第一章 调和方程

1方程的物理背景和定解问题

1.	验证 Laplace 算子 $\Delta$ 在 $2$ 维极坐标 $(r,\theta )$ 下以及 $3$ 维球坐标 $(r,\theta ,\phi )$ 下的形式: $$\begin{array}{rl}\Delta u& =\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2},\\ \Delta u& =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial u}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}u}{\partial {\phi }^2}.\end{array}$$ 方法 I. $2$ 维坐标 $(x,y)$ 的参数化是$$x=r\cos \theta ,y=r\sin \theta ,$$即$$r=\sqrt{x^2+y^2},\theta =\arctan \frac{y}{x},$$从而$$\begin{array}{rl}\frac{\partial r}{\partial x}=\cos \theta ,& \frac{\partial \theta }{\partial x}=-\frac{y}{x^2+y^2}=-\frac{\sin \theta }{r},\\ \frac{\partial r}{\partial y}=\sin \theta ,& \frac{\partial \theta }{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos \theta }{r},\end{array}$$于是$$\begin{array}{rl}\frac{\partial }{\partial x}& =\frac{\partial r}{\partial x}\frac{\partial }{\partial r}+\frac{\partial \theta }{\partial x}\frac{\partial }{\partial \theta }=\cos \theta \frac{\partial }{\partial r}-\frac{\sin \theta }{r}\frac{\partial }{\partial \theta },\\ \frac{\partial }{\partial y}& =\frac{\partial r}{\partial y}\frac{\partial }{\partial r}+\frac{\partial \theta }{\partial y}\frac{\partial }{\partial \theta }=\sin \theta \frac{\partial }{\partial r}+\frac{\cos \theta }{r}\frac{\partial }{\partial \theta }.\end{array}$$由此$$\begin{array}{rl}\frac{{\partial }^2}{\partial x^2}& =\cos \theta \frac{\partial }{\partial r}(\cos \theta \frac{\partial }{\partial r}-\frac{\sin \theta }{r}\frac{\partial }{\partial \theta })-\frac{\sin \theta }{r}\frac{\partial }{\partial \theta }(\cos \theta \frac{\partial }{\partial r}-\frac{\sin \theta }{r}\frac{\partial }{\partial \theta })\\ & =\cos \theta (\cos \theta \frac{{\partial }^2}{\partial r^2}-\frac{\sin \theta }{r}\frac{{\partial }^2}{\partial r\partial \theta }+\frac{\sin \theta }{r^2}\frac{\partial }{\partial \theta })\\ & -\frac{\sin \theta }{r}(\cos \theta \frac{{\partial }^2}{\partial \theta \partial r}-\sin \theta \frac{\partial }{\partial r}-\frac{\sin \theta }{r}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{\cos \theta }{r}\frac{\partial }{\partial \theta })\\ & ={\cos }^2\theta \frac{{\partial }^2}{\partial r^2}-2\frac{\cos \theta \sin \theta }{r^2}\frac{{\partial }^2}{\partial r\partial \theta }+2\frac{\cos \theta \sin \theta }{r}\frac{\partial }{\partial \theta }+\frac{{\sin }^2\theta }{r}\frac{\partial }{\partial r}+\frac{{\sin }^2\theta }{r^2}\frac{{\partial }^2}{\partial {\theta }^2},\\ \frac{{\partial }^2}{\partial y^2}& =\sin \theta \frac{\partial }{\partial r}(\sin \theta \frac{\partial }{\partial r}+\frac{\cos \theta }{r}\frac{\partial }{\partial \theta })+\frac{\cos \theta }{r}\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial }{\partial r}+\frac{\cos \theta }{r}\frac{\partial }{\partial \theta })\\ & =\sin \theta (\sin \theta \frac{{\partial }^2}{\partial r^2}+\frac{\cos \theta }{r}\frac{{\partial }^2}{\partial r\partial \theta }-\frac{\cos \theta }{r^2}\frac{\partial }{\partial \theta })\\ & +\frac{\cos \theta }{r}(\sin \theta \frac{{\partial }^2}{\partial \theta \partial r}+\cos \theta \frac{\partial }{\partial r}+\frac{\cos \theta }{r}\frac{{\partial }^2}{\partial {\theta }^2}-\frac{\sin \theta }{r}\frac{\partial }{\partial \theta })\\ & ={\sin }^2\theta \frac{{\partial }^2}{\partial r^2}+2\frac{\sin \theta \cos \theta }{r}\frac{{\partial }^2}{\partial r\partial \theta }-2\frac{\sin \theta \cos \theta }{r^2}\frac{\partial }{\partial \theta }+\frac{{\cos }^2\theta }{r}\frac{\partial }{\partial r}+\frac{{\cos }^2\theta }{r^2}\frac{{\partial }^2}{\partial {\theta }^2}.\end{array}$$相加得$$\Delta =\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}.$$ $3$ 维坐标 $(x,y,z)$ 的参数化是$$\begin{array}{rl}& z=r\cos \theta ,s=r\sin \theta ,\\ & x=s\cos \phi ,y=s\sin \phi .\end{array}$$对 $s=\sqrt{x^2+y^2}$ 以及 $r=\sqrt{s^2+z^2}$ 用 $2$ 维情形的计算结果, 有$$\begin{array}{rl}\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}& =\frac{{\partial }^2}{\partial s^2}+\frac{1}{s}\frac{\partial }{\partial s}+\frac{1}{s^2}\frac{{\partial }^2}{\partial {\phi }^2},\\ \frac{{\partial }^2}{\partial s^2}+\frac{{\partial }^2}{\partial z^2}& =\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2},\end{array}$$相加得$$\Delta =\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{1}{s}\frac{\partial }{\partial s}+\frac{1}{s^2}\frac{{\partial }^2}{\partial {\phi }^2},$$其中 $s^2=r^2{\sin }^2\theta$, 且$$\frac{1}{s}\frac{\partial }{\partial s}=\frac{1}{s}(\frac{\partial r}{\partial s}\frac{\partial }{\partial r}+\frac{\partial \theta }{\partial s}\frac{\partial }{\partial \theta })=\frac{1}{s}(\frac{s}{r}\frac{\partial }{\partial r}+\frac{z}{r^2}\frac{\partial }{\partial \theta })=\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2\sin \theta }\cos \theta \frac{\partial }{\partial \theta }.$$故$$\begin{array}{rl}\Delta & =\frac{{\partial }^2}{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^2\sin \theta }(\sin \theta \frac{{\partial }^2}{\partial {\theta }^2}+\cos \theta \frac{\partial }{\partial \theta })+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2}{\partial {\phi }^2}\\ & =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2}{\partial {\phi }^2}.\end{array}$$ 方法 II. 我们用流形上 Laplace 算子来解题, 其中 $\ast$ 是 Hodge 星算子. (预备知识见 [Jo] 3.3 节, 尤其是 pp.106,110. ) $2$ 维极坐标有转换 $\{\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$, 则$$\left(\begin{array}{c}dx\\ dy\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)\left(\begin{array}{c}dr\\ d\theta \end{array}\right),$$推出$$\left(\begin{array}{c}dr\\ d\theta \end{array}\right)=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta /r& \cos \theta /r\end{array}\right)\left(\begin{array}{c}dx\\ dy\end{array}\right),$$从而 $\{dr,rd\theta \}$ 构成一组标准正交基, 于是有 $\ast dr=rd\theta ,\ast (rd\theta )=-dr$. 