定义1：隐式方程
设\(D \subset \mathbb{R}^2\)是一开集，\(F: D \to \mathbb{R}\)是一个含有两个自变量\(x,y\)的函数，对于\(D\)中的点\((x,y)\)满足方程
\[ F(x,y) = 0 \]
的点的全体组成\(D\)内的一条曲线，而方程就称为该曲线的隐式方程。
定理1：隐函数定理
设开集\(D \subset \mathbb{R}^2\)，函数\(F: D \to \mathbb{R}\)满足条件：
（a）\(F \in C^1(D)\)；
（b）点\((x_0,y_0) \in D\)使得\(F(x_0,y_0) = 0\)；
（c）\(\dfrac{\partial F(x_0,y_0)}{\partial y} \ne 0\)
那么存在一个包含\((x_0,y_0)\)的开矩形\(I \times J \subset D\)，使得：
（1）对每一个\(x \in I\)，方程\(F(x,y)=0\)在\(J\)中有唯一的解\(f(x)\)；
（2）\(y_0 = f(x_0)\)；
（3）\(f \in C^1(I)\)；
（4）当\(x \in I\)时，有
\[ f^\prime(x) = -\frac{\frac{\partial F}{\partial x}(x,y)}{\frac{\partial F}{\partial y}(x,y)} \]
其中\(y = f(x)\)。
证：不妨设\(\displaystyle \frac{\partial
F(x_0,y_0)}{\partial y} > 0\)，由条件\((a)\)可知，存在一个包含\((x_0,y_0)\)的开矩形\(I^\prime \times J\)，满足\(I^\prime \times \bar J \subset
D\)，且在\(I^\prime \times \bar
J\)上有\(\displaystyle \frac{\partial
F}{\partial y} > 0\)。从而对任意给定的\(x \in I^\prime\)，\(F(x,y)\)在闭区间\(\bar J\)上是严格递增的连续函数。设\(J = (c,d)\)，由条件\((b)\)可知必有
\[
F(x_0, c) < 0, \quad F(x_0, d) > 0
\]
由条件\((a)\)能推出\(F \in C(D)\)，因此存在含\(x_0\)的开区间\(I
\subset I^\prime\)，使得当\(x \in
I\)时，
\[
F(x, c) < 0, \quad F(x, d) > 0
\]
由连续函数的零值定理和严格单调性可知，对每一个\(x \in I\)，存在唯一的一个数，记作\(f(x) \in (c, d) = J\)，使得\(F(x, f(x)) =
0\)，这就证明了（1），显然\(f\)满足（2）。
为了证明（3）和（4），先证明\(f\)在开区间\(I\)上连续。特别地，\(x_0 \in I\)，由上面的证明可知无论区间\(J\)取得多小，一定存在足够小的区间\(I\)使得对每一个\(x \in I\)时，有\(f(x) \in J\)；这时\(|f(x) - f(x_0)| < |J|\)，其中\(|J|\)表示区间\(J\)的长度。即证明了\(f\)在\(x_0\)处连续。现任取\(x_1 \in I\)，设\(y_1 = f(x_1)\)，则\((x_1,y_1) \in I \times J\)。因为有\(F(x_1,y_1)=0\)，\(\displaystyle \frac{\partial F(x_1,y_1)}{\partial
y} > 0\)，所以\(F\)在点\((x_1,y_1)\)处满足它在\((x_0,y_0)\)处的同样条件，所以\(f\)在\(x_1\)处也是连续的，从而\(f\)在整个区间\(I\)上连续。
再证（3）和（4）。设\(x \in
I\)，取\(h\)很小，使得\(x+h \in I\)。令\(y = f(x)\)，\(k =
f(x+h) - f(x)\)。由\(F\)的可微性，并利用函数导数八的定理1可得
\[
0 = F(x+h, y+k) - F(x, y) = \frac{\partial F}{\partial x}(x, y) h +
\frac{\partial F}{\partial y}(x, y) k + \alpha h + \beta k
\]
其中当\(h \to 0\)时，\(\alpha \to 0\)，当\(k \to 0\)时，\(\beta \to 0\)。又由于\(f \in C(I)\)，所以当\(h \to 0\)时，\(k
\to 0\)，从而\(\beta \to
0\)。从而
\[
\lim \limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim \limits_{h
\to 0} \frac{k}{h} = \lim\limits_{h \to 0} \frac{-\frac{\partial
F}{\partial x} + \alpha}{\frac{\partial F}{\partial y} + \beta}
\]
即
\[
f^\prime(x) = - \frac{\frac{\partial F}{\partial x} (x,
y)}{\frac{\partial F}{\partial y} (x,y)}
\]
其中\(x \in I\)且\(y = f(x)\)。由于\((a)\)知，\(f^\prime\)在\(I\)上连续。
Q.E.D.
