雅可比 (Jacobi) 行列式与隐函数方程组

定理 6 (隐函数存在定理 3 ) 设函数 $F (x, y, u, v) 、 G (x, y, u, v)$ 在点 $P_{0} (x_{0}, y_{0}, u_{0}, v_{0})$ 的某个邻域内具有对各个变量的一阶连续偏导数，又 $F (x_{0}, y_{0}, u_{0}, v_{0}) = 0 ， G (x_{0}, y_{0}, u_{0}, v_{0}) = 0$ ，且在点 $P_{0} (x_{0}, y_{0}, u_{0}, v_{0})$ 处偏导数所组成的函数行列式 (也称雅可比 (Jacobi) 行列式) $J = \frac{\partial (F, G)}{\partial (u, v)} = | \begin{array}{ll} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v} \end{array} | \neq 0$ ，则方程组 ${\begin{cases} F (x, y, u, v) = 0 \\ G (x, y, u, v) = 0 \end{cases}$ 在点 $P_{0} (x_{0}, y_{0}, u_{0}, v_{0})$ 的某个邻域内恒能唯一确定一组连续且具有连续偏导数的函数 $u = u (x, y) 、 v = v (x, y)$ ，它们满足条件 $u_{0} = u (x_{0}, y_{0}) ， v_{0} = v (x_{0}, y_{0}) ＼ m a t h r m ，并且$

\begin{aligned} \frac{\partial u}{\partial x} = - \frac{1}{J} \frac{\partial (F, G)}{\partial (x, v)} = - \frac{| \begin{array}{ll} F_{x} & F_{v} \\ G_{x} & G_{v} \end{array} |}{| \begin{array}{ll} F_{u} & F_{v} \\ G_{u} & G_{v} \end{array} |}, \frac{\partial v}{\partial x} = - \frac{1}{J} \frac{\partial (F, G)}{\partial (u, x)} = - \frac{| \begin{array}{ll} F_{u} & F_{x} \\ G_{u} & G_{x} \end{array} |}{| \begin{array}{ll} F_{u} & F_{v} \\ G_{u} & G_{v} \end{array} |}; \\ \frac{\partial u}{\partial y} = - \frac{1}{J} \frac{\partial (F, G)}{\partial (y, v)} = - \frac{| \begin{array}{ll} F_{y} & F_{v} \\ G_{y} & G_{v} \end{array} |}{| \begin{array}{ll} F_{u} & F_{v} \\ G_{u} & G_{v} \end{array} |}, \frac{\partial v}{\partial y} = - \frac{1}{J} \frac{\partial (F, G)}{\partial (u, y)} = - \frac{| \begin{array}{ll} F_{u} & F_{y} \\ G_{u} & G_{y} \end{array} |}{| \begin{array}{ll} F_{u} & F_{v} \\ G_{u} & G_{v} \end{array} |} . \end{aligned}

上述公式比较复杂，我们可以通过推导，注意它的形成过程，这样对记忆有帮助，比如可通过微分：

{\begin{matrix} F_{x} d x + F_{y} d y + F_{u} d u + F_{v} d v = 0 \\ G_{x} d x + G_{y} d y + G_{u} d u + G_{v} d v = 0 \end{matrix}

然后解出 $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ .
也可将函数 $u = u (x, y) 、 v = v (x, y)$ “代入" ${\begin{cases} F (x, y, u, v) = 0 \\ G (x, y, u, v) = 0 \end{cases}$ ，即得

方程组关于 $x$ 求偏导，得:

{\begin{cases} F_{x} + F_{u} \frac{\partial u}{\partial x} + F_{v} \frac{\partial v}{\partial x} = 0 \\ G_{x} + G_{u} \frac{\partial u}{\partial x} + G_{v} \frac{\partial v}{\partial x} = 0 \end{cases}

然后解出 $\frac{\partial u}{\partial x}, \frac{\partial v}{\partial x}$ ;
同理关于 $y$ 求偏导，得:

得: {\begin{cases} F_{y} + F_{u} \frac{\partial u}{\partial y} + F_{v} \frac{\partial v}{\partial y} = 0 \\ G_{y} + G_{u} \frac{\partial u}{\partial y} + G_{v} \frac{\partial v}{\partial y} = 0 \end{cases} 然后解出 \frac{\partial u}{\partial y} ， \frac{\partial v}{\partial y} .

