{2\}\right\]^3\}\\\\ =&\sup\_\{\mathbb\{R\}^2\}|\varphi| \cdot \frac\{z\}\{2\pi\} \cdot 8\int\_\{\delta\_1\}^\infty \frac\{1\}\{r^3\}\cdot 2\pi r\mathrm\{ d\} r =\frac\{8z\}\{\delta\_1\}\sup\_\{\mathbb\{R\}^2\}|\varphi|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle \delta=\min\left\\{\frac\{\delta\_1\}\{2\}, \frac\{\varepsilon \delta\_1\}\{16\left(\sup\_\{\mathbb\{R\}^2\}|\varphi|+1\right)\}\right\\}$, 则当 $\displaystyle |P\_0M\_0|<\delta$ 时,
\begin\{aligned\} |u(P\_0)-\varphi(M\_0)|=I\_1+I\_2<\frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \lim\_\{\mathbb\{R\}^3\_+\ni P\_0\to M\_0\}u(P\_0)=\varphi(M\_0)$. 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)/
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2、 试求 $\displaystyle \mathbb\{R\}^2$ 中调和函数在两平行线间的 Green 函数.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 通过旋转和平移, 不妨设 $\displaystyle \varOmega$ 是带型区域
\begin\{aligned\} \left\\{(x,y)\in\mathbb\{R\}^2; x\in\mathbb\{R\}, 0<y<l\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 $\displaystyle l$ 为正常数.
\begin\{aligned\} G(x,y)=&\frac\{1\}\{2\pi\}\sum\_\{k=-\infty\}^\infty \left\[\ln \frac\{1\}\{\sqrt\{(x-x\_0)^2+^2\}\}-\ln \left(|k|+1\right)\right\]\\\\ &-\frac\{1\}\{2\pi\}\sum\_\{k=-\infty\}^\infty \left\[\ln \frac\{1\}\{\sqrt\{(x-x\_0)^2+^2\}\}-\ln\left(|k|+1\right)\right\]\\\\ =&\frac\{1\}\{2\pi\}\sum\_\{k=-\infty\}^\infty \ln \frac\{\sqrt\{(x-x\_0)^2+^2\}\} \{\sqrt\{(x-x\_0)^2+^2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle G$ 在 $\displaystyle \varOmega$ 除了 $\displaystyle P\_0$ 外是调和的, 且 $\displaystyle G|\_\{\partial\varOmega\}=0$. 注:这里 $\displaystyle -\ln\left(|k|+1\right)$ 是为了保证收敛. 这个带状区域的 Green 函数只能写成函数项级数的形式了. 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)/
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3、 若 $\displaystyle \varphi(\theta)$ 满足 $\displaystyle \int\_0^\{2\pi\}\varphi(\theta)\mathrm\{ d\} \theta=0$, 求圆上第二边值问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\Delta\_2u=0,\\\\ \left.\frac\{\partial u\}\{\partial r\}\right|\_\{r=R\}=\varphi(\theta)\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\}
的解的表达式.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由习题 7.1 题 1 知
\begin\{aligned\} &0=\Delta\_2u=u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\theta\} =\frac\{1\}\{r\}(ru\_r)\_r+\frac\{1\}\{r^2\}u\_\{\theta\theta\}\\\\ \Rightarrow&(ru\_r)\_r=-\frac\{1\}\{r\}u\_\{\theta\theta\} \Rightarrow (ru\_r)\_\{rr\}=\frac\{1\}\{r^2\}u\_\{\theta\theta\} -\frac\{1\}\{r\}u\_\{r\theta\theta\} =-\frac\{1\}\{r\}(ru\_r)\_r -\frac\{1\}\{r^2\}(ru\_r)\_\{\theta\theta\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle v=ru\_r$, 则 $\displaystyle \Delta v=0$. 由 Poisson 公式知