定理2
设开集\(D \subset \mathbb{R}^{n+1}\)，\(F: D \to \mathbb{R}\)，满足条件：
（a）\(F \in C^1(D)\)；
（b）\(F(\boldsymbol{x}_0, y_0) = 0\)，这里\(\boldsymbol{x}_0 \in \mathbb{R}^n,y_0 \in \mathbb{R}\)且\((\boldsymbol{x}_0, y_0) \in D\)；
（c）\(\dfrac{\partial F(\boldsymbol{x}_0, y_0)}{\partial y} \ne 0\)。
那么存在\((\boldsymbol{x}_0, y_0)\)的一个邻域\(G \times J\)，其中\(G\)是\(\boldsymbol{x_0}\)在\(\mathbb{R}^n\)的一个邻域，\(J\)是\(\mathbb{R}\)中的一个开区间，使得：
（1）对每一个\(\boldsymbol{x} \in G\)，方程
\[ F(\boldsymbol{x}, y) = 0 \]
在\(J\)中存在唯一的解，记为\(f(\boldsymbol{x})\)；
（2）\(y_0 = f(\boldsymbol{x}_0)\)；
（3）\(f \in C^1(G)\)；
（4）当\(\boldsymbol{x} \in G\)时，
\[ \frac{\partial f}{\partial x_i} = - \frac{\frac{\partial F}{\partial x_i}(\boldsymbol{x}, y)}{\frac{\partial F}{\partial y}(\boldsymbol{x}, y)} \quad (i=1,2,\cdots,n) \]
其中\(y = f(\boldsymbol{x})\)。
证：由定理1的证明，可知（1）（2）的证明方式一模一样；而在证明（3）（4）时，只需令\(\boldsymbol{h} = (0, \cdots, h_i, \cdots, 0)^T\)，固定\(x_i\)，其他证明过程一样，即可证得（3）和（4）。
Q.E.D.
设有\(m\)个方程组成的方程组
\[
\left\{ \begin{aligned}
& F_1(x_1,x_2,\cdots,x_n, y_1,\cdots, y_m) = 0 \\
& \cdots, \\
& F_m(x_1,x_2,\cdots,x_n, y_1,\cdots, y_m) = 0
\end{aligned} \right.
\]
按照隐函数的想法，是否可以解出
\[
\left\{ \begin{aligned}
& y_1 = f_1(x_1,\cdots,x_n) \\
& \cdots \\
& y_m = f_m(x_1,\cdots,x_n)
\end{aligned} \right.
\]
为了缩短记号，可令
\[
\boldsymbol{F} = \left[ \begin{matrix}
F_1 \\
\vdots \\
F_m
\end{matrix} \right], \quad
\boldsymbol{f} = \left[ \begin{matrix}
f_1 \\
\vdots \\
f_m
\end{matrix} \right]
\]
从而可以把方程改写为
\[
\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}) = \boldsymbol{0}
\]
而解出式可改写为
\[
\boldsymbol{y} = \boldsymbol{f}(\boldsymbol{x})
\]
需要再定义几个记号，设\(\boldsymbol{F}\)定义在开集\(D \subset \mathbb{R}^{m+n}\)上，在\(m \times (m+n)\)矩阵
\[
J\boldsymbol{F} = \left[
\begin{matrix}
\frac{\partial F_1}{\partial x_1} & \cdots &
\frac{\partial F_1}{\partial x_n} & \frac{\partial F_1}{\partial
y_1} & \cdots & \frac{\partial F_1}{\partial y_m} \\
\vdots & & \vdots & \vdots & & \vdots \\
\frac{\partial F_m}{\partial x_1} & \cdots &
\frac{\partial F_m}{\partial x_n} & \frac{\partial F_m}{\partial
y_1} & \cdots & \frac{\partial F_m}{\partial y_m}
\end{matrix}
\right]
\]
中做分块：\(J\boldsymbol{F} =
(J_{\boldsymbol{x}}\boldsymbol{F},
J_{\boldsymbol{y}}\boldsymbol{F})\)，其中
\[
J_{\boldsymbol{x}}\boldsymbol{F} = \left[
\begin{matrix}
\frac{\partial F_1}{\partial x_1} & \cdots &
\frac{\partial F_1}{\partial x_n} \\
\vdots & & \vdots\\
\frac{\partial F_m}{\partial x_1} & \cdots &
\frac{\partial F_m}{\partial x_n}
\end{matrix}
\right], \quad
J_{\boldsymbol{y}}\boldsymbol{F} = \left[
\begin{matrix}
\frac{\partial F_1}{\partial y_1} & \cdots &
\frac{\partial F_1}{\partial y_m} \\
