📝 题目
例 27 设 $D : {x}^{2} + {y}^{2} < 1.f\left( {x,y}\right)$ 为有界正值函数,在 $D$ 上有二阶连续偏导数, 且满足
$\Delta \ln f\left( {x,y}\right) \geq {f}^{2}\left( {x,y}\right) \;\left( {\Delta \text{ 为拉普拉斯算符,即 }\Delta = \frac{{\partial }^{2}}{\partial {x}^{2}} + \frac{{\partial }^{2}}{\partial {y}^{2}}}\right) .$
💡 答案与解析
证明
$$ f\left( {x,y}\right) \leq \frac{2}{1 - {x}^{2} - {y}^{2}}\;\left( {\forall \left( {x,y}\right) \in D}\right) . $$
证 令 $g\left( {x,y}\right) \overset{\text{ 定义 }}{ = }\frac{2}{1 - {x}^{2} - {y}^{2}}$ ,则
$$ \Delta \ln g\left( {x,y}\right) = \frac{4}{{\left( 1 - {x}^{2} - {y}^{2}\right) }^{2}} = {g}^{2}\left( {x,y}\right) , $$
所以
$$ \Delta \left( {\ln g\left( {x,y}\right) - \ln f\left( {x,y}\right) }\right) \leq {g}^{2}\left( {x,y}\right) - {f}^{2}\left( {x,y}\right) . \tag{7.48} $$
记函数 $F\left( {x,y}\right) = \ln g\left( {x,y}\right) - \ln f\left( {x,y}\right) = \ln \frac{g\left( {x,y}\right) }{f\left( {x,y}\right) }$ ,由条件可得出
$$ \mathop{\lim }\limits_{{\left( {x,y}\right) \rightarrow \partial D}}F\left( {x,y}\right) = + \infty . $$
所以函数 $F\left( {x,y}\right)$ 在 $D$ 内某一点达到最小值,设最小值点为 $\left( {{x}_{0},{y}_{0}}\right)$ ,则
$$ \ln \frac{g\left( {x,y}\right) }{f\left( {x,y}\right) } = F\left( {x,y}\right) \geq F\left( {{x}_{0},{y}_{0}}\right) = \ln \frac{g\left( {{x}_{0},{y}_{0}}\right) }{f\left( {{x}_{0},{y}_{0}}\right) } $$
$$ \left( {\forall \left( {x,y}\right) \in D}\right) \text{ , } \tag{7.49} $$
且在该点处 $\frac{{\partial }^{2}F\left( {{x}_{0},{y}_{0}}\right) }{\partial {x}^{2}} \geq 0,\frac{{\partial }^{2}F\left( {{x}_{0},{y}_{0}}\right) }{\partial {y}^{2}} \geq 0$ (否则与 $\left( {{x}_{0},{y}_{0}}\right)$ 是 $F$ 最小值矛盾). 相加即得 $\Delta \left( {\ln g\left( {{x}_{0},{y}_{0}}\right) - \ln f\left( {{x}_{0},{y}_{0}}\right) }\right) \geq 0$ . 再由 (7.48) 式得出 ${g}^{2}\left( {{x}_{0},{y}_{0}}\right) - {f}^{2}\left( {{x}_{0},{y}_{0}}\right) \geq 0$ ,进而可得 $g\left( {{x}_{0},{y}_{0}}\right) /f\left( {{x}_{0},{y}_{0}}\right) \geq$ 1. 把结果代入 (7.49) 式得到
$$ \ln \frac{g\left( {x,y}\right) }{f\left( {x,y}\right) } \geq \ln \frac{g\left( {{x}_{0},{y}_{0}}\right) }{f\left( {{x}_{0},{y}_{0}}\right) } \geq 0\;\left( {\forall \left( {x,y}\right) \in D}\right) , $$
即有
$$ f\left( {x,y}\right) \leq g\left( {x,y}\right) = \frac{2}{1 - {x}^{2} - {y}^{2}}\;\left( {\forall \left( {x,y}\right) \in D}\right) . $$