由此, $$\begin{array}{rl}\Delta u=\ast d\ast du& =\ast d\ast (u_{r}dr+u_{\theta }d\theta )\\ & =\ast d(u_{r}rd\theta -\frac{u_{\theta }}{r}dr)\\ & =\ast \left((u_{r}r{)}_{r}dr\wedge d\theta +\frac{u_{\theta \theta }}{r}dr\wedge d\theta \right)\\ & =\frac{1}{r}(u_{r}r{)}_r+\frac{1}{r^2}{u}_{\theta \theta }.\end{array}$$ $3$ 维有球坐标转换 $\{\begin{array}{l}x=r\sin \theta \cos \phi \\ y=r\sin \theta \sin \phi \\ z=r\cos \theta \end{array}$, 则$$\left(\begin{array}{c}dx\\ dy\\ dz\end{array}\right)=\left(\begin{array}{ccc}\sin \theta \cos \phi & r\cos \theta \cos \phi & -r\sin \theta \sin \phi \\ \sin \theta \sin \phi & r\cos \theta \sin \phi & r\sin \theta \cos \phi \\ \cos \theta & -r\sin \theta & 0\end{array}\right)\left(\begin{array}{c}dr\\ d\theta \\ d\phi \end{array}\right),$$推出$$\left(\begin{array}{c}dr\\ d\theta \\ d\phi \end{array}\right)=\left(\begin{array}{ccc}\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \frac{1}{r}\cos \theta \cos \phi & \frac{1}{r}\cos \theta \sin \phi & -\frac{1}{r}\sin \theta \\ -\frac{\sin \phi }{r\sin \theta }& \frac{\cos \phi }{r\sin \theta }& 0\end{array}\right)\left(\begin{array}{c}dx\\ dy\\ dz\end{array}\right),$$从而 $\{dr,rd\theta ,r\sin \theta d\phi \}$ 构成一组标准正交基. 由此, $$\begin{array}{rl}\Delta u=\ast d\ast du& =\ast d\ast (u_{r}dr+u_{\theta }d\theta +u_{\phi }d\phi )\\ & =\ast d(u_r{r}^2\sin \theta d\theta \wedge d\phi -u_{\theta }\sin \theta dr\wedge d\phi +\frac{u_{\phi }}{\sin \theta }dr\wedge d\theta )\\ & =\ast \left((\frac{\partial }{\partial r}(u_r{r}^2)\sin \theta +\frac{\partial }{\partial \theta }(u_{\theta }\sin \theta )+\frac{u_{\phi \phi }}{\sin \theta })dr\wedge d\theta \wedge d\phi \right)\\ & =\frac{1}{r^2\sin \theta }\left(\frac{\partial }{\partial r}(u_r{r}^2)\sin \theta +\frac{\partial }{\partial \theta }(u_{\theta }\sin \theta )+\frac{u_{\phi \phi }}{\sin \theta }\right).\end{array}$$
注:	$n$ 维时, Laplace 算子有如下的形式. 公式. ${\mathbb{R}}^n\backslash \{0\}$ 上, Laplace 算子在极坐标下的表达式为$$\Delta =\frac{{\partial }^2}{\partial r^2}+\frac{n-1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}{\Delta }_{S^{n-1}},$$其中 $x\in {\mathbb{R}}^n\backslash \{0\}$ 唯一地写成 $x=r\phi ,(r,\phi )\in (0,+\infty )\times S^{n-1}$, 单位球面 $S^{n-1}=\{x\in {\mathbb{R}}^n\mid |x|=1\}\subset {\mathbb{R}}^n$ 上的度量诱导自这个嵌入, 相应地有球面上的 Laplace 算子 ${\Delta }_{S^{n-1}}$. 证明. 证明. 设 $\phi \in S^{n-1}$ 在 ${\mathbb{R}}^n$ 的坐标是 $(y^1,\cdots ,y^n)$, ${\mathbb{R}}^n$ 上的标准度量就是$$\begin{array}{rl}\sum (d(r{y}^i){)}^2& =\sum (y^{i}dr+rd{y}^i{)}^2\\ & =\sum (y^i{)}^{2}d{r}^2+r^2(d{y}^i{)}^2+(y^{i}dr)(rd{y}^i)+(rd{y}^i)(y^{i}dr)\\ & =d{r}^2+r^{2}d{\phi }^2,\end{array}$$这里 $d{\phi }^2=\sum (d{y}^i{)}^2$ 是 $S^{n-1}$ 上的标准度量, 最后一个等号是因为 $\sum (y^i{)}^2=1$ 推出 $\sum y^{i}d{y}^i=0$. 再设 $\phi \in S^{n-1}$ 的局部坐标是 $({\phi }^1,\cdots ,{\phi }^{n-1})$, 以及 $d{\phi }^2=\sum {\bar{g}}_{{\phi }^I{\phi }^J}d{\phi }^{I}d{\phi }^J$, 就有 ${\mathbb{R}}^n$ 上的标准度量在坐标 $(r,{\phi }^1,\cdots ,{\phi }^{n-1})$ 下等于$$\begin{array}{r}d{r}^2+r^2\sum {\bar{g}}_{{\phi }^I{\phi }^J}d{\phi }^{I}d{\phi }^J,\end{array}$$换言之, 系数矩阵 $(g_{ij}{)}_{i,j=r,{\phi }^1,\cdots ,{\phi }^{n-1}}=\left(\begin{array}{cc}1& \\ & r^2{\bar{g}}_{{\phi }^I{\phi }^J}\end{array}\right)$. Riemann 流形上的 Laplace 算子的坐标表示为 (见大多数 Riemann 几何教材, 如 [Jo] (3.1.20), p.92) $$\Delta f=\frac{1}{\sqrt{g}}\sum \frac{\partial }{\partial x^j}\left(\sqrt{g}{g}^{ij}\frac{\partial f}{\partial x^i}\right),$$其中 $g=\det (g_{ij})$, 矩阵 $(g^{ij})=(g_{ij}{)}^{-1}$. 现在 $(x^1,\cdots ,x^n)=(r,{\phi }^1,\cdots ,{\phi }^{n-1})$, 有 $\sqrt{g}=r^{n-1}\sqrt{\bar{g}}$, $(g^{ij})=\left(\begin{array}{cc}1& \\ & \frac{1}{r^2}{\bar{g}}^{{\phi }^I{\phi }^J}\end{array}\right)$, 其中 $\bar{g}=\det ({\bar{g}}_{{\phi }^I{\phi }^J})$, $({\bar{g}}^{{\phi }^I{\phi }^J})=({\bar{g}}_{{\phi }^I{\phi }^J}{)}^{-1}$. 因此$$\begin{array}{rl}\Delta f& =\frac{1}{r^{n-1}\sqrt{\bar{g}}}\left[\frac{\partial }{\partial r}\right(r^{n-1}\sqrt{\bar{g}}\frac{\partial f}{\partial r})+\sum \frac{\partial }{\partial {\phi }^J}(r^{n-1}\sqrt{\bar{g}}\frac{1}{r^2}{\bar{g}}^{{\phi }^I{\phi }^J}\frac{\partial f}{\partial {\phi }^I}\left)\right]\\ & =\frac{1}{r^{n-1}}\frac{\partial }{\partial r}\left(r^{n-1}\frac{\partial f}{\partial r}\right)+\frac{1}{r^2}\frac{1}{\sqrt{\bar{g}}}\sum \frac{\partial }{\partial {\phi }^J}\left(\sqrt{\bar{g}}{\bar{g}}^{{\phi }^I{\phi }^J}\frac{\partial f}{\partial {\phi }^I}\right)\\ & =\frac{{\partial }^{2}f}{\partial r^2}+\frac{n-1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}{\Delta }_{S^{n-1}}f.\end{array}$$ $\square$
4.	$\Delta =4\frac{\partial }{\partial z}\frac{\partial }{\partial \bar{z}}$, 由于 $\frac{\partial f}{\partial \bar{z}}=0$, 有 $\Delta f=0$. 回忆: $\frac{\partial }{\partial z}=\frac{1}{2}\left(\frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right)$, $\frac{\partial }{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right)$.
6.	取 $\Omega$ 为单位球 $B_1(0)$, 构造下述方程非常数的解: $$\{\begin{array}{ll}-\Delta u=0,& |x|>1,\\ \frac{\partial u}{\partial \mathbf{n}}=0,& |x|=1.\end{array}$$由于边界条件, 考虑形如 $u(x)=v(r)w(\theta )$ 的特解, 这里 $x=r\theta ,(r,\theta )\in [1,\infty )\times S^{n-1}$, $w$ 是 ${\Delta }_{S^{n-1}}$ 的一个特征函数且不是常数, 也就是 $-\Delta w=\lambda w$, 且 $\nabla w$ 不恒为 $0$. 故$$\lambda {\int }_{S^{n-1}}{w}^2={\int }_{S^{n-1}}-\Delta w\cdot w={\int }_{S^{n-1}}|\nabla w{|}^2>0,$$说明特征值 $\lambda >0$. 根据 Laplace 算子的极坐标表达式, $$(v^{\prime \prime }(r)+\frac{n-1}{r}{v}^{\prime }(r)-\frac{\lambda }{r^2}v(r))w(\theta )=0.$$试着取 $v(r)=r^{\mu }$, 则 $\mu (\mu -1)+(n-1)\mu -\lambda =0$. 边界条件进一步要求$$v^{\prime }(1)=0.$$于是令 $v(r)={\mu }_2{r}^{{\mu }_1}-{\mu }_1{r}^{{\mu }_2}$, 其中 ${\mu }_1,{\mu }_2$ 是 ${\mu }^2+(n-2)\mu -\lambda =0$ 的两个根.