\begin\{aligned\} v(r,\theta)=&\frac\{1\}\{2\pi R\} \int\_0^\{2\pi\} \frac\{R^2-r^2\}\{(r\cos\theta-R\cos\tau)^2+(r\sin\theta-R\sin\tau)^2\} R\varphi(\tau)\cdot R\mathrm\{ d\} \tau\\\\ =&\frac\{R\}\{2\pi\} \int\_0^\{2\pi\} \frac\{R^2-r^2\}\{R^2+r^2-2Rr\cos (\tau-\theta)\}\varphi(\tau)\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\}
于是相差一个常数下,
\begin\{aligned\} u(\rho,\theta)=&\int\_0^\rho \frac\{v(r,\theta)\}\{r\}\mathrm\{ d\} r\\\\ =&\frac\{R\}\{2\pi\}\int\_0^\{2\pi\} \varphi(\tau) \mathrm\{ d\} \tau \int^\rho \frac\{R^2-r^2\}\{r\}\mathrm\{ d\} r\\\\ =&\frac\{R\}\{2\pi\}\int\_0^\{2\pi\} g(\tau)\mathrm\{ d\} \tau \int^\rho \left\[\frac\{1\}\{r\}-\frac\{2r-2R\cos(\tau-\theta)\}\{R^2+r^2-2Rr\cos(\tau-\theta)\} \right\]\mathrm\{ d\} r\\\\ \stackrel\{\int\_0^\{2\pi\}\varphi(\tau)\mathrm\{ d\} \tau=0\}\{=\}& -\frac\{R\}\{2\pi\} \int\_0^\{2\pi\} g(\tau) \ln \left\\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\}
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4、 设 $\displaystyle \varOmega$ 为四分之一空间:
\begin\{aligned\} x\in\mathbb\{R\}, 0<y<\infty, 0<z<\infty, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
试对 $\displaystyle P\_0(x\_0,y\_0,z\_0)\in\varOmega$ 求出 Green 函数 $\displaystyle G(P,P\_0)$ 的表达式, 并由此求解 Dirichlet 问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_3u=0, x\in\mathbb\{R\}, 0<y<\infty, 0<z<\infty,\\\\ u|\_\{y=0\}=\varphi(x,z), x\in\mathbb\{R\}, 0<z<\infty,\\\\ u|\_\{z=0\}=\psi(x,y), x\in\mathbb\{R\}, 0<y<\infty.\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)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle P\_0(x\_0,y\_0,z\_0), P\_1(x\_0,-y\_0,z\_0), P\_2(x\_0,-y\_0,-z\_0), P\_3(x\_0,y\_0,-z\_0)$, 令
\begin\{aligned\} G(P,P\_0)=\frac\{1\}\{4\pi r\_\{PP\_0\}\}-\frac\{1\}\{4\pi r\_\{PP\_1\}\} +\frac\{1\}\{4\pi r\_\{PP\_2\}\} -\frac\{1\}\{4\pi r\_\{PP\_3\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle G$ 在 $\displaystyle \varOmega\backslash \left\\{P\_0\right\\}$ 上调和, 且
\begin\{aligned\} P=(x,0,z)\Rightarrow&r\_\{PP\_0\}=r\_\{PP\_1\}, r\_\{PP\_2\}=r\_\{PP\_3\},\\\\ P=(x,y,0)\Rightarrow&r\_\{PP\_0\}=r\_\{PP\_3\}, r\_\{PP\_1\}=r\_\{PP\_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 $\displaystyle P\in \partial\varOmega\Rightarrow G(P,P\_0)=0$. 进而题中 Dirichlet 问题的解为
