江南大学 2026年数学分析第0题

设 $u = x - z$, $v = y - z$，则原方程 $F(u, v) = 0$。对 $x$ 求偏导：$F_u \cdot (1 - z_x) + F_v \cdot (-z_x) = 0$，整理得 $F_u - (F_u + F_v)z_x = 0$，故 $z_x = \frac{F_u}{F_u + F_v}$。对 $y$ 求偏导：$F_u \cdot (-z_y) + F_v \cdot (1 - z_y) = 0$，整理得 $F_v - (F_u + F_v)z_y = 0$，故 $z_y = \frac{F_v}{F_u + F_v}$。记分母 $D = F_u + F_v$。

由 $z_x = \frac{F_u}{D}$ 可得 $1 - z_x = 1 - \frac{F_u}{D} = \frac{D - F_u}{D} = \frac{F_v}{D}$，以及 $-z_x = -\frac{F_u}{D}$。同理，由 $z_y = \frac{F_v}{D}$ 可得 $1 - z_y = \frac{F_u}{D}$，$-z_y = -\frac{F_v}{D}$。这些关系将用于后续二阶偏导的计算。

对 $z_x = \frac{F_u}{D}$ 再对 $x$ 求导，使用商的求导法则：$z_{xx} = \frac{(F_{uu}u_x + F_{uv}v_x)D - F_u[(F_{uu}+F_{uv})u_x + (F_{uv}+F_{vv})v_x]}{D^2}$。代入 $u_x = \frac{F_v}{D}$，$v_x = -\frac{F_u}{D}$，分子化简为 $\frac{F_v^2 F_{uu} - 2F_u F_v F_{uv} + F_u^2 F_{vv}}{D}$，再除以 $D^2$ 得 $z_{xx} = \frac{F_v^2 F_{uu} - 2F_u F_v F_{uv} + F_u^2 F_{vv}}{D^3}$。

由对称性，将 $z_{xx}$ 表达式中的 $u$ 与 $v$ 互换，同时注意 $u_y = -\frac{F_v}{D}$，$v_y = \frac{F_u}{D}$，代入计算可得 $z_{yy} = \frac{F_u^2 F_{vv} - 2F_u F_v F_{uv} + F_v^2 F_{uu}}{D^3}$，与 $z_{xx}$ 表达式完全相同。

对 $z_x = \frac{F_u}{D}$ 再对 $y$ 求导：$z_{xy} = \frac{(F_{uu}u_y + F_{uv}v_y)D - F_u[(F_{uu}+F_{uv})u_y + (F_{uv}+F_{vv})v_y]}{D^2}$。代入 $u_y = -\frac{F_v}{D}$，$v_y = \frac{F_u}{D}$，分子化简为 $\frac{-F_u F_v F_{uu} + F_u^2 F_{uv} + F_v^2 F_{uv} - F_u F_v F_{vv}}{D}$，整理得 $z_{xy} = \frac{-F_u F_v F_{uu} + (F_u^2 + F_v^2)F_{uv} - F_u F_v F_{vv}}{D^3}$。

将 $z_{xx}$、$z_{xy}$、$z_{yy}$ 的表达式代入：$z_{xx} + 2z_{xy} + z_{yy} = \frac{1}{D^3}\left[ (F_v^2 F_{uu} - 2F_u F_v F_{uv} + F_u^2 F_{vv}) + 2(-F_u F_v F_{uu} + (F_u^2+F_v^2)F_{uv} - F_u F_v F_{vv}) + (F_v^2 F_{uu} - 2F_u F_v F_{uv} + F_u^2 F_{vv}) \right]$。合并同类项：$F_{uu}$ 项系数为 $F_v^2 - 2F_u F_v + F_v^2 = 2F_v^2 - 2F_u F_v$；$F_{vv}$ 项系数为 $F_u^2 - 2F_u F_v + F_u^2 = 2F_u^2 - 2F_u F_v$；$F_{uv}$ 项系数为 $-2F_u F_v + 2(F_u^2+F_v^2) - 2F_u F_v = 2F_u^2 + 2F_v^2 - 4F_u F_v$。总和为 $\frac{2(F_u^2 + F_v^2 - 2F_u F_v)(F_{uu} + F_{uv} + F_{vv})?}{D^3}$，实际化简后分子为 $2(F_u - F_v)^2 (F_{uu} + 2F_{uv} + F_{vv})$？需重新仔细合并：将三项相加，$F_{uu}$ 总系数：$F_v^2 - 2F_u F_v + F_v^2 = 2F_v^2 - 2F_u F_v$；$F_{vv}$ 总系数：$F_u^2 - 2F_u F_v + F_u^2 = 2F_u^2 - 2F_u F_v$；$F_{uv}$ 总系数：$-2F_u F_v + 2F_u^2 + 2F_v^2 - 2F_u F_v = 2F_u^2 + 2F_v^2 - 4F_u F_v$。注意到 $2F_v^2 - 2F_u F_v = 2F_v(F_v - F_u)$，$2F_u^2 - 2F_u F_v = 2F_u(F_u - F_v)$，$2F_u^2 + 2F_v^2 - 4F_u F_v = 2(F_u - F_v)^2$。因此总和为 $\frac{2(F_u - F_v)^2 (F_{uu} + F_{uv} + F_{vv})?}{D^3}$，但实际 $F_{uu}$ 和 $F_{vv}$ 系数不同，不能直接提出公因子。正确合并：将表达式写为 $\frac{2}{D^3}[ (F_v^2 - F_u F_v)F_{uu} + (F_u^2 - F_u F_v)F_{vv} + (F_u^2 + F_v^2 - 2F_u F_v)F_{uv} ]$，注意到 $F_v^2 - F_u F_v = F_v(F_v - F_u)$，$F_u^2 - F_u F_v = F_u(F_u - F_v)$，$F_u^2 + F_v^2 - 2F_u F_v = (F_u - F_v)^2$。因此 $z_{xx} + 2z_{xy} + z_{yy} = \frac{2(F_u - F_v)}{D^3} [ -F_v F_{uu} + F_u F_{vv} + (F_u - F_v)F_{uv} ]$。进一步观察，若令 $A = F_u - F_v$，则表达式可写为 $\frac{2A}{D^3}( -F_v F_{uu} + F_u F_{vv} + A F_{uv})$。但题目通常期望结果为0？检查：若 $F$ 是线性函数，则二阶导为0，结果为0。但一般情形下，该表达式不一定为0。然而，注意到原题可能隐含 $F$ 是齐次或特殊形式？实际上，由 $F(x-z, y-z)=0$ 可推出 $z = x + y - \text{常数}$？不，更常见的是通过隐函数定理，$z_x + z_y = 1$，因为 $F_u(1-z_x) - F_v z_x = 0$ 和 $F_u(-z_y) + F_v(1-z_y)=0$，两式相加得 $F_u(1-z_x-z_y) + F_v(1-z_x-z_y)=0$，即 $(F_u+F_v)(1-z_x-z_y)=0$，若 $F_u+F_v \neq 0$，则 $z_x+z_y=1$。于是 $z_{xx}+2z_{xy}+z_{yy} = (z_x+z_y)_x + (z_x+z_y)_y = 1_x + 1_y = 0$。因此结果为0。