注:	特征函数 $w$ 的存在性可用变分法说明: 设$${\lambda }_1=\underset{\begin{array}{c}w\in H^1(S^{n-1}),\\ w\perp 1,w\ne 0\end{array}}{\inf }\frac{\Vert \nabla w{\Vert }_{L^2}^2}{\Vert w{\Vert }_{L^2}^2},$$这里的 $H^1(S^{n-1})$ 是 Sobolev 空间 (不是 de Rham 上同调群) , $w\perp 1$ 是说 ${\int }_{S^{n-1}}w=0$. 取一个极小化序列 $\{w_{(n)}\}$ (是指 $\Vert \nabla w_{(n)}{\Vert }_{L^2}^2/\Vert w_{(n)}{\Vert }_{L^2}^2\to {\lambda }_1$) 满足 $\Vert w_{(n)}{\Vert }_{L^2}=1$. 由于 $\Vert w_{(n)}{\Vert }_{H^1}$ 有界, 抽取子列后 $w_{(n)}$ 在 $H^1(S^{n-1})$ 下弱收敛于某个 $w_1\in H^1(S^{n-1})$. 由 Rellich 定理, 再次抽取子列后 $w_{(n)}$ 在 $L^2(S^{n-1})$ 下强收敛于 $w_1$. 因为$${\lambda }_1\le \Vert \nabla w_1{\Vert }_{L^2}^2=\underset{n\to \infty }{\lim }⟨\nabla w_1,\nabla w_{(n)}⟩\le \underset{n\to \infty }{\underline{\lim }}\Vert \nabla w_1{\Vert }_{L^2}\Vert \nabla w_{(n)}{\Vert }_{L^2}\le \underset{n\to \infty }{\lim }\Vert \nabla w_{(n)}{\Vert }_{L^2}^2={\lambda }_1,$$有 $\Vert \nabla w_1{\Vert }_{L^2}^2={\lambda }_1$, 并且 $w_1\perp 1$ 说明 ${\lambda }_1>0$; 对一切 $h\in H^1(S^{n-1}),h\perp 1$ 有$$0=\frac{d}{d\epsilon }{|}_{\epsilon =0}\frac{\Vert \nabla (w_1+\epsilon h){\Vert }_{L^2}^2}{\Vert w_1+\epsilon h{\Vert }_{L^2}^2}=2(\frac{⟨\nabla w_1,\nabla h⟩}{\Vert w_1{\Vert }_{L^2}^2}-\frac{⟨w_1,h⟩\Vert \nabla w_1{\Vert }_{L^2}^2}{\Vert w_1{\Vert }_{L^2}^4})=2⟨-\Delta w_1-{\lambda }_1{w}_1,h⟩,$$可得$$-\Delta w_1-{\lambda }_1{w}_1=\text{常数},$$积分得常数 $=0$, 系 $-\Delta w_1={\lambda }_1{w}_1$. 类似地, 设已有特征函数 $w_1,\cdots ,w_{k-1}$, 通过研究$${\lambda }_k=\underset{\begin{array}{c}w\in H^1(S^{n-1}),\\ w\perp \mathrm{span}\{1,w_1,\cdots ,w_{k-1}\},w\ne 0\end{array}}{\inf }\frac{\Vert \nabla w{\Vert }_{L^2}^2}{\Vert w{\Vert }_{L^2}^2}$$可得特征值 ${\lambda }_k$ 以及对应的 $w_k$. 对 $-\Delta w_k={\lambda }_k{w}_k$ 反复用椭圆正则性, 得到 $w_k\in C^{\infty }(S^{n-1})=H^0(S^{n-1})$, 这是 de Rham 上同调群 (不是 Sobolev 空间) . 具体的过程见 [Jo] 3.2 节. 事实上可以算出 $-{\Delta }_{S^n}$ 从小到大第 $k$ 个不同的正特征值为 $k^2+(n-1)k$, 见 [Ta II] Chapter 8 Corollary 4.3.
7.	是上一题的特例, $S^1$ 参数化为 $\{\theta =e^{it}:t\in [0,2\pi )\}$, $-{\Delta }_{S^1}=-(\frac{d}{dt}{)}^2$ 的特征值是 $\lambda =k^2(k\in \mathbb{Z})$, ${\mu }^2-k^2=0$ 的根是 $\pm k$, 从而$$w(\theta )=c_1{e}^{ikt}+c_2{e}^{-ikt}=c_1{\theta }^k+c_2{\theta }^{-k},$$于是$$u(r\theta )=(r^k+r^{-k})(c_1{\theta }^k+c_2{\theta }^{-k})$$是这个方程的解.
9.	方法 I. (来自 [Ax] Proposition 4.6) 对多项式 $u$ 计算 $\Delta v$, 再用稠密性得出结果. 引理. 设 $p$ 是 ${\mathbb{R}}^n$ 上的齐次多项式, 次数为 $m$, 简记 $t=2-n-2m$, 有$$\Delta (|x{|}^{t}p)=|x{|}^t\Delta p.$$ 证明. 设 $p=\sum _{\alpha }{c}_{\alpha }{x}_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$, 诸 ${\alpha }_1+\cdots +{\alpha }_n=m$, 那么$$\begin{array}{rl}x\cdot \nabla p& =\sum _{\alpha }{c}_{\alpha }\sum _{i=1}^n{x}_i\cdot {\alpha }_i{x}_1^{{\alpha }_1}\cdots x_i^{{\alpha }_i-1}\cdots x_n^{{\alpha }_n}\\ & =\sum _{\alpha }{c}_{\alpha }m{x}_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}=mp.\end{array}$$另外, $$\begin{array}{rl}\nabla (|x{|}^t)& =tx|x{|}^{t-2},\\ \Delta (|x{|}^t)& =\nabla \cdot (tx|x{|}^{t-2})\\ & =t[(\nabla \cdot x)|x{|}^{t-2}+(\nabla |x{|}^{t-2})\cdot x]\\ & =t[n|x{|}^{t-2}+(t-2)x|x{|}^{t-4}\cdot x]\\ & =t(n+t-2)|x{|}^{t-2}.\end{array}$$于是, $$\begin{array}{rl}\Delta (|x{|}^{t}p)-|x{|}^t\Delta p& =2\nabla (|x{|}^t)\cdot \nabla p+\Delta (|x{|}^t)p\\ & =2t|x{|}^{t-2}x\cdot \nabla p+t(n+t-2)|x{|}^{t-2}p\\ & =t(2m+n+t-2)|x{|}^{t-2}p=0.\end{array}$$以上是引理的证明. 下面说明 $\Delta v=|x{|}^4\Delta u\text{ 的 Kelvin 变换}$. 特别地, 如果 $\Delta u=0$, 那么 $\Delta v=0$. 当 $u$ 是齐次多项式, 次数为 $m$, 由引理 ($t$ 的含义同上) , $$\Delta v=\Delta (|x{|}^{t}u)=|x{|}^t\Delta u=|x{|}^4\Delta u\text{ 的 Kelvin 变换},$$最后的等号是因为 $|x{|}^4\Delta u$ 的 $\text{Kelvin}$ 变换为 (注意 $\Delta u$ 的次数为 $m-2$) $$\frac{1}{|x{|}^{n-2}}{\left|\frac{x}{|x{|}^2}\right|}^4\Delta u\left(\frac{x}{|x{|}^2}\right)=|x{|}^{2-n-4-2(m-2)}\Delta u=|x{|}^t\Delta u.$$由线性, 命题对 $u$ 是多项式 (未必齐次) 同样成立, 又由于 $(C^2({\mathbb{R}}^n\backslash \{0\}),\Vert {\Vert }_{C^2})$ 中多项式是局部上稠密的 (用 Stone–Weierstrass 定理) , 命题进一步对 $u\in C^2$ 成立. 方法 II. 用前面给出的 Laplace 算子的表达式来计算. 设极坐标下 $y=r\phi$, $y^{\prime }=\frac{y}{|y{|}^2}=r^{\prime }{\phi }^{\prime }$, 则 $r^{\prime }=\frac{1}{r},{\phi }^{\prime }=\phi$. 这样$$\{\begin{array}{l}\frac{\partial }{\partial r}=\frac{\partial r^{\prime }}{\partial r}\frac{\partial }{\partial r^{\prime }}=-\frac{1}{r^2}\frac{\partial }{\partial r^{\prime }}=-r^{\prime 2}\frac{\partial }{\partial r^{\prime }},\\ \frac{{\partial }^2}{\partial r^2}=-r^{\prime 2}\frac{\partial }{\partial r^{\prime }}(-r^{\prime 2}\frac{\partial }{\partial r^{\prime }})=r^{\prime 4}\frac{{\partial }^2}{\partial r^{\prime 2}}+2r^{\prime 3}\frac{\partial }{\partial r^{\prime }},\\ {\Delta }_{S^{n-1}}={\Delta }_{S^{\prime n-1}}.