河南师范大学 2025年数学分析第0题

已知 $x = u e^w$, $y = v e^w$, $z = \omega e^w$。由于题目最终要求 $w(u,v)$ 的方程，且条件中出现 $w_u, w_v$，合理的假设是 $\omega = w$，即 $z = w e^w$。

对 $x = u e^w$ 两边对 $x$ 求偏导（$y$ 固定）：$1 = e^w u_x + u e^w w_x$。对 $y = v e^w$ 两边对 $x$ 求偏导：$0 = e^w v_x + v e^w w_x$，得 $v_x = -v w_x$。类似地，对 $x$ 对 $y$ 求偏导得 $u_y = -u w_y$，对 $y$ 对 $y$ 求偏导得 $1 = e^w v_y + v e^w w_y$，即 $v_y = 1 - v w_y$。

由 $w_x = w_u u_x + w_v v_x$，代入 $u_x = e^{-w} - u w_x$ 和 $v_x = -v w_x$，得 $w_x = w_u(e^{-w} - u w_x) + w_v(-v w_x)$，整理得 $w_x(1 + u w_u + v w_v) = w_u e^{-w}$，故 $w_x = \frac{w_u e^{-w}}{1 + u w_u + v w_v}$。同理，$w_y = w_u u_y + w_v v_y = w_u(-u w_y) + w_v(1 - v w_y)$，得 $w_y(1 + u w_u + v w_v) = w_v$，即 $w_y = \frac{w_v}{1 + u w_u + v w_v}$。

由 $z = w e^w$ 得 $z_x = e^w(1+w) w_x$, $z_y = e^w(1+w) w_y$。代入原方程 $(x z_x)^2 + (y z_y)^2 = z^2 z_x z_y$，左边 $= (u e^w \cdot e^w(1+w) w_x)^2 + (v e^w \cdot e^w(1+w) w_y)^2 = e^{4w}(1+w)^2 (u^2 w_x^2 + v^2 w_y^2)$，右边 $= w^2 e^{2w} \cdot e^{2w}(1+w)^2 w_x w_y = w^2 e^{4w}(1+w)^2 w_x w_y$。约去 $e^{4w}(1+w)^2$ 得 $u^2 w_x^2 + v^2 w_y^2 = w^2 w_x w_y$。

将 $w_x = \frac{w_u e^{-w}}{A}$, $w_y = \frac{w_v}{A}$ 代入 $u^2 w_x^2 + v^2 w_y^2 = w^2 w_x w_y$，得 $u^2 \left(\frac{w_u e^{-w}}{A}\right)^2 + v^2 \left(\frac{w_v}{A}\right)^2 = w^2 \cdot \frac{w_u e^{-w}}{A} \cdot \frac{w_v}{A}$。两边乘以 $A^2$ 得 $u^2 w_u^2 e^{-2w} + v^2 w_v^2 = w^2 w_u w_v e^{-w}$。再乘以 $e^{2w}$ 得 $u^2 w_u^2 + v^2 w_v^2 e^{2w} = w^2 w_u w_v e^{w}$。