设 ${\bf u}$ 为满足第 3 题中条件的解. 证明 ${\bf u}$ 为如下变分问题 $$\bex \min_{{\bf w}\in A}\cfrac{1}{2}\int_\Omega |{\bf w}|^2\rd x \eex$$ 的解, 其中 $$\bex A=\sed{{\bf w}\in C^1(\Omega)\cap C^0(\bar \Omega);\ \Div{\bf w}=0,\mbox{in }\Omega;\ {\bf w}\cdot {\bf n}={\bf u}_B\cdot{\bf n}\mbox{ on }\vGa}. \eex$$
证明: 对 $\forall\ {\bf w}\in A$, 令 ${\bf v}={\bf w}-{\bf u}$, 则 $$\bex \Div{\bf v}=0,\mbox{in }\Omega;\quad {\bf v}\cdot{\bf n}=0,\mbox{on }\vGa. \eex$$ 于是 $$\beex \bea \cfrac{1}{2}\int_\Omega |{\bf w}|^2\rd x &=\cfrac{1}{2}\int_\Omega |{\bf u}+{\bf v}|^2\rd x\\ &=\cfrac{1}{2}\int_\Omega |{\bf u}|^2+2{\bf u}\cdot{\bf v}+|{\bf v}|^2\rd x\\ &=\cfrac{1}{2}\int_\Omega |{\bf u}|^2 -2\n\phi\cdot{\bf v}+|{\bf v}|^2\rd x\\ &=\cfrac{1}{2}\int_\Omega |{\bf u}|^2+2\phi \cdot\Div{\bf v}+|{\bf v}|^2\rd x -\int_\vGa \phi{\bf v}\cdot{\bf n}\rd S\\ &=\cfrac{1}{2}\int_\Omega |{\bf u}|^2+|{\bf v}|^2\rd x\\ &\geq \cfrac{1}{2}\int_\Omega |{\bf u}|^2\rd x. \eea \eeex$$