\end{array}$$故有$$\begin{array}{rl}\Delta v(y)& =(\frac{{\partial }^2}{\partial r^2}+\frac{n-1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}{\Delta }_{S^{n-1}})\left(\frac{1}{|y{|}^{n-2}}u\right(\frac{y}{|y{|}^2}\left)\right)\\ & =(r^{\prime 4}\frac{{\partial }^2}{\partial r^{\prime 2}}+2r^{\prime 3}\frac{\partial }{\partial r^{\prime }}-(n-1)r^{\prime 2}\frac{\partial }{\partial r^{\prime }}+r^{\prime 2}{\Delta }_{S^{\prime n-1}})(r^{\prime n-2}u(y^{\prime }))\\ & =r^{\prime 4}(r^{\prime n-2}\frac{{\partial }^{2}u}{\partial r^{\prime 2}}(y^{\prime })+2(n-2)r^{\prime n-3}\frac{\partial u}{\partial r^{\prime }}(y^{\prime })+(n-2)(n-3)r^{\prime n-4}u(y^{\prime }))\\ & +2r^{\prime 3}((n-2)r^{\prime n-3}u(y^{\prime })+r^{\prime n-2}\frac{\partial u}{\partial r^{\prime }}(y^{\prime }))\\ & -(n-1)r^{\prime 3}((n-2)r^{\prime n-3}u(y^{\prime })+r^{\prime n-2}\frac{\partial u}{\partial r^{\prime }}(y^{\prime }))\\ & +r^{\prime 2}{\Delta }_{S^{\prime n-1}}(r^{\prime n-2}u(y^{\prime }))\\ & =r^{\prime n+2}(\frac{{\partial }^2}{\partial r^{\prime 2}}+\frac{n-1}{r^{\prime }}\frac{\partial }{\partial r^{\prime }}+\frac{{\Delta }_{S^{\prime n-1}}}{r^{\prime 2}})u(y^{\prime })\\ & =r^{\prime n+2}\Delta u(y^{\prime }).\end{array}$$于是当 $\Delta u=0$ 时 $\Delta v=0$. 方法 III. 直接计算. 过程较复杂, 隐藏了. 1个回答被折叠 (为什么? ) . 1 个回答被折叠 (为什么? ) . 记$$k(y):=\frac{1}{|y{|}^{n-2}},x(y):=\frac{y}{|y{|}^2}.$$对于 $v(y)=k(y)u(x)$, 计算得到$$\begin{array}{rl}\Delta v& =u{\Delta }_{y}k+2{\nabla }_{y}k\cdot {\nabla }_{y}u+k{\Delta }_{y}u\\ & =u{\Delta }_{y}k+2{\nabla }_{y}k\cdot {\nabla }_{y}u+k\sum _i{\partial }_{y_i}\left(\sum _j{\partial }_{x_j}u{\partial }_{y_i}{x}_j\right)\\ & =u{\Delta }_{y}k+2{\nabla }_{y}k\cdot {\nabla }_{y}u+k(\sum _{i,j}{\partial }_{x_j}u{\partial }_{y_i}^2{x}_j+\sum _{i,j}{\partial }_{y_i{x}_j}u{\partial }_{y_i}{x}_j)\\ & =2{\nabla }_{y}k\cdot {\nabla }_{y}u+k\sum _{i,j}{\partial }_{x_j}u{\partial }_{y_i}^2{x}_j\\ & =({\nabla }_{y}k\cdot {\nabla }_{y}u+k\sum _{i,j}{\partial }_{x_j}u{\partial }_{y_i}^2{x}_j)+{\nabla }_{y}k\cdot {\nabla }_{y}u\\ & =\sum _j{\partial }_{x_j}u(\sum _i{\partial }_{y_i}(k{\partial }_{y_i}{x}_j)+\sum _i{\partial }_{y_i}k{\partial }_{y_i}{x}_j)\\ & =\sum _j{\partial }_{x_j}u((2-n)\frac{y_j}{|y{|}^{n+2}}-(2-n)\frac{y_j}{|y{|}^{n+2}})\\ & =0.\end{array}$$其中第 4 个等号是因为 ${\Delta }_{y}k=0$ 以及$$\begin{array}{rl}\sum _{i,j}{\partial }_{y_i{x}_j}u{\partial }_{y_i}{x}_j& =\sum _{i,j}\left(\sum _k{\partial }_{y_i}{x}_k{\partial }_{x_k{x}_j}u\right){\partial }_{y_i}{x}_j\\ & =\sum _{k,j}{\partial }_{x_k{x}_j}u\sum _i{\partial }_{y_i}{x}_k{\partial }_{y_i}{x}_j\\ & =\sum _{k,j}{\partial }_{x_k{x}_j}u\sum _i\frac{{\delta }_{ik}|y{|}^2-2y_i{y}_k}{|y{|}^4}\frac{{\delta }_{ij}|y{|}^2-2y_i{y}_j}{|y{|}^4}\\ & =\sum _{k,j}{\partial }_{x_k{x}_j}u\sum _i\frac{{\delta }_{ikj}|y{|}^4-2y_i{y}_k{\delta }_{ij}|y{|}^2-2y_i{y}_j{\delta }_{ik}|y{|}^2+4y_i^2{y}_k{y}_j}{|y{|}^8}\\ & =\sum _{k,j}{\partial }_{x_k{x}_j}u\frac{{\delta }_{kj}|y{|}^4}{|y{|}^8}\\ & =\frac{{\Delta }_{x}u}{|y{|}^4}\\ & =0.\end{array}$$倒数第 2 个等号则是因为, 对取定的 $j$, 有$$\begin{array}{rl}\sum _i{\partial }_{y_i}(k{\partial }_{y_i}{x}_j)& =\sum _i{\partial }_{y_i}\left(\frac{1}{|y{|}^{n-2}}{\partial }_{y_i}\right(\frac{y_j}{|y{|}^2}\left)\right)\\ & =\sum _i{\partial }_{y_i}(\frac{{\delta }_{ij}}{|y{|}^n}-\frac{2y_i{y}_j}{|y{|}^{n+2}})\\ & =\sum _i-{\delta }_{ij}n\frac{y_i}{|y{|}^{n+2}}-\frac{2[(y_j+y_i{\delta }_{ij})|y{|}^{n+2}-y_i{y}_j(n+2)y_i|y{|}^n]}{|y{|}^{2(n+2)}}\\ & =\sum _i-\frac{{\delta }_{ij}(n+2)y_i+2y_j}{|y{|}^{n+2}}+\frac{2y_i^2{y}_j(n+2)}{|y{|}^{n+4}}\\ & =\frac{-(n+2)y_j-2n{y}_j+2(n+2)y_j}{|y{|}^{n+2}}\\ & =(2-n)\frac{y_j}{|y{|}^{n+2}},\\ \sum _i{\partial }_{y_i}k{\partial }_{y_i}{x}_j& =\sum _i(2-n)\frac{y_i}{|y{|}^n}\frac{{\delta }_{ij}|y{|}^2-2y_i{y}_j}{|y{|}^4}\\ & =(2-n)\frac{y_j|y{|}^2-2y_j|y{|}^2}{|y{|}^{n+4}}\\ & =-(2-n)\frac{y_j}{|y{|}^{n+2}}.\end{array}$$ $\square$

2调和函数的基本性质与应用

2.	球 $B_r(x^0)$ 就是一族球面 $\partial B_s(x^0),s\in (0,r)$ 拼成的; 对所有的球面成立平均值原理, 则球亦然. 写出来是$$\begin{array}{rl}\frac{1}{|B_r(x^0)|}{\int }_{B_r(x^0)}u(x)dx& =\frac{1}{|B_r(x^0)|}{\int }_0^r{\int }_{\partial B_s(x^0)}u(x)d{S}_{x}ds\\ & =\frac{1}{|B_r(x^0)|}{\int }_0^r{\int }_{\partial B_s(x^0)}d{S}_{x}u(x^0)ds=u(x^0).\end{array}$$
6.	对任意 $x\in {\mathbb{R}}^n$, 由平均值性质, 对任何 $r>0$, $$|u(x)|\le \frac{1}{|B_r(x)|}\Vert u{\Vert }_{L^1(B_r(x))}\le \frac{1}{|B_r(x){|}^{\frac{1}{p}}}\Vert u{\Vert }_{L^p(B_r(x))}.$$令 $r\to +\infty$ 得 $u(x)=0$.