\vdots & & \vdots \\
\frac{\partial F_m}{\partial y_1} & \cdots &
\frac{\partial F_m}{\partial y_m}
\end{matrix}
\right]
\]
\(J_{\boldsymbol{x}}\boldsymbol{F}\)是一个\(m \times n\)矩阵，\(J_{\boldsymbol{y}}\boldsymbol{F}\)是一个\(m\)阶方阵。
定理3：隐映射定理
设开集\(D \subset \mathbb{R}^{n+m}\)，\(\boldsymbol{F}: D \to \mathbb{R}^m\)，满足下列条件：
（a）\(\boldsymbol{F} \in C^1(D)\)；
（b）有一点\((\boldsymbol{x}_0, \boldsymbol{y}_0) \in D\)，使得\(\boldsymbol{F}(\boldsymbol{x}_0, \boldsymbol{y}_0) = \boldsymbol{0}\)；
（c）行列式\(\det J_{\boldsymbol{y}} \boldsymbol{F}(\boldsymbol{x}_0, \boldsymbol{y}_0) \ne 0\)
那么存在\((\boldsymbol{x}_0, \boldsymbol{y}_0)\)的一个邻域\(G \times H\)，使得：
（1）对每一个\(\boldsymbol{x} \in G\)，方程\(\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y})=\boldsymbol{0}\)在\(H\)中有唯一的解，记为\(\boldsymbol{f}(\boldsymbol{x})\)；
（2）\(\boldsymbol{y}_0 = \boldsymbol{f}(\boldsymbol{x}_0)\)
（3）\(\boldsymbol{f} \in C^1(D)\)
（4）当\(\boldsymbol{x} \in G\)时，
\[ J\boldsymbol{f}(\boldsymbol{x}) = -(J_{\boldsymbol{y}}\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y})^{-1})J_{\boldsymbol{x}}\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}) \]
其中\(\boldsymbol{y} = \boldsymbol{f}(\boldsymbol{x})\)
证：对方程组的个数\(m\)进行归纳。当\(m=1\)时，即为该定理定理2。先设方程组个数为\(m-1\)时该定理成立，再证明在\(m\)时依然成立即可。
由于\(\det
J_{\boldsymbol{y}}\boldsymbol{F}(\boldsymbol{x}_0, \boldsymbol{y}_0) \ne
0\)，且\(\boldsymbol{F} \in
C^1(D)\)，所以总可以找到一个包含\((\boldsymbol{x}_0,
\boldsymbol{y}_0)\)的开集\(D^\prime\)满足\((\boldsymbol{x}_0, \boldsymbol{y}_0) \in D^\prime
\subset D\)，且在\(D^\prime\)上的每一个点处都有\(\det J_{\boldsymbol{y}}\boldsymbol{F} \ne
0\)。由条件\(（c）\)可知\(m\)阶方阵\(J_{\boldsymbol{y}}\boldsymbol{F}(\boldsymbol{x}_0,
\boldsymbol{y}_0)\)的元素不全为\(0\)，不妨设
\[
\frac{\partial F_m}{\partial y_j}(\boldsymbol{x}_0,
\boldsymbol{y}_0) \ne 0 \tag{1}
\]
同时令
\[
\boldsymbol{u} = (y_1, \cdots, y_{m-1}), \quad t = y_m, \quad
\boldsymbol{y} = (\boldsymbol{u}, t)
\]
同样可以定义\(\boldsymbol{y}_0 =
(\boldsymbol{u}_0, t_0)\)来规定\(\boldsymbol{u}_0\)和\(t_0\)的意义，从而式(1)可写成
\[
\frac{\partial F_m}{\partial t}(\boldsymbol{x}_0, \boldsymbol{u}_0,
t_0) \ne 0
\]
又有
\[
F_m (\boldsymbol{x}_0, \boldsymbol{u}_0, t_0) =
F_m(\boldsymbol{x}_0, \boldsymbol{y}_0) = 0
\]
由定理2可知，存在\((\boldsymbol{x}_0, \boldsymbol{u}_0,
t_0)\)的一个邻域\((G_n \times G_{m-1})
\times J \subset D^\prime\)，使得：
（i）对每一点\((\boldsymbol{x},
\boldsymbol{u}) \in G_{n} \times G_{m=1}\)，方程
\[
F_m(\boldsymbol{x}, \boldsymbol{u}, t) = 0
\]
在\(J\)中有唯一的解\(t = \varphi(\boldsymbol{x},
\boldsymbol{u})\)，这里函数\(\varphi:
G_n \times G_{m-1} \to J\)；
（ii）\(\varphi(\boldsymbol{x}_0,
\boldsymbol{u}_0) = t_0\)；
（iii）\(\varphi \in C^1(G_{n} \times
G_{m-1})\)；
