华南理工大学 2025年数学分析第11题

设 $u = x + y$, $v = \frac{1}{x} + \frac{1}{y}$，则 $z = z(x, y)$ 可视为 $z = z(u, v)$。由链式法则： $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}$$ $$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}$$ 计算偏导：$\frac{\partial u}{\partial x} = 1$, $\frac{\partial u}{\partial y} = 1$, $\frac{\partial v}{\partial x} = -\frac{1}{x^2}$, $\frac{\partial v}{\partial y} = -\frac{1}{y^2}$。 因此： $$z_x = z_u - \frac{1}{x^2} z_v, \quad z_y = z_u - \frac{1}{y^2} z_v$$

对 $z_x = z_u - \frac{1}{x^2} z_v$ 再对 $x$ 求偏导： $$z_{xx} = \frac{\partial}{\partial x} z_u - \frac{\partial}{\partial x}\left(\frac{1}{x^2}\right) z_v - \frac{1}{x^2} \frac{\partial}{\partial x} z_v$$ 其中 $\frac{\partial}{\partial x} z_u = z_{uu} \cdot 1 + z_{uv} \cdot \left(-\frac{1}{x^2}\right)$，$\frac{\partial}{\partial x} z_v = z_{vu} \cdot 1 + z_{vv} \cdot \left(-\frac{1}{x^2}\right)$，且 $\frac{\partial}{\partial x}\left(\frac{1}{x^2}\right) = -\frac{2}{x^3}$。 代入并整理（利用混合偏导连续，$z_{uv}=z_{vu}$）： $$z_{xx} = z_{uu} - \frac{2}{x^2} z_{uv} + \frac{1}{x^4} z_{vv} + \frac{2}{x^3} z_v$$

类似地，对 $z_y = z_u - \frac{1}{y^2} z_v$ 对 $y$ 求偏导： $$z_{yy} = \frac{\partial}{\partial y} z_u - \frac{\partial}{\partial y}\left(\frac{1}{y^2}\right) z_v - \frac{1}{y^2} \frac{\partial}{\partial y} z_v$$ 其中 $\frac{\partial}{\partial y} z_u = z_{uu} \cdot 1 + z_{uv} \cdot \left(-\frac{1}{y^2}\right)$，$\frac{\partial}{\partial y} z_v = z_{vu} \cdot 1 + z_{vv} \cdot \left(-\frac{1}{y^2}\right)$，$\frac{\partial}{\partial y}\left(\frac{1}{y^2}\right) = -\frac{2}{y^3}$。 代入整理得： $$z_{yy} = z_{uu} - \frac{2}{y^2} z_{uv} + \frac{1}{y^4} z_{vv} + \frac{2}{y^3} z_v$$

对 $z_x = z_u - \frac{1}{x^2} z_v$ 对 $y$ 求偏导： $$z_{xy} = \frac{\partial}{\partial y} z_u - \frac{1}{x^2} \frac{\partial}{\partial y} z_v$$ 其中 $\frac{\partial}{\partial y} z_u = z_{uu} \cdot 1 + z_{uv} \cdot \left(-\frac{1}{y^2}\right)$，$\frac{\partial}{\partial y} z_v = z_{vu} \cdot 1 + z_{vv} \cdot \left(-\frac{1}{y^2}\right)$。 代入并整理： $$z_{xy} = z_{uu} - \frac{1}{y^2} z_{uv} - \frac{1}{x^2} \left( z_{uv} - \frac{1}{y^2} z_{vv} \right) = z_{uu} - \left(\frac{1}{x^2} + \frac{1}{y^2}\right) z_{uv} + \frac{1}{x^2 y^2} z_{vv}$$

原方程为：$x^2 z_{xx} - (x^2 + y^2) z_{xy} + y^2 z_{yy} = 0$。 分别计算各项： $$x^2 z_{xx} = x^2 z_{uu} - 2 z_{uv} + \frac{1}{x^2} z_{vv} + \frac{2}{x} z_v$$ $$-(x^2+y^2) z_{xy} = -(x^2+y^2) z_{uu} + (x^2+y^2)\left(\frac{1}{x^2}+\frac{1}{y^2}\right) z_{uv} - \frac{x^2+y^2}{x^2 y^2} z_{vv}$$ $$y^2 z_{yy} = y^2 z_{uu} - 2 z_{uv} + \frac{1}{y^2} z_{vv} + \frac{2}{y} z_v$$ 合并同类项： - $z_{uu}$ 系数：$x^2 - (x^2+y^2) + y^2 = 0$，消去。 - $z_{uv}$ 系数：$-2 + (x^2+y^2)\left(\frac{1}{x^2}+\frac{1}{y^2}\right) - 2 = \frac{(x^2+y^2)^2}{x^2 y^2} - 4$。 - $z_{vv}$ 系数：$\frac{1}{x^2} - \frac{x^2+y^2}{x^2 y^2} + \frac{1}{y^2} = \frac{y^2 - (x^2+y^2) + x^2}{x^2 y^2} = 0$，消去。 - $z_v$ 系数：$\frac{2}{x} + \frac{2}{y} = 2\left(\frac{1}{x} + \frac{1}{y}\right) = 2v$。 因此方程化为： $$\left[\frac{(x^2+y^2)^2}{x^2 y^2} - 4\right] z_{uv} + 2v z_v = 0$$

注意到 $u = x+y$，$v = \frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = \frac{u}{xy}$，所以 $xy = \frac{u}{v}$。 计算 $x^2 + y^2 = (x+y)^2 - 2xy = u^2 - \frac{2u}{v}$。 则 $\frac{(x^2+y^2)^2}{x^2 y^2} = \frac{(u^2 - 2u/v)^2}{(u/v)^2} = \frac{u^2 (u - 2/v)^2}{u^2/v^2} = v^2 \left(u - \frac{2}{v}\right)^2 = (uv - 2)^2$。 因此系数化为：$(uv - 2)^2 - 4 = u^2 v^2 - 4uv + 4 - 4 = u^2 v^2 - 4uv = uv(uv - 4)$。 代入方程得： $$uv(uv - 4) z_{uv} + 2v z_v = 0$$ 两边除以 $v$（假设 $v \neq 0$）： $$u(uv - 4) z_{uv} + 2 z_v = 0$$