方企勤 第六章 多元函数积分学 第8题
📝 题目
证 设 $u\left( {x,y}\right)$ 为 $D$ 上调和函数, $L$ 所围区域记为 $\Omega \subset D$ . 由方向导数公式和格林公式, 即得
$$ {\oint }_{L}\frac{\partial u}{\partial \mathbf{n}}\mathrm{d}s = {\oint }_{L}\left( {\frac{\partial u}{\partial x}\cos \langle n,\mathbf{i}\rangle + \frac{\partial u}{\partial y}\cos \langle n,\mathbf{j}\rangle }\right) \mathrm{d}s $$
$$ = {\iint }_{\Omega }\left( {\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y = 0. $$
反之,设 $\displaystyle{\oint }_{L}\frac{\partial u}{\partial \mathbf{n}}\mathrm{d}s = 0}$ ,要证 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} \equiv 0$ ,我们用反证法. 假设 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} ≢ 0$ ,则它在某一点 $\left( {{x}_{0},{y}_{0}}\right) \in D$ 不为零. 无妨设它的值 $a$ 为正的. 由连续性, $\exists \delta > 0$ ,使 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}$ 在 ${\left( x - {x}_{0}\right) }^{2} + {\left( y - {y}_{0}\right) }^{2} \leq {\delta }^{2}$ 上的值大于等于 $a/2$ (必要时可缩小 $\delta$ ,使圆包含在 $D$ 内). 取圆周 $L$ : ${\left( x - {x}_{0}\right) }^{2} + {\left( y - {y}_{0}\right) }^{2} = {\delta }^{2}$ ,则
$$ {\oint }_{L}\frac{\partial u}{\partial \mathbf{n}}\mathrm{d}s = {\iint }_{{\left( x - {x}_{0}\right) }^{2} + {\left( y - {y}_{0}\right) }^{2} \leq {\delta }^{2}}\left( {\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y $$
$$ \geq \frac{a}{2} \cdot \pi {\delta }^{2} > 0. $$
与已知条件矛盾. 这矛盾说明 $u\left( {x,y}\right)$ 为调和函数.
💡 答案解析
证 设 $u\left( {x,y}\right)$ 为 $D$ 上调和函数, $L$ 所围区域记为 $\Omega \subset D$ . 由方向导数公式和格林公式, 即得
$$ {\oint }_{L}\frac{\partial u}{\partial \mathbf{n}}\mathrm{d}s = {\oint }_{L}\left( {\frac{\partial u}{\partial x}\cos \langle n,\mathbf{i}\rangle + \frac{\partial u}{\partial y}\cos \langle n,\mathbf{j}\rangle }\right) \mathrm{d}s $$
$$ = {\iint }_{\Omega }\left( {\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y = 0. $$
反之,设 $\displaystyle{\oint }_{L}\frac{\partial u}{\partial \mathbf{n}}\mathrm{d}s = 0}$ ,要证 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} \equiv 0$ ,我们用反证法. 假设 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} ≢ 0$ ,则它在某一点 $\left( {{x}_{0},{y}_{0}}\right) \in D$ 不为零. 无妨设它的值 $a$ 为正的. 由连续性, $\exists \delta > 0$ ,使 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}$ 在 ${\left( x - {x}_{0}\right) }^{2} + {\left( y - {y}_{0}\right) }^{2} \leq {\delta }^{2}$ 上的值大于等于 $a/2$ (必要时可缩小 $\delta$ ,使圆包含在 $D$ 内). 取圆周 $L$ : ${\left( x - {x}_{0}\right) }^{2} + {\left( y - {y}_{0}\right) }^{2} = {\delta }^{2}$ ,则
$$ {\oint }_{L}\frac{\partial u}{\partial \mathbf{n}}\mathrm{d}s = {\iint }_{{\left( x - {x}_{0}\right) }^{2} + {\left( y - {y}_{0}\right) }^{2} \leq {\delta }^{2}}\left( {\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y $$
$$ \geq \frac{a}{2} \cdot \pi {\delta }^{2} > 0. $$
与已知条件矛盾. 这矛盾说明 $u\left( {x,y}\right)$ 为调和函数.