2019 Putnam B4

Let $\mathcal{F}$ be the set of functions $f(x,y)$ that are twice continuously differentiable for $x \geq 1$, $y \geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives):

\begin{gather*}

xf_x + yf_y = xy \ln(xy), \\

x^2 f_{xx} + y^2 f_{yy} = xy.

\end{gather*}

For each $f \in \mathcal{F}$, let

$$

m(f) = \min_{s \geq 1} \left(f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s) \right).

$$

Determine $m(f)$, and show that it is independent of the choice of $f$.

令 $\mathcal{F}$ 为函数集 $f(x,y)$，对于 $x \geq 1$、$y \geq 1$ 两次连续可微，并且满足以下两个方程(其中下标表示偏导数)：

\开始{收集*}

xf_x + yf_y = xy \ln(xy), \\

x^2 f_{xx} + y^2 f_{yy} = xy。

\结束{聚集*}

对于每个 $f \in \mathcal{F}$，令

$$

m(f) = \min_{s \geq 1} \left(f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s) \right)。

$$

确定$m(f)$，并证明它与$f$的选择无关。