同济高数 第9章 第9-4-*14题
📝 题目
＊14．设 $u=f(x, y)$ 的所有二阶偏导数连续，而
$$
x=\frac{s-\sqrt{3} t}{2}, \quad y=\frac{\sqrt{3} s+t}{2},
$$
证明
$$
\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial s}\right)^{2}+\left(\frac{\partial u}{\partial t}\right)^{2} \text { 及 } \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{2} u}{\partial s^{2}}+\frac{\partial^{2} u}{\partial t^{2}} \text {. }
$$
💡 答案解析
[AI解答]
[AI解答]
我们已知变换：
$$
x = \frac{s - \sqrt{3} t}{2}, \quad y = \frac{\sqrt{3} s + t}{2}.
$$
且 $u = f(x, y)$ 的所有二阶偏导数连续。
第一步：证明梯度模长不变性
首先计算一阶偏导变换。由链式法则：
$$
\frac{\partial u}{\partial s} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s},
$$
$$
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t}.
$$
计算偏导数：
$$
\frac{\partial x}{\partial s} = \frac{1}{2}, \quad \frac{\partial x}{\partial t} = -\frac{\sqrt{3}}{2},
$$
$$
\frac{\partial y}{\partial s} = \frac{\sqrt{3}}{2}, \quad \frac{\partial y}{\partial t} = \frac{1}{2}.
$$
因此：
$$
\frac{\partial u}{\partial s} = \frac{1}{2} u_x + \frac{\sqrt{3}}{2} u_y,
$$
$$
\frac{\partial u}{\partial t} = -\frac{\sqrt{3}}{2} u_x + \frac{1}{2} u_y.
$$
现在计算 $(\frac{\partial u}{\partial s})^2 + (\frac{\partial u}{\partial t})^2$：
$$
\left( \frac{1}{2} u_x + \frac{\sqrt{3}}{2} u_y \right)^2 + \left( -\frac{\sqrt{3}}{2} u_x + \frac{1}{2} u_y \right)^2.
$$
展开：
第一项：
$$
\frac{1}{4} u_x^2 + \frac{\sqrt{3}}{2} u_x u_y + \frac{3}{4} u_y^2.
$$
第二项：
$$
\frac{3}{4} u_x^2 - \frac{\sqrt{3}}{2} u_x u_y + \frac{1}{4} u_y^2.
$$
相加：
$$
\left( \frac{1}{4} + \frac{3}{4} \right) u_x^2 + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \right) u_x u_y + \left( \frac{3}{4} + \frac{1}{4} \right) u_y^2 = u_x^2 + u_y^2.
$$
因此第一个等式成立。
第二步：证明拉普拉斯算子不变性
计算二阶偏导。由链式法则，二阶偏导算子满足：
$$
\frac{\partial^2}{\partial s^2} = \left( \frac{\partial x}{\partial s} \frac{\partial}{\partial x} + \frac{\partial y}{\partial s} \frac{\partial}{\partial y} \right)^2,
$$
$$
\frac{\partial^2}{\partial t^2} = \left( \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial}{\partial y} \right)^2.
$$
代入系数：
$$
\frac{\partial}{\partial s} = \frac{1}{2} \frac{\partial}{\partial x} + \frac{\sqrt{3}}{2} \frac{\partial}{\partial y},
$$
$$
\frac{\partial}{\partial t} = -\frac{\sqrt{3}}{2} \frac{\partial}{\partial x} + \frac{1}{2} \frac{\partial}{\partial y}.
$$
计算 $\frac{\partial^2 u}{\partial s^2}$：
$$
\frac{\partial^2 u}{\partial s^2} = \left( \frac{1}{2} \frac{\partial}{\partial x} + \frac{\sqrt{3}}{2} \frac{\partial}{\partial y} \right) \left( \frac{1}{2} u_x + \frac{\sqrt{3}}{2} u_y \right).
$$
展开：
$$
= \frac{1}{4} u_{xx} + \frac{\sqrt{3}}{4} u_{xy} + \frac{\sqrt{3}}{4} u_{yx} + \frac{3}{4} u_{yy}.
$$
由于混合偏导相等，$u_{xy} = u_{yx}$，所以：
$$
\frac{\partial^2 u}{\partial s^2} = \frac{1}{4} u_{xx} + \frac{\sqrt{3}}{2} u_{xy} + \frac{3}{4} u_{yy}.
$$
计算 $\frac{\partial^2 u}{\partial t^2}$：
$$
\frac{\partial^2 u}{\partial t^2} = \left( -\frac{\sqrt{3}}{2} \frac{\partial}{\partial x} + \frac{1}{2} \frac{\partial}{\partial y} \right) \left( -\frac{\sqrt{3}}{2} u_x + \frac{1}{2} u_y \right).
$$
展开：
$$
= \frac{3}{4} u_{xx} - \frac{\sqrt{3}}{4} u_{xy} - \frac{\sqrt{3}}{4} u_{yx} + \frac{1}{4} u_{yy}.
$$
合并：
$$
\frac{\partial^2 u}{\partial t^2} = \frac{3}{4} u_{xx} - \frac{\sqrt{3}}{2} u_{xy} + \frac{1}{4} u_{yy}.
$$
相加：
$$
\frac{\partial^2 u}{\partial s^2} + \frac{\partial^2 u}{\partial t^2} = \left( \frac{1}{4} + \frac{3}{4} \right) u_{xx} + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \right) u_{xy} + \left( \frac{3}{4} + \frac{1}{4} \right) u_{yy} = u_{xx} + u_{yy}.
$$
因此第二个等式也成立。
结论：该变换保持梯度模长和拉普拉斯算子不变，即该变换是正交变换（旋转加缩放，此处缩放因子为1，实际上是旋转）。
难度：★★☆☆☆