东华大学 2026年数学分析第0题

已知新自变量为 \( u = \frac{x}{x^2+y^2}, v = \frac{y}{x^2+y^2} \)。计算 \( u^2+v^2 = \frac{x^2+y^2}{(x^2+y^2)^2} = \frac{1}{x^2+y^2} \)，因此 \( x^2+y^2 = \frac{1}{u^2+v^2} \)。反解得到 \( x = \frac{u}{u^2+v^2}, y = \frac{v}{u^2+v^2} \)，说明变换是对合的。

先求 \( u, v \) 对 \( x, y \) 的偏导：\( u_x = \frac{y^2-x^2}{(x^2+y^2)^2} \), \( u_y = \frac{-2xy}{(x^2+y^2)^2} \), \( v_x = \frac{-2xy}{(x^2+y^2)^2} \), \( v_y = \frac{x^2-y^2}{(x^2+y^2)^2} \)。由链式法则：\( z_x = z_u u_x + z_v v_x \), \( z_y = z_u u_y + z_v v_y \)。

对 \( z_x = A z_u + B z_v \) 再对 \( x \) 求导：\( z_{xx} = A_x z_u + A (z_{uu} u_x + z_{uv} v_x) + B_x z_v + B (z_{vu} u_x + z_{vv} v_x) \)。代入 \( u_x = A, v_x = B \) 并利用混合偏导相等，得 \( z_{xx} = A_x z_u + B_x z_v + (A^2+B^2) z_{uu} + 2AB z_{uv} \)。计算 \( A_x = \frac{2x(x^2-3y^2)}{(x^2+y^2)^3}, B_x = \frac{2y(3x^2-y^2)}{(x^2+y^2)^3} \)。

由对称性，\( z_y = B z_u - A z_v \)，对 \( y \) 求导：\( z_{yy} = B_y z_u + B (z_{uu} u_y + z_{uv} v_y) - A_y z_v - A (z_{vu} u_y + z_{vv} v_y) \)。代入 \( u_y = B, v_y = -A \)，得 \( z_{yy} = B_y z_u - A_y z_v + (B^2+A^2) z_{uu} - 2AB z_{uv} \)。计算 \( A_y = \frac{2y(3x^2-y^2)}{(x^2+y^2)^3}, B_y = \frac{2x(x^2-3y^2)}{(x^2+y^2)^3} \)。

拉普拉斯方程为 \( z_{xx} + z_{yy} = 0 \)。将 \( z_{xx} \) 和 \( z_{yy} \) 相加：\( (A_x + B_y) z_u + (B_x - A_y) z_v + 2(A^2+B^2) z_{uu} + (2AB - 2AB) z_{uv} = 0 \)。由于 \( A_x + B_y = \frac{2x(x^2-3y^2)}{(x^2+y^2)^3} + \frac{2x(x^2-3y^2)}{(x^2+y^2)^3} = \frac{4x(x^2-3y^2)}{(x^2+y^2)^3} \)，\( B_x - A_y = \frac{2y(3x^2-y^2)}{(x^2+y^2)^3} - \frac{2y(3x^2-y^2)}{(x^2+y^2)^3} = 0 \)。又 \( A^2+B^2 = \frac{(y^2-x^2)^2 + 4x^2y^2}{(x^2+y^2)^4} = \frac{(x^2+y^2)^2}{(x^2+y^2)^4} = \frac{1}{(x^2+y^2)^2} \)。代入得：\( \frac{4x(x^2-3y^2)}{(x^2+y^2)^3} z_u + \frac{2}{(x^2+y^2)^2} z_{uu} = 0 \)。两边乘以 \( (x^2+y^2)^2 \)：\( \frac{4x(x^2-3y^2)}{x^2+y^2} z_u + 2 z_{uu} = 0 \)。

由 \( x = \frac{u}{u^2+v^2}, y = \frac{v}{u^2+v^2} \)，计算 \( x^2+y^2 = \frac{1}{u^2+v^2} \)，\( x^2-3y^2 = \frac{u^2-3v^2}{(u^2+v^2)^2} \)。代入系数：\( \frac{4x(x^2-3y^2)}{x^2+y^2} = 4 \cdot \frac{u}{u^2+v^2} \cdot \frac{u^2-3v^2}{(u^2+v^2)^2} \cdot (u^2+v^2) = \frac{4u(u^2-3v^2)}{(u^2+v^2)^2} \)。因此方程化为 \( \frac{4u(u^2-3v^2)}{(u^2+v^2)^2} z_u + 2 z_{uu} = 0 \)，即 \( z_{uu} + \frac{2u(u^2-3v^2)}{(u^2+v^2)^2} z_u = 0 \)。

原方程为 $z_{xx}+z_{yy}=0$，代入上式得 $(12u^2v - 4v^3)z_v + (u^2+v^2)^2(z_{uu}+z_{vv}) = 0$。整理得 $(u^2+v^2)^2(z_{uu}+z_{vv}) = 4v(3u^2 - v^2)z_v$。两边除以 $(u^2+v^2)^2$（假设 $u^2+v^2 \neq 0$），得到 $z_{uu}+z_{vv} = \frac{4v(3u^2 - v^2)}{(u^2+v^2)^2} z_v$。

因此，原拉普拉斯方程在 $(u,v)$ 坐标下的形式为 $\displaystyle \frac{\partial^2 z}{\partial u^2}+\frac{\partial^2 z}{\partial v^2} = \frac{4v(3u^2 - v^2)}{(u^2+v^2)^2} \frac{\partial z}{\partial v}$。