\begin\{aligned\} u(P\_0)=-\int\_\{\partial\varOmega\}\frac\{\partial G(P,P\_0)\}\{\partial \nu\} u(P)\mathrm\{ d\} S\_P. \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\} &\left. \frac\{\partial G(P,P\_0)\}\{\partial \nu\} \right|\_\{y=0\} =\left. -\frac\{\partial G(P,P\_0)\}\{\partial \nu\} \right|\_\{y=0\}\\\\ =&\frac\{1\}\{4\pi\}\left.\left\[ -\frac\{y-y\_0\}\{r\_\{PP\_0\}^3\} +\frac\{y+y\_0\}\{r\_\{PP\_1\}^3\} -\frac\{y+y\_0\}\{r\_\{PP\_2\}^3\} +\frac\{y-y\_0\}\{r\_\{PP\_3\}^3\}\right\] \right|\_\{y=0\}\\\\ =&\frac\{1\}\{4\pi\}\left\\{ \frac\{2y\_0\}\{[(x-x\_0)^2+y\_0^2+(z-z\_0)^2]^\frac\{3\}\{2\}\} -\frac\{2y\_0\}\{[(x-x\_0)^2+y\_0^2+(z+z\_0)^2]^\frac\{3\}\{2\}\}\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. \frac\{\partial G(P,P\_0)\}\{\partial \nu\} \right|\_\{z=0\} =\left.-\frac\{\partial G(P,P\_0)\}\{\partial z\}\right|\_\{z=0\}\\\\ =&\frac\{1\}\{4\pi\} \left.\left\[ -\frac\{z-z\_0\}\{r\_\{PP\_0\}^3\} +\frac\{z-z\_0\}\{r\_\{PP\_1\}^3\} -\frac\{z+z\_0\}\{r\_\{PP\_2\}^3\} +\frac\{z+z\_0\}\{r\_\{PP\_3\}^3\} \right\]\right|\_\{z=0\}\\\\ =&\frac\{1\}\{4\pi\}\left\\{ \frac\{2z\_0\}\{[(x-x\_0)^2+(y-y\_0)^2+z\_0^2]^\frac\{3\}\{2\}\} -\frac\{2z\_0\}\{[(x-x\_0)^2+(y+y\_0)^2+z\_0^2]^\frac\{3\}\{2\}\}\right\\} \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(P\_0)=&\frac\{y\_0\}\{2\pi\} \int\_\{\mathbb\{R\}^2\} \left\\{ \begin\{array\}\{c\} \frac\{1\}\{[(x-x\_0)^2+y\_0^2+(z+z\_0)^2]^\frac\{3\}\{2\}\}\\\\ -\frac\{1\}\{[(x-x\_0)^2+y\_0^2+(z-z\_0)^2]^\frac\{3\}\{2\}\}\end\{array\}\right\\}\varphi(x,z)\mathrm\{ d\} x\mathrm\{ d\} z\\\\ &+\frac\{z\_0\}\{2\pi\}\int\_\{\mathbb\{R\}^2\} \left\\{\begin\{array\}\{c\} \frac\{1\}\{[(x-x\_0)^2+(y+y\_0)^2+z\_0^2]^\frac\{3\}\{2\}\}\\\\ -\frac\{1\}\{[(x-x\_0)^2+(y-y\_0)^2+z\_0^2]^\frac\{3\}\{2\}\}\end\{array\}\right\\}\psi(x,y)\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\}
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5、 证明处处满足平均值公式 (1.14) 的连续函数一定是调和函数.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 对 $\displaystyle \varOmega$ 中的任一球 $\displaystyle K$, 由 Poisson 公式知
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta v=0,&x\in K,\\\\ v=u,&x\in \partial K\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\}
有解 $\displaystyle v$. 令 $\displaystyle w=u-v$, 则 $\displaystyle w$ 满足平均值公式, 而由极大模原理 (参见习题 7-1 题 9 及其证明) 知
\begin\{aligned\} 0=\min\_\{\partial K\}w=\min\_\{\bar\{K\}\}w \leq\max\_\{\bar\{K\}\}w=\max\_\{\partial K\}w=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle w\equiv 0, x\in K\Rightarrow u\equiv v, x\in K$. 因此 $\displaystyle u$ 在 $\displaystyle K$ 内是调和的. 由 $\displaystyle K$ 的任意性知 $\displaystyle u$ 在 $\displaystyle \varOmega$ 上是调和的. 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)/
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6、 若 $\displaystyle \left\\{u\_k(x,y,z)\right\\}$ 为区域 $\displaystyle \varOmega\subset \mathbb\{R\}^3$ 上的一个单调不减的调和函数序列, 试证:若 $\displaystyle \left\\{u\_k(x,y,z)\right\\}$ 在$\varOmega$ 内某点 $\displaystyle (x\_0,y\_0,z\_0)$ 收敛, 则它就在 $\displaystyle \varOmega$ 上处处收敛, 且其极限为一个调和函数.