这时将\(t = \varphi(\boldsymbol{x},
\boldsymbol{u})\)代入到原始方程中，即将\(y_m\)用\(x_1,\cdots,x_n,y_1,\cdots,y_{m-1}\)代入，
\[
\Phi_i(\boldsymbol{x}, \boldsymbol{u}) = F_i(\boldsymbol{x},
\boldsymbol{u}, \varphi(\boldsymbol{x}, \boldsymbol{u})) = 0 \quad
(i=1,2,\cdots,m-1) \tag{2}
\]
考虑映射
\[
\boldsymbol{\Phi} = \left[\begin{matrix}
\Phi_1 \\
\vdots \\
\Phi_{m-1}
\end{matrix}\right]: G_n \times G_{m-1} \to \mathbb{R}^{m-1}
\]
若能证明\(\boldsymbol{\Phi}\)满足定理的三个条件，便可使用归纳假设了。显然\(\boldsymbol{\Phi} \in C^1\)，并且
\[
\phi_i(\boldsymbol{x}_0, \boldsymbol{u}_0) = F_i(\boldsymbol{x}_0,
\boldsymbol{u}_0, \varphi(\boldsymbol{x}_0, \boldsymbol{u}_0)) = 0 \quad
(i=1,2\cdots,m-1)
\]
所以\(\boldsymbol{\Phi}(\boldsymbol{x}_0,
\boldsymbol{y}_0)=\boldsymbol{0}\)。对式(2)两边同时关于\(u_j\)（即\(y_j\)）\((j=1,2,\cdots,m-1)\)求导，得
\[
\frac{\partial \Phi_i}{\partial u_j} = \frac{\partial F_i}{\partial
y_j} + \frac{\partial F_i}{\partial y_m} \frac{\partial
\varphi}{\partial u_j} \quad (i,j=1,2,\cdots,m-1)
\]
又由（i）可知
\[
F_m(\boldsymbol{x}, \boldsymbol{u}, \varphi(\boldsymbol{x},
\boldsymbol{u})) = 0
\]
对上式也关于\(u_j\)（即\(y_j\)）\((j=1,2,\cdots,m-1)\)求导得
\[
\frac{\partial F_m}{\partial y_j} + \frac{\partial F_m}{\partial
y_m} \frac{\partial \varphi}{\partial u_j} = 0 \quad (j=1,2,\cdots,m-1)
\]
从而由
\[
\begin{aligned}
\left|
\begin{matrix}
\frac{\partial F_1}{\partial y_1} & \cdots &
\frac{\partial F_1}{\partial y_m} \\
\vdots & & \vdots \\
\frac{\partial F_m}{\partial y_1} & \cdots &
\frac{\partial F_m}{\partial y_m}
\end{matrix}
\right|
& =
\left|
\begin{matrix}
\frac{\partial F_1}{\partial y_1} + \frac{\partial
F_1}{\partial y_m}\frac{\partial \varphi}{\partial u_1} &
\frac{\partial F_1}{\partial y_2} + \frac{\partial F_1}{\partial
y_m}\frac{\partial \varphi}{\partial u_2} & \cdots &
\frac{\partial F_1}{\partial y_m} \\
\vdots & \vdots & & \vdots \\
\frac{\partial F_m}{\partial y_1} + \frac{\partial
F_m}{\partial y_m}\frac{\partial \varphi}{\partial u_1} &
\frac{\partial F_m}{\partial y_2} + \frac{\partial F_m}{\partial
y_m}\frac{\partial \varphi}{\partial u_2} & \cdots &
\frac{\partial F_m}{\partial y_m}
\end{matrix}
\right| \\
& =
\left|
\begin{matrix}
\frac{\partial \Phi_1}{\partial u_1} & \cdots &
\frac{\partial \Phi_1}{\partial u_{m-1}} & \frac{\partial
F_1}{\partial y_m} \\
\vdots & & \vdots & \vdots \\
\frac{\partial \Phi_{m-1}}{\partial u_1} & \cdots
& \frac{\partial \Phi_{m-1}}{\partial u_{m-1}} & \frac{\partial
F_{m-1}}{\partial y_m} \\
0 & \cdots & 0 & \frac{\partial
F_m}{\partial y_m}
\end{matrix}
\right| \\
& = \frac{\partial F_m}{\partial y_m}(\boldsymbol{x}_0,
\boldsymbol{u}_0, t_0) \det(J_{\boldsymbol{u}}
