西南财经大学 2020年数学分析第0题
📝 题目
七、 $\displaystyle \left(20\right.$ 分）已知函数 $\displaystyle z=z(x, y)$ 满足 $\displaystyle x^{2} \frac{\partial z}{\partial x}+y^{2} \frac{\partial z}{\partial y}=z^{2}$ ，作变量替换 $\displaystyle u=x, v=\frac{1}{y}-\frac{1}{x}, w=\frac{1}{z}-\frac{1}{x}$ ，证明：$\displaystyle \frac{\partial w}{\partial u}=0$ ．
💡 答案解析
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📋 详细解题步骤
步骤 1/6
目标：明确已知方程和变量替换关系
已知函数 $z=z(x,y)$ 满足方程 $x^{2} \frac{\partial z}{\partial x}+y^{2} \frac{\partial z}{\partial y}=z^{2}$。作变量替换 $u=x, v=\frac{1}{y}-\frac{1}{x}, w=\frac{1}{z}-\frac{1}{x}$，目标是证明 $\frac{\partial w}{\partial u}=0$。
公式：原方程：$x^{2} \frac{\partial z}{\partial x}+y^{2} \frac{\partial z}{\partial y}=z^{2}$
提示：注意 $w$ 是新的因变量，由 $z$ 和 $x$ 定义，$u,v$ 是新的自变量。
步骤 2/6
目标：将 $z$ 和 $y$ 用新变量表示
由 $w = \frac{1}{z} - \frac{1}{x}$ 得 $\frac{1}{z} = w + \frac{1}{x}$，所以 $z = \frac{1}{w + \frac{1}{x}} = \frac{x}{1 + x w}$。由 $v = \frac{1}{y} - \frac{1}{x}$ 得 $\frac{1}{y} = v + \frac{1}{x}$，所以 $y = \frac{1}{v + \frac{1}{x}} = \frac{x}{1 + x v}$。
公式：$z = \frac{x}{1 + x w}, \quad y = \frac{x}{1 + x v}$
提示：注意 $w$ 是 $u,v$ 的函数，这里只是形式表达，后续需用链式法则处理。
步骤 3/6
目标：用 $w$ 的偏导数表示 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$
对 $w = \frac{1}{z} - \frac{1}{x}$ 两边关于 $x$ 求偏导（$y$ 固定）：$\frac{\partial w}{\partial x} = -\frac{1}{z^2} \frac{\partial z}{\partial x} + \frac{1}{x^2}$，解得 $\frac{\partial z}{\partial x} = z^2 \left( \frac{1}{x^2} - \frac{\partial w}{\partial x} \right)$。对 $y$ 求偏导（$x$ 固定）：$\frac{\partial w}{\partial y} = -\frac{1}{z^2} \frac{\partial z}{\partial y}$，解得 $\frac{\partial z}{\partial y} = -z^2 \frac{\partial w}{\partial y}$。
公式：$\frac{\partial z}{\partial x} = z^2 \left( \frac{1}{x^2} - \frac{\partial w}{\partial x} \right), \quad \frac{\partial z}{\partial y} = -z^2 \frac{\partial w}{\partial y}$
提示：求偏导时注意 $z$ 是 $x,y$ 的函数，$w$ 也是 $x,y$ 的函数，需使用复合函数求导法则。
步骤 4/6
目标：代入原方程并化简
将 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 代入原方程：$x^2 \cdot z^2 \left( \frac{1}{x^2} - \frac{\partial w}{\partial x} \right) + y^2 \cdot \left( -z^2 \frac{\partial w}{\partial y} \right) = z^2$。左边化简得 $z^2 \left( 1 - x^2 \frac{\partial w}{\partial x} \right) - y^2 z^2 \frac{\partial w}{\partial y} = z^2$。两边除以 $z^2$（假设 $z \neq 0$）得 $1 - x^2 \frac{\partial w}{\partial x} - y^2 \frac{\partial w}{\partial y} = 1$，即 $- x^2 \frac{\partial w}{\partial x} - y^2 \frac{\partial w}{\partial y} = 0$，所以 $x^2 \frac{\partial w}{\partial x} + y^2 \frac{\partial w}{\partial y} = 0$。
公式：$x^2 \frac{\partial w}{\partial x} + y^2 \frac{\partial w}{\partial y} = 0$
提示：化简时注意 $z^2$ 项抵消，得到关于 $w$ 的线性偏微分方程。
步骤 5/6
目标：将偏导数转换到新变量 $(u,v)$
由于 $u = x$，$v = \frac{1}{y} - \frac{1}{x}$，计算：$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial w}{\partial v} \cdot \frac{\partial v}{\partial x} = \frac{\partial w}{\partial u} + \frac{1}{x^2} \frac{\partial w}{\partial v}$（因为 $\frac{\partial u}{\partial x}=1$，$\frac{\partial v}{\partial x} = \frac{1}{x^2}$）。$\frac{\partial w}{\partial y} = \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial w}{\partial v} \cdot \frac{\partial v}{\partial y} = -\frac{1}{y^2} \frac{\partial w}{\partial v}$（因为 $\frac{\partial u}{\partial y}=0$，$\frac{\partial v}{\partial y} = -\frac{1}{y^2}$）。
公式：$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} + \frac{1}{x^2} \frac{\partial w}{\partial v}, \quad \frac{\partial w}{\partial y} = -\frac{1}{y^2} \frac{\partial w}{\partial v}$
提示：注意 $u$ 与 $y$ 无关，所以 $\frac{\partial u}{\partial y}=0$；$v$ 对 $x$ 和 $y$ 的偏导需仔细计算。
步骤 6/6
目标：代入化简后的方程并得出结论
将上一步结果代入 $x^2 \frac{\partial w}{\partial x} + y^2 \frac{\partial w}{\partial y} = 0$ 得：$x^2 \left( \frac{\partial w}{\partial u} + \frac{1}{x^2} \frac{\partial w}{\partial v} \right) + y^2 \left( -\frac{1}{y^2} \frac{\partial w}{\partial v} \right) = 0$。化简得 $x^2 \frac{\partial w}{\partial u} + \frac{\partial w}{\partial v} - \frac{\partial w}{\partial v} = 0$，即 $x^2 \frac{\partial w}{\partial u} = 0$。由于 $x = u$ 不一定恒为零，故必有 $\frac{\partial w}{\partial u} = 0$。证毕。
公式：$x^2 \frac{\partial w}{\partial u} = 0 \Rightarrow \frac{\partial w}{\partial u} = 0$
提示：最后一步需注意 $x$ 可能为零的情况，但在一般讨论中假设 $x \neq 0$，否则需单独考虑。
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