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(1)、 先证明一般的 Harnack 不等式. 设 $\displaystyle u$ 在 $\displaystyle \varOmega$ 上是非负调和函数, 则对任意 $\displaystyle \varOmega$ 的子区域 $\displaystyle \varOmega'\subset \varOmega$,
\begin\{aligned\} \exists\ C(n,\varOmega') > 0,\mathrm\{ s.t.\} \sup\_\{\varOmega'\}u\leq C\inf\_\{\varOmega'\}u. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上,
(1-1)、 对 $\displaystyle \forall\ y\in\varOmega, B\_\{4R\}(y)\subset \varOmega$, 对 $\displaystyle \forall\ x\_1,x\_2\in B\_R(y)$,
\begin\{aligned\} u(x\_1)&=\frac\{1\}\{\frac\{\omega\_n\}\{n\}R^n\}\int\_\{B\_R(x\_1)\}u\mathrm\{ d\} x \leq \frac\{1\}\{\frac\{\omega\_n\}\{n\}R^n\}\int\_\{B\_\{2R\}(y)\}u\mathrm\{ d\} x,\\\\ u(x\_2)&=\frac\{1\}\{\frac\{\omega\_n\}\{n\}(3R)^n\}\int\_\{B\_\{3R\}(x\_2)\}u\mathrm\{ d\} x \geq \frac\{1\}\{\frac\{\omega\_n\}\{n\}(3R)^n\}\int\_\{B\_\{2R\}(y)\}u\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\}
故
\begin\{aligned\} \sup\_\{B\_R(y)\}u\leq 3^n \inf\_\{B\_R(y)\}u. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 对 $\displaystyle \forall\ \varOmega'\subset\subset \varOmega$ (也即 $\displaystyle \overline\{\varOmega'\}\subset \varOmega$), 设
\begin\{aligned\} x\_1,x\_2\in\overline\{\varOmega'\}, u(x\_1)=\max\_\{\overline\{\varOmega'\}\}u, u(x\_2)=\min\_\{\overline\{\varOmega'\}\}u. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则取一含于 $\displaystyle \overline\{\varOmega'\}$ 的折线 $\displaystyle \varGamma$ 连接 $\displaystyle x\_1,x\_2$, 再取 $\displaystyle 4R<\mathrm\{dist\}(\varGamma,\partial\varOmega)$, 则由
\begin\{aligned\} \varGamma\subset \bigcup\_\{y\in \varGamma\} B(y,4R) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及有限覆盖定理知
\begin\{aligned\} \varGamma\subset \bigcup\_\{i=1\}^N B(y\_i,4R). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在每个 $\displaystyle B(y\_i,4R)$ 上应用第 i 步知
\begin\{aligned\} u(x\_1)\leq 3^\{nN\}u(x\_2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 由 $\displaystyle \left\\{u\_k(x\_0)\right\\}$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ K,\mathrm\{ s.t.\} \forall\ l > k\geq K, 0\leq u\_l(x\_0)-u\_k(x\_0)<\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle \varOmega'\subset \varOmega$, 由第 1 步之 Harnack 不等式知
\begin\{aligned\} \sup\_\{\varOmega'\}\left\ \leq C\left\<C\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \left\\{u\_k(x)\right\\}$ 在 $\displaystyle \varOmega'$ 上一致收敛. 设极限函数为 $\displaystyle u(x)$, 则由一致收敛性知 $\displaystyle u$ 在 $\displaystyle \varOmega$ 上连续. 对 $\displaystyle \forall\ x\in\varOmega$, $\displaystyle 0<r<\mathrm\{dist\}(x,\partial\varOmega)$, 有
\begin\{aligned\} u(x)=&\lim\_\{k\to\infty\}u\_k(x) =\lim\_\{k\to\infty\}\frac\{1\}\{\frac\{\omega\_n\}\{n\}R^n\}\int\_\{B\_R(x)\}u\_k\mathrm\{ d\} x =\frac\{1\}\{\frac\{\omega\_n\}\{n\}R^n\}\int\_\{B\_R(x)\}u\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\}
由题 6 即知 $\displaystyle u$ 在 $\displaystyle \varOmega$ 上是调和的.
