方企勤 第七章 典型综合题分析 第22题
📝 题目
例 22 设 $u = u\left( {x,y}\right)$ 在平面区域 $D$ 上二阶连续可微. 求证:
$$ \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} \geq 0\;\left( {\forall \left( {x,y}\right) \in D}\right) $$
成立的充要条件为
$$ u\left( {{x}_{0},{y}_{0}}\right) \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {{x}_{0} + r\cos \theta ,{y}_{0} + r\sin \theta }\right) \mathrm{d}\theta \;\left( {\forall \left( {{x}_{0},{y}_{0}}\right) \in D}\right) , $$
其中 $0 \leq r < d\left( {{x}_{0},{y}_{0}}\right) ,d$ 是点 $\left( {{x}_{0},{y}_{0}}\right)$ 到 $D$ 的边界 $\partial D$ 的距离.
💡 答案解析
证 设 ${c}_{r}\left( {{x}_{0},{y}_{0}}\right)$ 和 ${\Delta }_{r}\left( {{x}_{0},{y}_{0}}\right)$ 分别表示以 $\left( {{x}_{0},{y}_{0}}\right)$ 为中心,以 $r$ 为半径的圆周和圆盘. 由格林公式
$$ {\iint }_{{\Delta }_{r}\left( {{x}_{0},{y}_{0}}\right) }\left( {\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y = {\int }_{{c}_{r}\left( {{x}_{0},{y}_{0}}\right) }\frac{\partial u}{\partial n}\mathrm{\;d}s, \tag{7.40} $$
其中 $n$ 为圆 ${c}_{r}\left( {{x}_{0},{y}_{0}}\right)$ 的外法线方向.
如果 $\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} \geq 0\left( {\forall \left( {x,y}\right) \in D}\right)$ ,对 $\forall \left( {{x}_{0},{y}_{0}}\right) \in D$ ,由 (7.40) 式有
$$ {\int }_{{c}_{r}\left( {{x}_{0},{y}_{0}}\right) }\frac{\partial u}{\partial n}\mathrm{\;d}s = {\int }_{{c}_{r}}\frac{\partial u}{\partial r}\mathrm{\;d}s = r\frac{\partial }{\partial r}{\int }_{0}^{2\pi }u\left( {{x}_{0} + r\cos \theta ,{y}_{0} + r\sin \theta }\right) \mathrm{d}\theta \geq 0, $$
从而
$$ f\left( r\right) \overset{\text{ 定义 }}{ = }{\int }_{0}^{2\pi }u\left( {{x}_{0} + r\cos \theta ,{y}_{0} + r\sin \theta }\right) \mathrm{d}\theta $$
在 $\lbrack 0,d)$ 上单调增加,于是 $f\left( r\right) \geq f\left( 0\right)$ ,即
$$ {2\pi u}\left( {{x}_{0},{y}_{0}}\right) \leq {\int }_{0}^{2\pi }u\left( {{x}_{0} + r\cos \theta ,{y}_{0} + r\sin \theta }\right) \mathrm{d}\theta . \tag{7.41} $$
反之,如果 (7.41) 式对 $\forall \left( {{x}_{0},{y}_{0}}\right) \in D$ 成立,要证
$$ \frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}} \geq 0\;\left( {\forall \left( {x,y}\right) \in D}\right) , \tag{7.42} $$
用反证法. 假设 $\exists \left( {{x}_{1},{y}_{1}}\right) \in D$ ,使得
$$ {u}_{xx}^{\prime \prime }\left( {{x}_{1},{y}_{1}}\right) + {u}_{yy}^{\prime \prime }\left( {{x}_{1},{y}_{1}}\right) < 0. $$
由二阶偏导数连续性, $\exists 0 < \delta < d\left( {{x}_{1},{y}_{1}}\right)$ ,使得
$$ {u}_{xx}^{\prime \prime }\left( {x,y}\right) + {u}_{yy}^{\prime \prime }\left( {x,y}\right) < 0\;\left( {\forall \left( {x,y}\right) \in {\Delta }_{\delta }\left( {{x}_{1},{y}_{1}}\right) }\right) . $$
取 $0 < r < \delta$ ,则由格林公式有
$$ {\int }_{{c}_{r}\left( {{x}_{1},{y}_{1}}\right) }\frac{\partial u}{\partial n}\mathrm{\;d}s = {\int }_{{c}_{r}\left( {{x}_{1},{y}_{1}}\right) }\frac{\partial u}{\partial r}\mathrm{\;d}s < 0, $$
即得
$$ \frac{\partial }{\partial r}{\int }_{0}^{2\pi }u\left( {{x}_{1} + r\cos \theta ,{y}_{1} + r\sin \theta }\right) \mathrm{d}\theta < 0. $$
从而 ${f}_{1}\left( r\right) = {\int }_{0}^{2\pi }u\left( {{x}_{1} + r\cos \theta ,{y}_{1} + r\sin \theta }\right) \mathrm{d}\theta$ 在 $\left\lbrack {0,\delta }\right\rbrack$ 上严格单调下降, 于是 ${f}_{1}\left( r\right) < {f}_{1}\left( 0\right) \left( {0 < r < \delta }\right)$ ,即
$$ {2\pi u}\left( {{x}_{1},{y}_{1}}\right) > {\int }_{0}^{2\pi }u\left( {{x}_{1} + r\cos \theta ,{y}_{1} + r\sin \theta }\right) \mathrm{d}\theta , $$
这与假设矛盾. 从而 (7.42) 式成立.