📝 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
$$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial s}\right)^{2}+\left(\frac{\partial u}{\partial t}\right)^{2} \text { 及 } \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{2} u}{\partial s^{2}}+\frac{\partial^{2} u}{\partial t^{2}} \text {. } $$
💡 答案与解析
[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：★★☆☆☆