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---
7、 若 $\displaystyle u$ 为全空间 $\displaystyle \mathbb\{R\}^3$ 上不恒等于零的调和函数, 证明:积分
\begin\{aligned\} \iiint\_\{\mathbb\{R\}^3\}u^2(x,y,z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
必发散.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若 $\displaystyle \iiint\_\{\mathbb\{R\}^3\} u^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z<\infty$, 则对 $\displaystyle \forall\ (x\_0,y\_0,z\_0)\in\mathbb\{R\}^3$,
\begin\{aligned\} |u(x\_0,y\_0,z\_0)|&=\left|\frac\{1\}\{\frac\{4\pi R^3\}\{3\}\} \iiint\_\{(x-x\_0)^2+(y-y\_0)^2+(z-z\_0)^2<R^2\} u\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\right|\\\\ &\stackrel\{\tiny\mbox\{Holder\}\}\{\leq\} \left\[\frac\{1\}\{\frac\{4\pi R^3\}\{3\}\} \iiint\_\{(x-x\_0)^2+(y-y\_0)^2+(z-z\_0)^2<R^2\} u^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\right\]^\frac\{1\}\{2\}\\\\ &\leq \sqrt\{\frac\{3\}\{4\pi R^3\}\} \sqrt\{\iiint\_\{\mathbb\{R\}^3\} u^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\}\xrightarrow\{R\to\infty\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle u\equiv 0$, 与题设矛盾. 故有结论. 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、 求 $\displaystyle \mathbb\{R\}^3$ 中单位球外部的 Dirichlet 外问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_3u=0, r > 1,\\\\ u|\_\{r=1\}=\frac\{\sqrt\{2\}\}\{\sqrt\{3+2(y-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\}
的有界解, 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)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 我们选取 $\displaystyle (x\_0,y\_0,z\_0)\in B\_1(0)$, 则
\begin\{aligned\} u(x,y,z)=\frac\{1\}\{\sqrt\{(x-x\_0)^2+(y-y\_0)^2+(z-z\_0)^2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle r > 1$ 上调和. 为使边界条件成立, 须当 $\displaystyle x^2+y^2+z^2=1$ 时,
\begin\{aligned\} &\frac\{1\}\{\sqrt\{\frac\{3\}\{2\}+y-x\}\} =\frac\{1\}\{\sqrt\{1+x\_0^2+y\_0^2+z\_0^2-2x\_0x-2y\_0y-2z\_0z\}\}\\\\ \Rightarrow&x\_0^2+y\_0^2+z\_0^2=1, -2x\_0=-1, -2y\_0+1, -2z\_0=0\\\\ \Rightarrow&x\_0=\frac\{1\}\{2\}, y\_0=-\frac\{1\}\{2\}, z\_0=0\\\\ \Rightarrow&u(x,y,z)=\frac\{1\}\{\sqrt\{\left(x-\frac\{1\}\{2\}\right)^2+\left(y+\frac\{1\}\{2\}\right)^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\}
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---
9、 求解三维半空间上的 Dirichlet 问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\Delta\_3u=0,&(x\_1,x\_2,x\_3)\in\left\\{(x\_1,x\_2,x\_3); (x\_1,x\_2)\in\mathbb\{R\}^2, x\_3 > 0\right\\},\\\\ u|\_\{x\_3\}=3x\_1+2x\_2^2,&(x\_1,x\_2)\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\}
[抱歉, 题目有问题, 书第 179 页注 7.7 前的最后一个积分是发散的!]
