kaoyan1basic 高等数学 第93题
📝 题目
### 第93题 设函数 $f(u, v)$ 具有二阶连续偏导数，且满足 $\displaystyle 4 \frac{\partial^{2} f}{\partial u^{2}}-\frac{\partial^{2} f}{\partial v^{2}}=1$ ，又 $g(x, y)=f\left(x^{2}+\right. \left.y^{2}, x y\right)$ ，则 $\displaystyle \frac{\partial^{2} g}{\partial x^{2}}-\frac{\partial^{2} g}{\partial y^{2}}=$ $\_\_\_\_$ ．
💡 答案解析
**答案**：$x^2 + y^2$ **解析**： 步骤1：$g(x,y) = f(x^2+y^2, xy)$，令$u = x^2+y^2$，$v = xy$。 步骤2：$\displaystyle \frac{\partial g}{\partial x} = f_u \cdot 2x + f_v \cdot y$，$\displaystyle \frac{\partial^2 g}{\partial x^2} = 2f_u + 4x^2 f_{uu} + 2xy f_{uv} + y^2 f_{vv}$。 步骤3：$\displaystyle \frac{\partial g}{\partial y} = f_u \cdot 2y + f_v \cdot x$，$\displaystyle \frac{\partial^2 g}{\partial y^2} = 2f_u + 4y^2 f_{uu} + 2xy f_{uv} + x^2 f_{vv}$。 步骤4：$\displaystyle \frac{\partial^2 g}{\partial x^2} - \frac{\partial^2 g}{\partial y^2} = (4x^2 - 4y^2) f_{uu} + (y^2 - x^2) f_{vv} = (x^2 - y^2)(4f_{uu} - f_{vv})$。 步骤5：由已知$4f_{uu} - f_{vv} = 1$，故原式$= x^2 - y^2$。 **难度**：★★★★★