证明: $f(x,y)\leq \frac{2}{1-x^2-y^2}, \forall\ (x,y)\in B$.
本帖最后由 flyy 于 2023-5-13 07:53 编辑令 $B=\left\{(x,y); x^2+y^2\leq 1\right\}$. 设$f\in C^2(B)$ 为正值函数, 满足
$\frac{\partial^2\ln f}{\partial x^2}+\frac{\partial^2\ln f}{\partial y^2}\geq f^2(x,y), \forall\ (x,y)\in B. $
证明:$f(x,y)\leq \frac{2}{1-x^2-y^2}, \forall\ (x,y)\in B$.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}, g=2\ln f=\ln (f^2)$, 则
$$\begin{aligned} \Delta g=2\Delta \ln f\geq 2f^2=2\mathrm{e}^g. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
再设 $\displaystyle h=2\ln \frac{2}{1-x^2-y^2}$, 则 $\displaystyle \Delta h=2\mathrm{e}^h$. 令 $\displaystyle w=g-h$, 则
$$\begin{aligned} \Delta w=2(\mathrm{e}^g-\mathrm{e}^h)\xlongequal[\tiny\mbox{中值}]{\tiny\mbox{Lagrange}} 2\mathrm{e}^\xi (g-h) =2\mathrm{e}^\xi w. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由 $\displaystyle \lim_{x^2+y^2\to 1^-}w=-\infty$ 知 $\displaystyle M=\max_{\overline{B}}w$ 必在 $\displaystyle B$ 内某 $\displaystyle (x_0,y_0)$ 处取得. 由
$$\begin{aligned} \frac{1}{2}w_{xx}(x_0,y_0) =\lim_{h\to 0}\frac{w(x_0+h,y_0)+w(x_0-h,y_0)-2w(x_0,y_0)}{h^2}\leq 0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} &0\geq \Delta w(x_0,y_0)=2\mathrm{e}^{\xi(x_0,y_0)}w(x_0,y_0)\\ \Rightarrow& g-h=w\leq M=w(x_0,y_0)\leq 0 \Rightarrow f(x,y)\leq \frac{2}{1-x^2-y^2}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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