📝 题目
7．设 $\Phi(u, v)$ 具有连续偏导数，证明由方程 $\Phi(c x-a z, c y-b z)=0$ 所确定的函数 $z=f(x, y)$ 满足 $a \frac{\partial z}{\partial x}+b \frac{\partial z}{\partial y}=c$.
💡 答案与解析
[AI解答]
[AI解答] 我们设 $$ u = c x - a z,\quad v = c y - b z, $$ 其中 $z = f(x, y)$ 由方程 $$ \Phi(u, v) = 0 $$ 隐式确定，且 $\Phi$ 具有连续偏导数，且 $\frac{\partial \Phi}{\partial u}, \frac{\partial \Phi}{\partial v}$ 不全为零（以保证隐函数存在）。
对方程 $\Phi(u, v) = 0$ 两边关于 $x$ 求偏导，注意 $u, v$ 均依赖于 $x$ 和 $y$：
$$ \frac{\partial \Phi}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial \Phi}{\partial v} \cdot \frac{\partial v}{\partial x} = 0. $$
而 $$ \frac{\partial u}{\partial x} = c - a \frac{\partial z}{\partial x},\quad \frac{\partial v}{\partial x} = 0 - b \frac{\partial z}{\partial x} = -b \frac{\partial z}{\partial x}. $$
代入得 $$ \frac{\partial \Phi}{\partial u} \left( c - a \frac{\partial z}{\partial x} \right) + \frac{\partial \Phi}{\partial v} \left( -b \frac{\partial z}{\partial x} \right) = 0. $$ 整理为 $$ c \frac{\partial \Phi}{\partial u} = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \frac{\partial z}{\partial x}. \tag{1} $$
再对 $y$ 求偏导： $$ \frac{\partial \Phi}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial \Phi}{\partial v} \cdot \frac{\partial v}{\partial y} = 0, $$ 其中 $$ \frac{\partial u}{\partial y} = 0 - a \frac{\partial z}{\partial y} = -a \frac{\partial z}{\partial y},\quad \frac{\partial v}{\partial y} = c - b \frac{\partial z}{\partial y}. $$
代入得 $$ \frac{\partial \Phi}{\partial u} \left( -a \frac{\partial z}{\partial y} \right) + \frac{\partial \Phi}{\partial v} \left( c - b \frac{\partial z}{\partial y} \right) = 0, $$ 整理为 $$ c \frac{\partial \Phi}{\partial v} = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \frac{\partial z}{\partial y}. \tag{2} $$
将 (1) 和 (2) 两式相加： $$ c \left( \frac{\partial \Phi}{\partial u} + \frac{\partial \Phi}{\partial v} \right) = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \left( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \right). $$
但我们实际需要的是 $a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y}$。 将 (1) 式乘以 $a$，(2) 式乘以 $b$，再相加：
$$ a c \frac{\partial \Phi}{\partial u} + b c \frac{\partial \Phi}{\partial v} = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \left( a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} \right). $$
左边提取公因子 $c$： $$ c \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) = \left( a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v} \right) \left( a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} \right). $$
由于 $\Phi$ 有连续偏导且隐函数存在，括号 $a \frac{\partial \Phi}{\partial u} + b \frac{\partial \Phi}{\partial v}$ 一般不为零，可约去，得到 $$ a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} = c. $$
证毕。
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