例 15 设 $u = u (x, y) 、 v = v (x, y)$ 由方程组 ${\begin{matrix} u + v = x + y \\ x u + y v = 1 \end{matrix}$ 确定，求 $\frac{\partial u}{\partial x} ， \frac{\partial u}{\partial y} ， \frac{\partial v}{\partial x} ， \frac{\partial v}{\partial y}$ .
这道题目可以用公式做，也可以用推导法解得，我们采取推导法.
解一在方程组 ${\begin{matrix} u + v = x + y \\ x u + y v = 1 \end{matrix}$ 两端微分: ${\begin{matrix} d u + d v = d x + d y \\ x d u + u d x + y d v + v d y = 0 \end{matrix}$ ，整理得

d u = - \frac{(u + y) d x + (v + y) d y}{x - y}, d v = \frac{(u + x) d x + (v + x) d y}{x - y_{- \dots}}

即 $\frac{\partial u}{\partial x} = - \frac{u + y}{x - y} ， \frac{\partial u}{\partial y} = - \frac{v + y}{x - y} ， \frac{\partial v}{\partial x} = \frac{u + x}{x - y} ， \frac{\partial v}{\partial y} = \frac{v + x}{x - y}$ ;
例 15 设 $u = u (x, y) 、 v = v (x, y)$ 由方程组 ${\begin{matrix} u + v = x + y \\ x u + y v = 1 \end{matrix}$ 确定，求 $\frac{\partial u}{\partial x} ， \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ .
解二在方程组 ${\begin{matrix} u + v = x + y \\ x u + y v = 1 \end{matrix}$ 两端关于 $x$ 求导，得 ${\begin{matrix} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} = 1 \\ u + x \frac{\partial u}{\partial x} + y \frac{\partial v}{\partial x} = 0 \end{matrix}$ ，
整理得 $\frac{\partial u}{\partial x} = - \frac{u + y}{x - y}, \frac{\partial v}{\partial x} = \frac{u + x}{x - y}$ ;
类似得 $\frac{\partial u}{\partial y} = - \frac{v + y}{x - y} ， \frac{\partial v}{\partial y} = \frac{v + x}{x - y}$ .

例 16 设函数 $y = y (x) 、 z = z (x)$ 由方程组 ${\begin{matrix} x + y + z + z^{2} = 1 \\ x + y^{2} + z + z^{3} = 2 \end{matrix}$ 确定，试求 $\frac{d y}{d x} ， \frac{d z}{d x}$ . 解将方程组微分 ${\begin{matrix} d x + d y + d z + 2 z d z = 0 \\ d x + 2 y d y + d z + 3 z^{2} d z = 0 \end{matrix}$ ，
消去 $d z$ ，得

\begin{aligned} d y = \frac{3 z^{2} - 2 z}{2 y + 4 y z - 1 - 3 z^{2}} d x ，即 \frac{d y}{d x} = \frac{3 z^{2} - 2 z}{2 y + 4 y z - 1 - 3 z^{2}} ； \\ d z = \frac{1 - 2 y}{2 y + 4 y z - 1 - 3 z^{2}} d x ，即 \frac{d z}{d x} = \frac{1 - 2 y}{2 y + 4 y z - 1 - 3 z^{2}} ； \end{aligned}