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [抱歉, 题目有问题, 书第 179 页注 7.7 前的最后一个积分是发散的!] 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)/
---
10、 求二维边值问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{xx\}+u\_\{yy\}=0, (x,y)\in\varOmega,\\\\ u|\_\{\partial\varOmega\}=g(x,y)\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 Green 函数, 其中
(1)、 $\displaystyle \varOmega$ 是第一象限;
(2)、 $\displaystyle \varOmega$ 是带型区域
\begin\{aligned\} \left\\{(x,y)\in\mathbb\{R\}^2; x\in\mathbb\{R\}, 0<y<l\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 $\displaystyle l$ 为正常数.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
(1)、 设 $\displaystyle P\_0 (x\_0,y\_0), P\_1 (-x\_0,y\_0), P\_2(-x\_0,-y\_0), P\_3 (x\_0,-y\_0)$, 则
\begin\{aligned\} G(x,y;x\_0,y\_0)=\frac\{1\}\{2\pi\}\ln \frac\{1\}\{r\_\{PP\_0\}\} -\frac\{1\}\{2\pi\}\ln \frac\{1\}\{r\_\{PP\_1\}\} +\frac\{1\}\{2\pi\}\ln \frac\{1\}\{r\_\{PP\_2\}\} -\frac\{1\}\{2\pi\}\ln \frac\{1\}\{r\_\{PP\_3\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, $\displaystyle G$ 在第一象限内除了 $\displaystyle P\_0$ 外是调和的, 且由对称性知
\begin\{aligned\} x=0\mbox\{或\} y=0\Rightarrow G=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、
\begin\{aligned\} G(x,y)=&\frac\{1\}\{2\pi\}\sum\_\{k=-\infty\}^\infty \left\[\ln \frac\{1\}\{\sqrt\{(x-x\_0)^2+^2\}\}-\ln \left(|k|+1\right)\right\]\\\\ &-\frac\{1\}\{2\pi\}\sum\_\{k=-\infty\}^\infty \left\[\ln \frac\{1\}\{\sqrt\{(x-x\_0)^2+^2\}\}-\ln\left(|k|+1\right)\right\]\\\\ =&\frac\{1\}\{2\pi\}\sum\_\{k=-\infty\}^\infty \ln \frac\{\sqrt\{(x-x\_0)^2+^2\}\} \{\sqrt\{(x-x\_0)^2+^2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle G$ 在 $\displaystyle \varOmega$ 除了 $\displaystyle P\_0$ 外是调和的, 且 $\displaystyle G|\_\{\partial\varOmega\}=0$. 注:这里 $\displaystyle -\ln\left(|k|+1\right)$ 是为了保证收敛. 这个带状区域的 Green 函数只能写成函数项级数的形式了.
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---
11、 求解 Dirichlet 问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{xx\}+u\_\{yy\}=-2, (x,y)\in\varOmega,\\\\ u|\_\{\partial\varOmega\}=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\}
其中 $\displaystyle \varOmega$ 是等腰三角形, 其顶点为 $\displaystyle (-1,0), (1,0)$ 和 $\displaystyle (0,\sqrt\{3\})$.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 等腰三角形的三条边的方程分别为
\begin\{aligned\} y=0, y+\sqrt\{3\}x-\sqrt\{3\}=0, y-\sqrt\{3\}x-\sqrt\{3\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故可设方程的解具有形式
\begin\{aligned\} u=cy\left(y+\sqrt\{3\}x-\sqrt\{3\}\right)\left(y-\sqrt\{3\}x-\sqrt\{3\}\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 $\displaystyle c$ 待定. 自然有 $\displaystyle u|\_\{\partial \varOmega\}=0$. 算出
\begin\{aligned\} &-2=\Delta\_2u=-4\sqrt\{3\}c \Rightarrow c=\frac\{\sqrt\{3\}\}\{6\}\\\\ \Rightarrow& u=\frac\{\sqrt\{3\}\}\{6\} y\left(y+\sqrt\{3\}x-\sqrt\{3\}\right)\left(y-\sqrt\{3\}x-\sqrt\{3\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
