我们考虑上半圆盘$D$上最简单的Laplace方程:
\begin{equation}\label{eq:n}
\begin{cases}
\Delta u=f\in L^2(D),&x\in D\\
\frac{\partial u}{\partial \nu}=0,&x\in\partial D
\end{cases}
\end{equation}
与
\begin{equation}\label{eq:d}
\begin{cases}
\Delta u=f\in L^2(D),&x\in D\\
u=0,&x\in\partial D.
\end{cases}
\end{equation}
事实上, 对\eqref{eq:n}只需验证, 对$\varphi\in C^\infty(B)$, 成立
\begin{equation}\label{eq:wn}
\int_B\nabla w\nabla\varphi+w\varphi=0.
\end{equation}
而对\eqref{eq:d}只需验证, 对$\varphi\in C_0^\infty(B)$, 成立
\begin{equation}\label{eq:wd}
\int_B\nabla w\nabla\varphi+w\varphi=0.
\end{equation}
首先, 对\eqref{eq:n}的解来验证。对$\varphi\in C^\infty(B)$, 令
$$
\varphi_e=\frac{1}{2}(\varphi(x)+\varphi(x^*)),\quad
\varphi_o=\frac{1}{2}(\varphi(x)-\varphi(x^*))
$$
分别为其偶部分与奇部分。直接计算得到：
\begin{align*}
\int_B\nabla w\nabla\varphi+w\varphi&=\int_B\nabla w\nabla\varphi_e+w\varphi_e+
\int_B\nabla w\nabla\varphi_o+w\varphi_o\\
\int_B\nabla w\nabla\varphi_e+w\varphi_e&=\int_D\nabla u\nabla\varphi_e+u\varphi_e+\int_{D^-}\nabla u(x^*)\nabla\varphi_e+u(x^*)\varphi_e\\
\int_{D^-}\nabla u(x^*)\nabla\varphi_e+u(x^*)\varphi_edx&=\int_{D}\nabla u\nabla\varphi_e(x^*)+u\varphi_e(x^*)\\
&=\int_D\nabla u\nabla\varphi_e+u\varphi_e\\
\int_B\nabla w\nabla\varphi_o+w\varphi_o&=\int_D\nabla w\nabla\varphi_o+w\varphi_o+\int_{D^-}\nabla w\nabla\varphi_o+w\varphi_o\\
&=\int_D\nabla u\nabla\varphi_o+u\varphi_o+\int_{D^-}\nabla u(x^*)\nabla\varphi_o+u(x^*)\varphi_o\\
\int_{D^-}\nabla u(x^*)\nabla\varphi_o+u(x^*)\varphi_o&=\int_D\nabla u\nabla\varphi_o(x^*)+u\varphi_o(x^*)\\
&=-\int_D\nabla u\nabla\varphi_o+u\varphi_o
\end{align*}
由于$\varphi\in C^\infty(B)$故$\varphi_e\in C^\infty(D)$. 而$u$是\eqref{eq:n}在$D$上的$W^{1,2}$弱解, 于是
$$
\int_D\nabla u\nabla\varphi_e+u\varphi_e=0.
$$
这样, 我们就证明了
$$
\int_B\nabla w\nabla\varphi+w\varphi=0,\quad\varphi\in C^\infty(B).
$$
对\eqref{eq:d}的解的验证是类似的. 只需注意到, 对$\varphi=\varphi_e+\varphi_o\in C_0^\infty(B)$, 和前面对Neumann情形的验证一样, 我们仍然可以得到(注意此时$w$是奇延拓),
\begin{align*}
\int_B\nabla w\nabla\varphi+w\varphi&=\int_B\nabla w\nabla\varphi_e+w\varphi_e+
\int_B\nabla w\nabla\varphi_o+w\varphi_o\\
\int_B\nabla w\nabla\varphi_e+w\varphi_e&=\int_D\nabla u\nabla\varphi_e+u\varphi_e\color{red}{-}\int_{D^-}\nabla u(x^*)\nabla\varphi_e+u(x^*)\varphi_e\\
&=0\\
\int_B\nabla w\nabla\varphi_o+w\varphi_o&=\int_D\nabla w\nabla\varphi_o+w\varphi_o+\int_{D^-}\nabla w\nabla\varphi_o+w\varphi_o\\
&=\int_D\nabla u\nabla\varphi_o+u\varphi_o\color{red}{-}\int_{D^-}\nabla u(x^*)\nabla\varphi_o+u(x^*)\varphi_o\\
&=2\int_D\nabla u\nabla\varphi_o+u\varphi_o.
\end{align*}
由于$\varphi\in C_0^\infty(B)$, 我们知道$\varphi_o\in C_0^\infty(B)$且$\varphi_o|_{\partial^0D}=0$. 这里, $\partial^0D=\{x=(x^1,x^2)\in D:x_2=0\}$, $\partial^+D=\{x\in\partial D:x^2>0\}$. 因此, 实际上$\varphi_o\in C_0^\infty(D)$. 故由$u$是\eqref{eq:d}在$D$上的弱解知道
$$
\int_D\nabla u\nabla\varphi_o+u\varphi_o=0.
$$
这样, 我们就证明了
$$
\int_B\nabla w\nabla\varphi+w\varphi=0,\quad\varphi\in C_0^\infty(B).
$$