例 17 设函数 $x = x (u, v) ， y = y (u, v)$ 在点 $(u, v)$ 的某个邻域内连续且具有一阶
连续偏导数， $J = \frac{\partial (x, y)}{\partial (u, v)} = | \begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | \neq 0$ ，试求 $\frac{\partial u}{\partial x} 、 \frac{\partial u}{\partial y} 、 \frac{\partial v}{\partial x} 、 \frac{\partial v}{\partial y}$ ;
解设 ${\begin{cases} F (x, y, u, v) = x - x (u, v) = 0 \\ G (x, y, u, v) = y - y (u, v) = 0 \end{cases} ， F ， G$ 在点 $(x, y, u, v)$ 的某个邻域内连续且具有一阶连续偏导数，且 $J = \frac{\partial (F, G)}{\partial (u, v)} = | \begin{array}{cc} - \frac{\partial x}{\partial u} & - \frac{\partial x}{\partial v} \\ - \frac{\partial y}{\partial u} & - \frac{\partial y}{\partial v} \end{array} | = | \begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | = \frac{\partial (x, y)}{\partial (u, v)} \neq 0$ ，

故在点 $(x, y)$ 的某个邻域内存在连续且具有一阶连续偏导数的反函数 $u = u (x, y) 、 v = v (x, y)$ .
由 ${\begin{cases} d x = x_{u} d u + x_{v} d v \\ d y = y_{u} d u + y_{v} d v^{'} \end{cases}$ 得 $d u = \frac{y_{v} d x - x_{v} d y}{J} ， d v = \frac{- y_{u} d x + x_{u} d y}{J}$ ，
因此 $\frac{\partial u}{\partial x} = \frac{y_{v}}{J} ， \frac{\partial u}{\partial y} = - \frac{x_{v}}{J}, \frac{\partial v}{\partial x} = - \frac{y_{u}}{J} ， \frac{\partial v}{\partial y} = \frac{x_{u}}{J}$ .

## 雅可比 (Jacobi) 行列式与隐函数方程组
定理 6 (隐函数存在定理 3 ) 设函数 $F(x, y, u, v) 、 G(x, y, u, v)$ 在点 $P_0\left(x_0, y_0, u_0, v_0\right)$ 的某个邻域内具有对各个变量的一阶连续偏导数，又 $F\left(x_0, y_0, u_0, v_0\right)=0 ， G\left(x_0, y_0, u_0, v_0\right)=0$ ， 且在点 $P_0\left(x_0, y_0, u_0, v_0\right)$ 处偏导数所组成 的函数行列式 (也称雅可比 (Jacobi) 行列式) $J=\frac{\partial(F, G)}{\partial(u, v)}=\left|\begin{array}{ll}\frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v}\end{array}\right| \neq 0$ ， 则方程组 $\left\{\begin{array}{l}F(x, y, u, v)=0 \\ G(x, y, u, v)=0\end{array}\right.$ 在点 $P_0\left(x_0, y_0, u_0, v_0\right)$ 的某个邻域内恒能唯一确定一 组连续且具有连续偏导数的函数 $u=u(x, y) 、 v=v(x, y)$ ，它们满足条件 $u_0=u\left(x_0, y_0\right) ， v_0=v\left(x_0, y_0\right) ＼mathrm{~ ， 并 且 ~}$

$$
\begin{aligned}
& \frac{\partial u}{\partial x}=-\frac{1}{J} \frac{\partial(F, G)}{\partial(x, v)}=-\frac{\left|\begin{array}{ll}
F_x & F_v \\
G_x & G_v
\end{array}\right|}{\left|\begin{array}{ll}
F_u & F_v \\
G_u & G_v
\end{array}\right|}, \frac{\partial v}{\partial x}=-\frac{1}{J} \frac{\partial(F, G)}{\partial(u, x)}=-\frac{\left|\begin{array}{ll}
F_u & F_x \\
G_u & G_x
\end{array}\right|}{\left|\begin{array}{ll}
F_u & F_v \\
G_u & G_v
\end{array}\right|} ; \\
& \frac{\partial u}{\partial y}=-\frac{1}{J} \frac{\partial(F, G)}{\partial(y, v)}=-\frac{\left|\begin{array}{ll}
F_y & F_v \\
G_y & G_v
\end{array}\right|}{\left|\begin{array}{ll}
F_u & F_v \\
G_u & G_v
\end{array}\right|}, \frac{\partial v}{\partial y}=-\frac{1}{J} \frac{\partial(F, G)}{\partial(u, y)}=-\frac{\left|\begin{array}{ll}
F_u & F_y \\
G_u & G_y
\end{array}\right|}{\left|\begin{array}{ll}
F_u & F_v \\
G_u & G_v
\end{array}\right|} . \\
&
\end{aligned}
$$