\boldsymbol{\Phi}(\boldsymbol{x}_0, \boldsymbol{u}_0))
\end{aligned} \tag{3}
\]
根据条件（c），式子(3)的左边不等于0，因为有
\[
\det(J_{\boldsymbol{u}} \boldsymbol{\Phi}(\boldsymbol{x}_0,
\boldsymbol{u}_0)) \ne 0
\]
从而证明了\(\boldsymbol{\Phi}\)满足本定理中的三个条件，从而对\(\boldsymbol{\Phi}\)使用归纳假设，可知定理中的结论（1），（2）和（3）对\(\boldsymbol{\Phi}\)都成立。即存在点\((\boldsymbol{x}_0,
\boldsymbol{y}_0)\)的邻域\(G \times
H_{m-1} \subset G_n \times G_{m-1}\)使得：
（aa）当\(\boldsymbol{x} \in
G\)时，方程\(\boldsymbol{\phi}(\boldsymbol{x}, \boldsymbol{u})
= \boldsymbol{0}\)在\(H_{m-1}\)中有唯一解\(\boldsymbol{u} =
\boldsymbol{g}(\boldsymbol{x})\)，其中映射\(\boldsymbol{g}: G \to H_{m-1}\)；
（ab）\(\boldsymbol{\boldsymbol{x}_0} =
\boldsymbol{u}_0\)；
（ac）\(\boldsymbol{g} \in
C^1(G)\)
令
\[
\boldsymbol{f}(\boldsymbol{x}) = (\boldsymbol{g}(\boldsymbol{x}),
\varphi(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x}))) \quad
(\boldsymbol{x} \in G)
\]
\[
H = H_{m-1} \times J
\]
于是\(\boldsymbol{f}: G \to
H\)。我们要证明\(\boldsymbol{f}\)满足条件（1），（2）和（3）。当\(\boldsymbol{x} \in G\)，\((\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x}))
\in G \times H_{m-1} \subset G_n \times
G_{m-1}\)，从而由（aa）可得
\[
F_i(\boldsymbol{x}, \boldsymbol{f}(\boldsymbol{x})) =
F_i(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x}),
\varphi(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x}))) =
\Phi_i(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x})) = 0 \quad
(i=1,2,\cdots,m-1)
\]
另外由（i）可知
\[
F_m(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x})) =
F_m(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x}),
\varphi(\boldsymbol{x}, \boldsymbol{g}(\boldsymbol{x}))) = 0
\]
从而\(\boldsymbol{f}\)满足（1）。由（ab）和（ii）可知
\[
\boldsymbol{f}(\boldsymbol{x}_0) =
(\boldsymbol{g}(\boldsymbol{x}_0), \varphi(\boldsymbol{x}_0,
\boldsymbol{g}(\boldsymbol{x}_0))) = (\boldsymbol{u}_0,
\varphi(\boldsymbol{x}_0, \boldsymbol{u}_0)) = (\boldsymbol{u}_0, t_0) =
\boldsymbol{y}_0
\]
所以\(\boldsymbol{f}\)满足（2）。再由（ac）和（iii）即知\(\boldsymbol{f}\)满足（3）。再由恒等式
\[
\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{f}(\boldsymbol{x})) =
\boldsymbol{0}
\]
对上式复合求导，得
\[
J_{\boldsymbol{x}} \boldsymbol{F}(\boldsymbol{x},
\boldsymbol{f}(\boldsymbol{x})) + J_y \boldsymbol{F}(\boldsymbol{x},
\boldsymbol{f}(\boldsymbol{x})) J\boldsymbol{f}(\boldsymbol{x}) =
\boldsymbol{0}
\]
由于在\(D^\prime\)上\(\det
J_{\boldsymbol{y}}\boldsymbol{F}\)处处不为0，所以\(J_{\boldsymbol{y}}\boldsymbol{F}\)是可逆方阵，在上式中取逆方阵，得出
\[
J\boldsymbol{f}(\boldsymbol{x}) =
-(J_{\boldsymbol{y}}\boldsymbol{F}(\boldsymbol{x},
\boldsymbol{f}(\boldsymbol{x})))^{-1}J_{\boldsymbol{x}}\boldsymbol{F}(\boldsymbol{x},
\boldsymbol{f}(\boldsymbol{x}))
\]
即表明\(\boldsymbol{f}\)满足（4）。
Q.E.D.