12、 写出球的外部区域的 Green 函数, 并由此导出对调和方程求解球的 Dirichlet 外问题的 Poisson 公式.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 球的外部区域的 Dirichlet 问题为
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\Delta u=0, x\in K,\\\\ u|\_\{\partial K\}=\varphi(x), \lim\_\{|x|\to+\infty\}u=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\}
其中 $\displaystyle K=\mathbb\{R\}^3\backslash \overline\{B\_R(0)\}$. 对 $\displaystyle P\_0\in K$, 设 $\displaystyle P\_1$ 为 $\displaystyle P_0$ 关于 $\displaystyle \partial K$ 的对称点, 则对 $\displaystyle \forall\ P\in \partial K$,
\begin\{aligned\} &\triangle OP\_1P\sim \triangle OPP\_0 \Rightarrow \frac\{OP\_1\}\{OP\}=\frac\{P\_1P\}\{PP\_0\} \Rightarrow \frac\{OP\_1\}\{R\}\cdot\frac\{1\}\{r\_\{PP\_1\}\} =\frac\{1\}\{r\_\{PP\_0\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 Green 函数为
\begin\{aligned\} G(P,P\_0)=\frac\{1\}\{4\pi r\_\{PP\_0\}\}-\frac\{OP\_1\}\{4\pi R r\_\{PP\_1\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle \forall\ P\_0\in K$, 由基本积分公式 (由 $\displaystyle \lim\_\{|x|\to+\infty\}u=0$ 知其仍然成立) 知
\begin\{aligned\} u(P\_0)=-\int\_\{\partial K\}\varphi(P)\frac\{\partial G(P,P\_0)\}\{\partial \nu\}\mathrm\{ d\} S\_P. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \rho=r\_\{OP\}, \rho\_0=r\_\{OP\_0\}, \rho\_1=r\_\{OP\_1\}, \angle P\_0OP=\alpha$, 则由余弦定理知
\begin\{aligned\} r\_\{PP\_0\}^2&=\rho\_0^2+\rho^2-2\rho\_0\rho \cos\alpha,\\\\ r\_\{PP\_1\}^2&=\rho\_1^2+\rho^2-2\rho\_1\rho \cos \alpha. \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\} G(P,P\_0)=&\frac\{1\}\{4\pi\}\left(\frac\{1\}\{\sqrt\{\rho\_0^2+\rho^2-2\rho\_0\rho \cos\alpha\}\} -\frac\{R\}\{\rho\_0 \sqrt\{\rho\_1^2+\rho^2-2\rho\_1\rho \cos\alpha\}\}\right)\\\\ =&\frac\{1\}\{4\pi\}\left( \frac\{1\}\{\sqrt\{\rho\_0^2+\rho^2-2\rho\_0\rho \cos\alpha\}\} -\frac\{R\}\{\sqrt\{R^4+\rho\_0^2\rho^2-2R^2\rho\_0\rho \cos \alpha\}\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} &\left. -\frac\{\partial G\}\{\partial \nu\} \right|\_\{\partial K\} =\left.\frac\{\partial G\}\{\partial \rho\}\right|\_\{\rho=R\}\\\\ =&\left. -\frac\{1\}\{4\pi\}\left\[ \frac\{\rho-\rho\_0\cos\alpha\}\{(\rho\_0^2+\rho^2-2\rho\_0\rho \cos\alpha)^\frac\{3\}\{2\}\} -\frac\{R(\rho\_0^2\rho-R^2\rho\_0\cos\alpha)\}\{(R^4+\rho\_0^2\rho^2-2R^2\rho\_0\rho \cos \alpha)^\frac\{3\}\{2\}\}\right\]\right|\_\{\rho=R\}\\\\ =&-\frac\{1\}\{4\pi R\}\frac\{R^2-\rho\_0^2\}\{(R^2+\rho\_0^2-2R\rho\_0\cos\alpha)^\frac\{3\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后,
\begin\{aligned\} u(P\_0)=&-\int\_\{\partial K\} \varphi(P) \frac\{1\}\{4\pi R\}\frac\{R^2-\rho\_0^2\}\{(R^2+\rho\_0^2-2R\rho\_0\cos\alpha)^\frac\{3\}\{2\}\}\mathrm\{ d\} S\_P\\\\ =&-\frac\{1\}\{4\pi R\}\int\_\{\partial K\}\frac\{R^2-\rho\_0^2\}\{r\_\{PP\_0\}^3\}\varphi(P)\mathrm\{ d\} S\_P. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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