方企勤 第五章 多元函数微分学 第3题
📝 题目
例 3 变换 $x + y = u,y = {uv}$ 把区域 $\{ \left( {u,v}\right) \mid u > 0,v > 0\}$ 变为区域 $\{ \left( {x,y}\right) \mid x + y > 0,y > 0\}$ . 试求雅可比行列式
$$ \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) },\;\frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) }. $$
💡 答案解析
解法 1 把 $x,y$ 写成 $u,v$ 的函数: $x = u\left( {1 - v}\right) ,y = {uv}$ ,所以
$$ \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) } = \left| \begin{matrix} {x}_{u}^{\prime } & {x}_{v}^{\prime } \\ {y}_{u}^{\prime } & {y}_{v}^{\prime } \end{matrix}\right| = \left| \begin{matrix} 1 - v & - u \\ v & u \end{matrix}\right| $$
$$ = u\left( {1 - v}\right) + {uv} = u. $$
逆变换的雅可比行列式为
$$ \frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) } = {\left\lbrack \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) }\right\rbrack }^{-1} = \frac{1}{u} = \frac{1}{x + y}. $$
解法 2 若变换不易解出 $x,y$ 或 $u,v$ 时,我们只能用隐函数求偏导数方法来求雅可比行列式,一般来说所得行列式可以含有变量 $x,y,u,v$ . 方程组先对 $u$ 求偏导数,得
$$ \left\{ {\begin{array}{l} {x}_{u}^{\prime } + {y}_{u}^{\prime } = 1, \\ {y}_{u}^{\prime } = v, \end{array}\text{ 解出 }\left\{ \begin{array}{l} {x}_{u}^{\prime } = 1 - v, \\ {y}_{u}^{\prime } = v. \end{array}\right. }\right. $$
再对 $v$ 求偏导数,得
$$ \left\{ {\begin{array}{l} {x}_{v}^{\prime } + {y}_{v}^{\prime } = 0, \\ {y}_{v}^{\prime } = u, \end{array}\text{ 解出 }\left\{ \begin{array}{l} {x}_{v}^{\prime } = - u, \\ {y}_{v}^{\prime } = u, \end{array}\right. }\right. $$
所以
$$ \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) } = \left| \begin{matrix} 1 - v & - u \\ v & u \end{matrix}\right| = u. $$