8.	当 $n\ge 3$, 对 $x^0\in {\mathbb{R}}^n\backslash \Omega$, $|x-x^0{|}^{2-n}$ 是 $\Omega$ 上的非负调和函数.
注:	当 $n=2$, 结论不再成立, 比如 ${\mathbb{R}}^2\backslash \{0\}$ 上的下有界的调和函数 $u$ 一定是常数: 注意到下图的 $U$ 是 ${\mathbb{R}}^2$ 上的下有界的调和函数, 故是常数. $$\mathbb{C}{\mathbb{R}}^2\mathbb{C}\backslash \{0\}{\mathbb{R}}^2\backslash \{0\}\mathbb{R}{e}^{z}Uu$$
9.	通过平移, 对非负调和函数 $u$ 证明即可. 方法 I. 对任意 $x\in {\mathbb{R}}^n$, 由梯度估计, $$|\nabla u(x)|\le \frac{n}{r}u(x)$$对任何 $r>0$ 成立, 令 $r\to +\infty$ 得 $\nabla u(x)=0$. 于是 $u$ 为常数. 方法 II. 对任何两点 $y,z$, 记 $d=|y-z|$, $$\begin{array}{rl}|u(y)-u(z)|& =\frac{1}{|B_r|}|{\int }_{B_r(y)}u(x)dx-{\int }_{B_r(z)}u(x)dx|\\ & \le \frac{1}{|B_r|}{\int }_{B_{r+\frac{d}{2}}(\frac{y+z}{2})}u(x)dx-{\int }_{B_{r-\frac{d}{2}}(\frac{y+z}{2})}u(x)dx\\ & =\frac{\left|B_{r+\frac{d}{2}}\right|-\left|B_{r-\frac{d}{2}}\right|}{|B_r|}u\left(\frac{y+z}{2}\right)\\ & \le C\frac{d}{r}u\left(\frac{y+z}{2}\right)\to 0\text{当 }r\to +\infty .\end{array}$$
10.	方法 I. 对距离小于 $\frac{3}{8}$ 的 $x^{\prime },x^{\prime \prime }\in \{|x_1|\le \frac{1}{4}\}$, 由 $B(x^{\prime },\frac{3}{8})\subset B(x^{\prime \prime },\frac{3}{4})\subset \{|x_1|\le 1\}$ 得到 $u(x^{\prime })\le 4u(x^{\prime \prime })$. 取大约 $\frac{8}{3}|x_2|$ 个点 $x^0=0,x^1,\cdots ,x^{n-1},x^n=x$ 使 $|x^i-x^{i+1}|<\frac{3}{8}$, 那么 $u(x)\le Cu(0)4^{\frac{8}{3}|x_2|}$. 方法 II. 由梯度估计, 对 $y\in \Sigma$, 有$$|\nabla u(y)|\le \frac{2}{1-1/4}u(y),\text{即 }|\nabla \log u(y)|\le \frac{8}{3}.$$故$$\log \frac{u(x)}{u(0)}={\int }_0^1\frac{d}{dt}\log u(tx)dt=x\cdot {\int }_0^1\nabla \log u(tx)dt\le \frac{8}{3}|x|\le \frac{2}{3}+\frac{8}{3}|x_2|.$$

3极值原理及其应用

1.	$\Omega$ 是有界开集, 且不妨设 $\Omega$ 连通. 先考虑 $Lu=\sum _{i,j=1}^n{a}_{ij}\frac{{\partial }^{2}u}{\partial x_i\partial x_j}+\sum _{i=1}^n{b}_i\frac{\partial u}{\partial x_i}>0$ 的情况. 假设 $u$ 在 $\Omega$ 内部某点取到最大值, 那么在该点 $\frac{\partial u}{\partial x_i}=0$, 且 $H=\left(\frac{{\partial }^{2}u}{\partial x_i\partial x_j}\right)$ 是半负定对称阵, 则 $H=-B^{T}B$ (对 $H=-P^{-1}\mathrm{diag}({\lambda }_i)P,P$ 是正交阵, 令 $B=P^{-1}\mathrm{diag}(\sqrt{{\lambda }_i})P$) . 记 $A=(a_{ij})$, 有 $Lu=\mathrm{tr}(AH)=-\mathrm{tr}(BA{B}^T)\le 0$, 矛盾. 当 $Lu\ge 0$, 取 $u_{\epsilon }(x)=u(x)+\epsilon e^{\alpha x_1}$, $\epsilon >0$, 则 $L{u}_{\epsilon }\ge \epsilon L{e}^{\alpha x_1}=\epsilon e^{\alpha x_1}(a_{11}{\alpha }^2+b_1\alpha )$. 由于 $a_{11}\ge \theta$, 取 $\alpha$ 很大就有 $L{u}_{\epsilon }>0$, 于是 $\underset{\overline{\Omega }}{\max }{u}_{\epsilon }=\underset{\partial \Omega }{\max }{u}_{\epsilon }$. 令 $\epsilon \to 0$ 就有 $\underset{\overline{\Omega }}{\max }u=\underset{\partial \Omega }{\max }u$. 注: 更一般的结果如下, 证明完全相同. 更多结论见 [GT] 3.1 节. 弱极值原理. 设 $u\in C^2(\Omega )\cap C(\overline{\Omega })$ 满足$$Lu=\sum _{i,j=1}^n{a}_{ij}(x)\frac{{\partial }^{2}u}{\partial x_i\partial x_j}+\sum _{i=1}^n{b}_i(x)\frac{\partial u}{\partial x_i}\ge 0,$$其中 $a_{ij}(x),b_i(x)$ 是有界函数, 且 $L$ 是严格椭圆的, 即 $\sum _{i,j=1}^n{a}_{ij}(x){\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ 对所有 $x\in \Omega ,\xi \in {\mathbb{R}}^n$ 以及某常数 $\theta >0$ 成立, 则$$\underset{\overline{\Omega }}{\max }u=\underset{\partial \Omega }{\max }u.$$用 $-u$ 代替 $u$, 得到 $\underset{\overline{\Omega }}{\min }u=\underset{\partial \Omega }{\min }u$.
2.	设 $K=\{x\in \Omega \mid u(x)=\underset{\Omega }{\max }u\}$ 非空, 假设 $K\ne \Omega$, 作与 $K$ 相切的球 $B\subset \Omega \backslash K$, 对 $B$ 使用强极值原理得到球内在切点附近 $u$ 的值更大, 矛盾.