上述公式比较复杂，我们可以通过推导，注意它的形成过程，这样对记忆有 帮助，比如可通过微分：

$$
\left\{\begin{array}{c}
F_x \mathrm{~d} x+F_y \mathrm{~d} y+F_u \mathrm{~d} u+F_v \mathrm{~d} v=0 \\
G_x \mathrm{~d} x+G_y \mathrm{~d} y+G_u \mathrm{~d} u+G_v \mathrm{~d} v=0
\end{array}\right.
$$

然后解出 $\quad \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$.
也可将函数 $u=u(x, y) 、 v=v(x, y)$ “代入" $\left\{\begin{array}{l}F(x, y, u, v)=0 \\ G(x, y, u, v)=0\end{array}\right.$ ，即得

方程组关于 $x$ 求偏导，得:

$$
\left\{\begin{array}{l}
F_x+F_u \frac{\partial u}{\partial x}+F_v \frac{\partial v}{\partial x}=0 \\
G_x+G_u \frac{\partial u}{\partial x}+G_v \frac{\partial v}{\partial x}=0
\end{array}\right.
$$

然后解出 $\frac{\partial u}{\partial x}, \frac{\partial v}{\partial x}$;
同理关于 $y$ 求偏导，得:

$$
\text { 得: }\left\{\begin{array}{l}
F_y+F_u \frac{\partial u}{\partial y}+F_v \frac{\partial v}{\partial y}=0 \\
G_y+G_u \frac{\partial u}{\partial y}+G_v \frac{\partial v}{\partial y}=0
\end{array} \quad \text { 然后解出 } \frac{\partial u}{\partial y} ， \frac{\partial v}{\partial y} .\right.
$$

例 15 设 $u=u(x, y) 、 v=v(x, y)$ 由方程组 $\left\{\begin{array}{c}u+v=x+y \\ x u+y v=1\end{array}\right.$ 确定，求 $\frac{\partial u}{\partial x} ， \frac{\partial u}{\partial y} ， \frac{\partial v}{\partial x} ， \frac{\partial v}{\partial y}$.
这道题目可以用公式做，也可以用推导法解得，我们采取推导法.
解一 在方程组 $\left\{\begin{array}{c}u+v=x+y \\ x u+y v=1\end{array}\right.$ 两端微分: $\left\{\begin{array}{c}\mathrm{d} u+\mathrm{d} v=\mathrm{d} x+\mathrm{d} y \\ x \mathrm{~d} u+u \mathrm{~d} x+y \mathrm{~d} v+v \mathrm{~d} y=0\end{array}\right.$ ，整理得

$$
\mathrm{d} u=-\frac{(u+y) \mathrm{d} x+(v+y) \mathrm{d} y}{x-y}, \mathrm{~d} v=\frac{(u+x) \mathrm{d} x+(v+x) \mathrm{d} y}{x-y_{-\ldots}}
$$

即 $\frac{\partial u}{\partial x}=-\frac{u+y}{x-y} ， \frac{\partial u}{\partial y}=-\frac{v+y}{x-y} ， \frac{\partial v}{\partial x}=\frac{u+x}{x-y} ， \frac{\partial v}{\partial y}=\frac{v+x}{x-y}$;
例 15 设 $u=u(x, y) 、 v=v(x, y)$ 由方程组 $\left\{\begin{array}{c}u+v=x+y \\ x u+y v=1\end{array}\right.$ 确定，求 $\frac{\partial u}{\partial x} ， \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$.
解二 在方程组 $\left\{\begin{array}{c}u+v=x+y \\ x u+y v=1\end{array}\right.$ 两端关于 $x$ 求导，得 $\left\{\begin{array}{c}\frac{\partial u}{\partial x}+\frac{\partial v}{\partial x}=1 \\ u+x \frac{\partial u}{\partial x}+y \frac{\partial v}{\partial x}=0\end{array}\right.$ ，
整理得 $\quad \frac{\partial u}{\partial x}=-\frac{u+y}{x-y}, \frac{\partial v}{\partial x}=\frac{u+x}{x-y}$;
类似得 $\frac{\partial u}{\partial y}=-\frac{v+y}{x-y} ， \frac{\partial v}{\partial y}=\frac{v+x}{x-y}$.