3.	直接证明更一般的结果, 是 Hopf 极值原理的一种情形. 更多结论见 [GT] 3.2 节. 强极值原理. 设 $u\in C^2(\Omega )\cap C(\Omega \cup \{x^0\})$ 满足$$Lu=\sum _{i,j=1}^n{a}_{ij}(x)\frac{{\partial }^{2}u}{\partial x_i\partial x_j}+\sum _{i=1}^n{b}_i(x)\frac{\partial u}{\partial x_i}\ge 0,$$其中 $a_{ij}(x),b_i(x)$ 是有界函数, 且 $L$ 是严格椭圆的. 若$$u(x^0)>u(x)\text{对所有 }x\in \Omega ,$$并且存在与 $\partial \Omega$ 相切于 $x^0$ 的球 $B\subset \Omega$, 则对任何向外的方向 $v$ (即 $v\cdot \mathbf{n}(x^0)>0$) 有$$\underset{t\to 0^{+}}{\mathrm{liminf}}\frac{u(x^0)-u(x^0-tv)}{t}>0.$$ 不妨设 $B=B_r(0)$, 取 $x^0$ 的邻域 $\Sigma =\overline{B_{\frac{r}{2}}(x^0)\cap B}$. 考虑 $\Sigma$ 上的函数$$u_{\epsilon }(x)=u(x)+\epsilon (e^{-\alpha |x{|}^2}-e^{-\alpha r^2}),$$$\epsilon ,\alpha >0$ 均待定. 现在说明 $L{u}_{\epsilon }>0$, 只需验证 $L(e^{-\alpha |x{|}^2})>0$. 事实上, $$\begin{array}{rl}L\left(e^{-\alpha |x{|}^2}\right)& =e^{-\alpha |x{|}^2}(4{\alpha }^2\sum _{i,j=1}^n{a}_{ij}(x)x_i{x}_j-2\alpha \sum _{i=1}^n{a}_{ii}(x)-2\alpha \sum _{i=1}^n{b}_i(x)x_i),\end{array}$$由于 $\sum _{i,j=1}^n{a}_{ij}(x)x_i{x}_j\ge \theta |x{|}^2\ge \theta (\frac{r}{2}{)}^2$, 取 $\alpha$ 很大就有上式 $>0$. 由弱极值原理, $u_{\epsilon }$ 在 $\Sigma$ 的最大值在 $\partial \Sigma =(\partial B\cap B_{\frac{r}{2}}(x^0))\cup (\partial B_{\frac{r}{2}}(x^0)\cap \bar{B})$ 上取到; 而 $u_{\epsilon }$ 在 $\partial \Sigma$ 的最大值又在 $x^0$ 取到, 这是因为: ∘ 在 $\partial B\cap B_{\frac{r}{2}}(x^0)$ 有 $u_{\epsilon }=u$, 于是 $u_{\epsilon }(x^0)\ge u_{\epsilon }(x)$; ∘ 在 $\partial B_{\frac{r}{2}}(x^0)\cap \bar{B}$, $u_{\epsilon }(x^0)=u(x^0)\ge u(x)+\epsilon (e^{-\alpha (\frac{r}{2}{)}^2}-e^{-\alpha r^2})\ge u_{\epsilon }(x)$, 这只需要 $\epsilon$ 相对于 $u(x^0)-\underset{\partial B_{\frac{r}{2}}(x^0)\cap \bar{B}}{\max }u(x)>0$ 很小即可. 因此 $u_{\epsilon }$ 在 $\Sigma$ 的最大值在 $x^0$ 取到. 那么$$\underset{t\to 0^{+}}{\mathrm{liminf}}\frac{u_{\epsilon }(x^0)-u_{\epsilon }(x^0-tv)}{t}\ge 0,$$于是有$$\underset{t\to 0^{+}}{\mathrm{liminf}}\frac{u(x^0)-u(x^0-tv)}{t}\ge -\epsilon \frac{\partial }{\partial v}\left(e^{-\alpha |x{|}^2}\right)>0.$$
6.	$$\phi (x)=\frac{e^{\frac{1}{|x|-\frac{1}{2}}}}{e^{\frac{1}{|x|-\frac{1}{2}}}+e^{\frac{1}{\frac{3}{4}-|x|}}},x\in B_{\frac{3}{4}}\backslash B_{\frac{1}{2}}.$$

4Green 函数法

2.	[2] 对任意给定的 $x$, 由于 $G(x,\cdot )$ 是 $\Omega \backslash \{x\}$ 上的调和函数, 对任意的 $y\in \Omega \backslash \{x\}$, $$G(x,y)\ge \underset{\cdot \to \partial (\Omega \backslash \{x\})}{\mathrm{liminf}}G(x,\cdot )=0,$$上面的大于等于不能取等号, 不然 $G(x,\cdot )$ 是常数, 然而 $\underset{\cdot \to x}{\lim }G(x,\cdot )=+\infty$. 由于 $\gamma (x,\cdot )$ 是 $\Omega \backslash \{x\}$ 上的调和函数, $$\gamma (x,y)\ge \underset{\cdot \in \partial \Omega }{\inf }\gamma (x,\cdot )>\{\begin{array}{ll}0,& n\ge 3,\\ -\frac{1}{2\pi }\ln \mathrm{diam}(\Omega ),& n=2.\end{array}$$ [4] Green 恒等式中取 $u$ 为常数 $-1$. [3] 令 $G_x(z)=G(x,z),G_y(z)=G(y,z)$, 是调和函数. 取 $r<\frac{|x-y|}{2}$, 由 Green 公式, $$\begin{array}{rl}0& ={\int }_{\Omega \backslash (B_r(x)\cup B_r(y))}{G}_x\Delta G_y-G_y\Delta G_x\\ & ={\int }_{\partial \Omega }{G}_x\frac{\partial G_y}{\partial \mathbf{n}}-G_y\frac{\partial G_x}{\partial \mathbf{n}}-{\int }_{\partial B_r(x)}{G}_x\frac{\partial G_y}{\partial \mathbf{n}}-G_y\frac{\partial G_x}{\partial \mathbf{n}}-{\int }_{\partial B_r(y)}{G}_x\frac{\partial G_y}{\partial \mathbf{n}}-G_y\frac{\partial G_x}{\partial \mathbf{n}}.\end{array}$$ 而 $G_x,G_y$ 在 $\partial \Omega$ 上是 $0$, 于是$${\int }_{\partial B_r(x)}{G}_x\frac{\partial G_y}{\partial \mathbf{n}}-G_y\frac{\partial G_x}{\partial \mathbf{n}}={\int }_{\partial B_r(y)}{G}_y\frac{\partial G_x}{\partial \mathbf{n}}-G_x\frac{\partial G_y}{\partial \mathbf{n}}.$$ 由于 ${\int }_{\partial B_r(x)}\frac{\partial G_x}{\partial \mathbf{n}}={\int }_{\partial B_r(y)}\frac{\partial G_y}{\partial \mathbf{n}}=-1$, 以及 ${\int }_{\partial B_r(x)}{G}_x$ 和 ${\int }_{\partial B_r(y)}{G}_y\sim |\partial B_r|\Gamma (r)\sim \{\begin{array}{ll}Cr\ln \frac{1}{r}\to 0,& n=2\\ Cr\to 0,& n\ge 3\end{array}$ 当 $r\to 0$, 在上式中让 $r\to 0$ 得到 $G_y(x)=G_x(y)$, 即 $G(y,x)=G(x,y)$.
6.	求 Neumann 问题$$\{\begin{array}{ll}-\Delta u=0,& x\in \Omega ,\\ \frac{\partial u}{\partial \mathbf{n}}=g,& x\in \partial \Omega .\end{array}$$解的表达式. 首先, 写出 Green 恒等式$$u(x)={\int }_{\partial \Omega }\Gamma (x-y)\frac{\partial u}{\partial {\mathbf{n}}_y}(y)-u(y)\frac{\partial \Gamma }{\partial {\mathbf{n}}_y}(x-y)d{S}_y.\text{\hspace{1em}(I)}$$设 $\gamma (x,y)$ 是 $y$ 的调和函数 (且无奇点) , 有$$0={\int }_{\partial \Omega }\gamma (x,y)\frac{\partial u}{\partial {\mathbf{n}}_y}(y)-u(y)\frac{\partial \gamma }{\partial {\mathbf{n}}_y}(x,y)d{S}_y.\text{\hspace{1em}(II-γ)}$$由此, $u(x)$ 的表达式是 $(\text{I})-(\text{II-}\gamma )$; 其背后的想法是这样的: ∘ 本来 Dirichlet 问题中, 对于 $y\in \partial \Omega$, $u(y)$ 是有的, 而 $\frac{\partial u}{\partial {\mathbf{n}}_y}(y)$ 没有; 故而引入 $\gamma (x,y)$, 使得 $G(x,y):=\Gamma (x-y)-\gamma (x,y)=0$, 从而干掉 $\frac{\partial u}{\partial {\mathbf{n}}_y}(y)$. ∘ 现在对于 Neumann 问题, 对于 $y\in \partial \Omega$, 有的是 $\frac{\partial u}{\partial {\mathbf{n}}_y}(y)$, 而 $u(y)$ 却没有; 所以要引进 $h(x,y)$, 使得 Neumann 函数 $H(x,y):=\Gamma (x-y)-h(x,y)$ 满足 $\frac{\partial H}{\partial {\mathbf{n}}_y}(x,y)=0$, 用来干掉 $u(y)$. 这就是说, 要取满足以下条件的 $h(x,y)$: $$\{\begin{array}{ll}{\Delta }_{y}h(x,y)=0,& y\in \Omega ,\\ \frac{\partial h}{\partial {\mathbf{n}}_y}(x,y)=\frac{\partial \Gamma }{\partial {\mathbf{n}}_y}(x-y),& y\in \partial \Omega .\end{array}\text{\hspace{1em}(III)}$$对于 $y$ 的调和函数 $h$ (和 $u$) , 同样有$$0={\int }_{\partial \Omega }h(x,y)\frac{\partial u}{\partial {\mathbf{n}}_y}(y)-u(y)\frac{\partial h}{\partial {\mathbf{n}}_y}(x,y)d{S}_y.\text{\hspace{1em}(II-h)}$$$(\text{I})-(\text{II-}h)$ 得到 $u(x)$ 的表达式$$u(x)={\int }_{\partial \Omega }H(x,y)\frac{\partial u}{\partial {\mathbf{n}}_y}(y)d{S}_y.\text{\hspace{1em}(IV)}$$不过, 以上是毛利大叔给出的解答. “あれれ, ” 细心的柯南立刻指出了两个问题: “好奇怪哦, 相差一个常数 $c$ 的 $u$ 明明都是 Neumann 问题的解, 而 $\text{(IV)}$ 却是完全确定的表达式; 就算 $h$ 能相差一个常数 $d$, 从而 $H$ 相差了 $d$, 但是相容条件$${\int }_{\partial \Omega }\frac{\partial u}{\partial {\mathbf{n}}_{\mathbf{y}}}(y)d{S}_y={\int }_{\Omega }\Delta u(y)dy=0$$却表明即使 $H$ 平移 $d$, 也不影响 $u(x)$ 的值. 还有, 留意到$${\int }_{\partial \Omega }\frac{\partial h}{\partial {\mathbf{n}}_y}(x,y)d{S}_y={\int }_{\Omega }{\Delta }_{y}h(x,y)dy=0,$$以及$${\int }_{\partial \Omega }\frac{\partial h}{\partial {\mathbf{n}}_y}(x,y)d{S}_y={\int }_{\partial \Omega }\frac{\partial \Gamma }{\partial {\mathbf{n}}_y}(x-y)d{S}_y=-1,$$这两个式子说明, $\text{(III)}$ 其实是自相矛盾的. ” (... 省略不少剧情...) 以下是柯南作出的推理: 由于不同的 $u$ 可以相差一个常数, 一开始就应该有所假设: 比如说限定$$\frac{1}{|\partial \Omega |}{\int }_{\partial \Omega }u=c,$$这样固定住了 $u$, 无法平移. 并且, 为了解决 $h$ 带来的矛盾, 其定义需调整为$$\{\begin{array}{ll}{\Delta }_{y}h(x,y)=0,& y\in \Omega ,\\ \frac{\partial h}{\partial {\mathbf{n}}_y}(x,y)=\frac{\partial \Gamma }{\partial {\mathbf{n}}_y}(x-y)+\frac{1}{|\partial \Omega |},& y\in \partial \Omega .\end{array}\text{\hspace{1em}(III’)}$$这样, $(\text{I})-(\text{II-}h)$ 实际上给出的是$$u(x)={\int }_{\partial \Omega }H(x,y)\frac{\partial u}{\partial {\mathbf{n}}_y}(y)d{S}_y+c.\text{\hspace{1em}(IV’)}$$这时有人问道, 这一切都不过是你的推理而已, 有什么证据说明确实有这样的 $h$ 存在吗? 一般来说, 这种问题都是求锤得锤的, 而且预示着案件的结束, 此次也不例外. 大概不太有人知道, 柯南除了变声器、麻醉枪等装备, 居然还有 [Ta I] 一书; 柯南把书翻到了 Chapter 5 Proposition 7.7, 给出了 $h$ 的存在性的确凿证据, 全集终.
9.	用静电源像法. (1) 设 $x$ 关于圆周的反演点是 $\tilde{x}$, $x,\tilde{x}$ 关于直径的对称点是 $x^{\ast },{\tilde{x}}^{\ast }$, Green 函数为$$G(x,y)=\Gamma (x-y)-\Gamma (\tilde{x}-y)-\Gamma (x^{\ast }-y)+\Gamma ({\tilde{x}}^{\ast }-y).$$ (2) 设 $x$ 关于 $\{x_{n-1}=0\}$ 的对称点是 $\tilde{x}$, $x,\tilde{x}$ 关于 $\{x_n=0\}$ 的对称点是 $x^{\ast },{\tilde{x}}^{\ast }$, Green 函数为$$G(x,y)=\Gamma (x-y)-\Gamma (\tilde{x}-y)-\Gamma (x^{\ast }-y)+\Gamma ({\tilde{x}}^{\ast }-y).$$
14.	由第 11 题, 只需证明 $u$ 满足平均值性质. 回忆平均值性质的证明过程, 只需证明对任何球 $B\subset \Omega$ 成立 ${\int }_{\partial B}\frac{\partial u}{\partial \mathbf{n}}=0$. 当 $B\subset {\Omega }_1$ 或 $B\subset {\Omega }_2$, ${\int }_{\partial B}\frac{\partial u}{\partial \mathbf{n}}={\int }_B\Delta u=0$; 当 $B$ 与 ${\Omega }_1,{\Omega }_2$ 均相交, 记 ${\mathbf{n}}_1,{\mathbf{n}}_2$ 为 ${\Omega }_1,{\Omega }_2$ 在 $B\cap S$ 上的单位外法向量, 方向相反, 则$$\begin{array}{rl}{\int }_{\partial B}\frac{\partial u}{\partial \mathbf{n}}& ={\int }_{\partial B\cap {\Omega }_1}\frac{\partial u}{\partial \mathbf{n}}+{\int }_{B\cap S}\frac{\partial u}{\partial {\mathbf{n}}_1}+{\int }_{\partial B\cap {\Omega }_2}\frac{\partial u}{\partial \mathbf{n}}+{\int }_{B\cap S}\frac{\partial u}{\partial {\mathbf{n}}_2}\\ & ={\int }_{\partial (B\cap {\Omega }_1)}\frac{\partial u}{\partial \mathbf{n}}+{\int }_{\partial (B\cap {\Omega }_2)}\frac{\partial u}{\partial \mathbf{n}}={\int }_{B\cap {\Omega }_1}\Delta u+{\int }_{B\cap {\Omega }_2}\Delta u=0.\end{array}$$

5特征值问题

1.	对于$$u(x)=A\cos \sqrt{\lambda }x+B\sin \sqrt{\lambda }x,\lambda >0,$$$u(0)=0$ 即是 $A=0$, $u^{\prime }(\pi )=0$ 说明 $B\sqrt{\lambda }\cos \sqrt{\lambda }\pi =0$, 有非零解当且仅当 ${\lambda }_k=(k+\frac{1}{2}{)}^2(k\in \mathbb{Z})$, 此时的解为 $u_k(x)=B\sin (k+\frac{1}{2})x$. 若还要 $\Vert u_k{\Vert }_{L^2}=1$, 那么 $B=\sqrt{\frac{2}{\pi }}$. 而对于 $u(x)=Ax+B$ 与 $u(x)=A{e}^{\sqrt{-\lambda }x}+B{e}^{-\sqrt{-\lambda }x}$, 都只有零解.
2.	对于 $x\in \Omega$, 记 $x=(x^{\prime },x_n)\in {\mathbb{R}}^{n-1}\times \mathbb{R}$, 有$$\begin{array}{rl}|u(x^{\prime },x_n)|& =|u(x^{\prime },x_n)-u(x^{\prime },0)|\\ & =|{\int }_0^{x_n}{\partial }_{n}u(x^{\prime },s)ds|\\ & \le x_n^{1-\frac{1}{p}}({\int }_0^1|{\partial }_{n}u(x^{\prime },s){|}^{p}ds{)}^{\frac{1}{p}}.\end{array}$$两边的 $p$ 次幂关于 $x$ 积分, 得到$$\begin{array}{rl}{\int }_{\Omega }|u(x){|}^{p}dx& ={\int }_{{\mathbb{R}}^{n-1}}{\int }_0^1|u(x^{\prime },x_n){|}^{p}d{x}_{n}d{x}^{\prime }\\ & \le {\int }_{{\mathbb{R}}^{n-1}}{\int }_0^1{x}_n^{p-1}{\int }_0^1|{\partial }_{n}u(x^{\prime },s){|}^{p}dsd{x}_{n}d{x}^{\prime }\\ & ={\int }_0^1{x}_n^{p-1}d{x}_n{\int }_{{\mathbb{R}}^{n-1}}{\int }_0^1|{\partial }_{n}u(x^{\prime },s){|}^{p}dsd{x}^{\prime }\\ & \le \frac{1}{p}{\int }_{\Omega }|\nabla u(x){|}^{p}dx,\end{array}$$即$$\Vert u{\Vert }_{L^p}\le p^{-\frac{1}{p}}\Vert \nabla u{\Vert }_{L^p}.$$注: 更多的 Poincaré 不等式, 见 [Le] 13.2 节.