例 16 设函数 $y=y(x) 、 z=z(x)$ 由方程组 $\left\{\begin{array}{c}x+y+z+z^2=1 \\ x+y^2+z+z^3=2\end{array}\right.$ 确定，试求 $\frac{\mathrm{d} y}{\mathrm{~d} x} ， \frac{\mathrm{d} z}{\mathrm{~d} x}$. 解 将方程组微分 $\left\{\begin{array}{c}\mathrm{d} x+\mathrm{d} y+\mathrm{d} z+2 z \mathrm{~d} z=0 \\ \mathrm{~d} x+2 y \mathrm{~d} y+\mathrm{d} z+3 z^2 \mathrm{~d} z=0\end{array}\right.$ ，
消去 $\mathrm{d} z$ ，得

$$
\begin{aligned}
& \mathrm{d} y=\frac{3 z^2-2 z}{2 y+4 y z-1-3 z^2} \mathrm{~d} x \text { ，即 } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 z^2-2 z}{2 y+4 y z-1-3 z^2} \text { ； } \\
& \mathrm{d} z=\frac{1-2 y}{2 y+4 y z-1-3 z^2} \mathrm{~d} x \text { ，即 } \frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{1-2 y}{2 y+4 y z-1-3 z^2} \text { ； } \\
&
\end{aligned}
$$

例 17 设函数 $x=x(u, v) ， y=y(u, v)$ 在点 $(u, v)$ 的某个邻域内连续且具有一阶
连续偏导数， $J=\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{ll}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{array}\right| \neq 0$ ，试求 $\frac{\partial u}{\partial x} 、 \frac{\partial u}{\partial y} 、 \frac{\partial v}{\partial x} 、 \frac{\partial v}{\partial y}$;
解 设 $\left\{\begin{array}{l}F(x, y, u, v)=x-x(u, v)=0 \\ G(x, y, u, v)=y-y(u, v)=0\end{array} ， \quad F ， G\right.$ 在点 $(x, y, u, v)$ 的某个邻域内连续 且具有一阶连续偏导数，且 $J=\frac{\partial(F, G)}{\partial(u, v)}=\left|\begin{array}{cc}-\frac{\partial x}{\partial u} & -\frac{\partial x}{\partial v} \\ -\frac{\partial y}{\partial u} & -\frac{\partial y}{\partial v}\end{array}\right|=\left|\begin{array}{ll}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{array}\right|=\frac{\partial(x, y)}{\partial(u, v)} \neq 0$ ，

故在点 $(x, y)$ 的某个邻域内存在连续且具有一阶连续偏导数的反函数 $u=u(x, y) 、 v=v(x, y)$.
由 $\left\{\begin{array}{l}\mathrm{d} x=x_u \mathrm{~d} u+x_v \mathrm{~d} v \\ \mathrm{~d} y=y_u \mathrm{~d} u+y_v \mathrm{~d} v^{\prime}\end{array}\right.$ 得 $\mathrm{d} u=\frac{y_v \mathrm{~d} x-x_v \mathrm{~d} y}{J} ， \mathrm{~d} v=\frac{-y_u \mathrm{~d} x+x_u \mathrm{~d} y}{J}$ ，
因此 $\frac{\partial u}{\partial x}=\frac{y_v}{J} ， \frac{\partial u}{\partial y}=-\frac{x_v}{J}, \frac{\partial v}{\partial x}=-\frac{y_u}{J} ， \frac{\partial v}{\partial y}=\frac{x_u}{J}$.

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