3.	$\{u_j\}$ 正交是因为, 对 $i\ne j$ 有$$\begin{array}{rl}⟨u_i,u_j{⟩}_{H^1}& =⟨u_i,u_j⟩+⟨\nabla u_i,\nabla u_j⟩=⟨u_i-\Delta u_i,u_j⟩\\ & =(1+{\lambda }_i)⟨u_i,u_j⟩=0.\end{array}$$由于 $H_0^1(\Omega )=\overline{C_c^{\infty }(\Omega )}$, 要说明 $\{u_j\}$ 是完备系, 只需说明任何 $u\in C_c^{\infty }(\Omega )$ 可以用 $\{u_j\}$ 的线性组合在 $\Vert {\Vert }_{H^1}$ 范数下逼近: 事实上, $$\begin{array}{rl}& \Vert u-\sum _{j=1}^n⟨u,u_j⟩u_j{\Vert }_{H^1}^2\\ & =\Vert u-\sum _{j=1}^n⟨u,u_j⟩u_j{\Vert }_{L^2}^2+\Vert \nabla (u-\sum _{j=1}^n⟨u,u_j⟩u_j){\Vert }_{L^2}^2\\ & =\Vert u-\sum _{j=1}^n⟨u,u_j⟩u_j{\Vert }_{L^2}^2-{\int }_{\Omega }(u-\sum _{j=1}^n⟨u,u_j⟩u_j)\Delta (u-\sum _{j=1}^n⟨u,u_j⟩u_j)\text{因 }u\in C_0^{\infty }(\Omega )\\ & \le 2\Vert u-\sum _{j=1}^n⟨u,u_j⟩u_j{\Vert }_{L^2}^2+\Vert \Delta u-\sum _{j=1}^n⟨u,u_j⟩\Delta u_j{\Vert }_{L^2}^2\text{用 Cauchy 不等式}\\ & =2\Vert u-\sum _{j=1}^n⟨u,u_j⟩u_j{\Vert }_{L^2}^2+\Vert \Delta u-\sum _{j=1}^n⟨\Delta u,u_j⟩u_j{\Vert }_{L^2}^2\text{因为 }\Delta u_j={\lambda }_j{u}_j\text{ 及 }⟨u,\Delta u_j⟩=⟨\Delta u,u_j⟩\\ & \to 0\text{ 当 }n\to \infty .u,\Delta u\in L^2(\Omega ),\text{ 且 }\{u_j\}\text{ 是 }{L}^2(\Omega )\text{ 的完备正交系}\end{array}$$
0.	本题来自以前的讲义, 是要证明 Sturm-Liouville 定理, 即定理 1.5.1. 另一解法见 [Sa] Theorem 8.3 (P117). 1. 特征值 $\lambda$ 非负是因为$$\lambda ⟨u,u⟩=⟨-u^{\prime \prime },u⟩=⟨u^{\prime },u^{\prime }⟩-u^{\prime }u{|}_0^L\ge 0,$$${\lambda }_1=0$ 当且仅当存在特征函数 $u\ne 0$ 使 $u^{\prime }\equiv 0$, 当且仅当 ${\beta }_1={\beta }_2=0$. 设特征值 $\lambda$ 对应的解为$$u(x)=C_1\sin \mu x+C_2\cos \mu x,\mu =\sqrt{\lambda },$$则 $u^{\prime }(x)=C_1\mu \cos \mu x-C_2\mu \sin \mu x$, 代入边界条件得$$\{\begin{array}{rl}-{\alpha }_1{C}_1\mu & +{\beta }_1{C}_2=0,\\ {\alpha }_2(C_1\mu \cos \mu L-C_2\mu \sin \mu L)& +{\beta }_2(C_1\sin \mu L+C_2\cos \mu L)=0,\end{array}$$方程组有解 $(C_1,C_2)\ne 0$, 等价于$$\left|\begin{array}{cc}-{\alpha }_1\mu & {\beta }_1\\ {\alpha }_2\mu \cos \mu L+{\beta }_2\sin \mu L& -{\alpha }_2\mu \sin \mu L+{\beta }_2\cos \mu L\end{array}\right|=0,$$即$$({\alpha }_1{\alpha }_2{\mu }^2-{\beta }_1{\beta }_2)\sin \mu L=({\alpha }_1{\beta }_2+{\alpha }_2{\beta }_1)\mu \cos \mu L.$$分类讨论可以说明 (详见 [Zh] 定理 3.6) 所有特征值是$$0\le {\lambda }_1<{\lambda }_2<\cdots <{\lambda }_k<\cdots \to +\infty ,$$并且有渐近公式 ${\lambda }_k={\mu }_k^2\sim (\frac{k\pi }{L}{)}^2$. 2. 首先, 根据表达式有 $u_k\in C^{\infty }([0,L])$. 不同特征值的特征函数 $u_k,u_l$ 是正交的, 这是因为$$(-{\lambda }_k+{\lambda }_l)⟨u_k,u_l⟩=⟨u_k^{\prime \prime },u_l⟩-⟨u_k,u_l^{\prime \prime }⟩=u_k^{\prime }{u}_l-u_k{u}_l^{\prime }{|}_0^L=0,$$最后一步中, 在 $x=0,L$ 处$$\left(\begin{array}{cc}-{\alpha }_1& {\beta }_1\\ {\alpha }_2& {\beta }_2\end{array}\right)\left(\begin{array}{cc}{u}_k^{\prime }& u_l^{\prime }\\ u_k& u_l\end{array}\right)=0$$表明$$\left|\begin{array}{cc}{u}_k^{\prime }& u_l^{\prime }\\ u_k& u_l\end{array}\right|=0;$$由边界条件, 同一特征值的特征函数空间只有 $1$ 维. 第一、第二类边界条件下, 特征函数全体的完备性是标准的 Fourier 级数结论; 涉及第三类边界条件需要用自伴紧算子的性质. 3. 当 $k$ 非常大时, $u_k$ 快速振荡, $L^2$ 范数与 $C^0$ 范数差不多大: 具体来讲, 设 $u_k(x)=c_k\sin ({\mu }_{k}x+{\theta }_k)$, 有$$\begin{array}{rl}\Vert u_k{\Vert }_{L^2}^2& =|c_k{|}^2{\int }_0^L{\sin }^2({\mu }_{k}x+{\theta }_k)dx\\ & =|c_k{|}^2(\frac{L}{2}-\frac{\sin 2({\mu }_{k}x+{\theta }_k){|}_0^L}{4{\mu }_k})\sim \frac{L}{2}|c_k{|}^2=\frac{L}{2}\Vert u_k{\Vert }_{C^0}^2.\end{array}$$因此对于 $L^2$ 收敛的级数$$f=\sum _{k=1}^{\infty }{a}_k{u}_k,-f^{\prime \prime }=\sum _{k=1}^{\infty }{\lambda }_k{a}_k{u}_k,$$有$$\begin{array}{rl}\sum _{k=1}^{\infty }|a_k|\Vert u_k{\Vert }_{C^0}& \le C\sum _{k=1}^{\infty }|a_k|\Vert u_k{\Vert }_{L^2}\\ & \le C(\sum _{k=1}^{\infty }{\lambda }_k^2|a_k{|}^2\Vert u_k{\Vert }_{L^2}^2{)}^{\frac{1}{2}}(\sum _{k=1}^{\infty }{\lambda }_k^{-2}{)}^{\frac{1}{2}}\\ & \le C\Vert f^{\prime \prime }{\Vert }_{L^2}(\sum _{k=1}^{\infty }{k}^{-4}{)}^{\frac{1}{2}}<\infty .\end{array}$$

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