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函数导数十一:逆映射定理

定理1: 局部逆映射定理

设开集\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^n\),满足:
(a)\(\boldsymbol{f} \in C^1(D)\)
(b)有\(\boldsymbol{x}_0 \in D\),使得
\[ \det J\boldsymbol{f}(\boldsymbol{x}_0) \ne 0 \]
\(\boldsymbol{y}_0 = \boldsymbol{f}(\boldsymbol{x}_0)\),那么存在\(\boldsymbol{x}_0\)的一个邻域\(U\)\(\boldsymbol{y}_0\)的一个邻域\(V\),使得
(1)\(\boldsymbol{f}(U) = V\),且\(\boldsymbol{f}\)\(U\)上是单射;
(2)记\(\boldsymbol{g}\)\(\boldsymbol{f}\)\(U\)上的逆映射,\(\boldsymbol{g} \in C^1(V)\)
(3)当\(\boldsymbol{y} \in V\)时,
\[ J\boldsymbol{g}(\boldsymbol{y}) = (J\boldsymbol{f}(\boldsymbol{x}))^{-1} \]
其中\(\boldsymbol{x} = \boldsymbol{g}(\boldsymbol{y})\)

证:令
\[ \boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}) = \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{y} \]
这个映射定义在\(D \times \mathbb{R}^n\)上,显然有\(\boldsymbol{F} \in C^1(D \times \mathbb{R}^n)\),并且、
\[ \boldsymbol{F}(\boldsymbol{x}_0, \boldsymbol{y}_0) = \boldsymbol{f}(\boldsymbol{x}_0) - \boldsymbol{y}_0 = \boldsymbol{0} \]
再由条件(b)可知,
\[ \det J_{\boldsymbol{x}} \boldsymbol{F} (\boldsymbol{x}_0, \boldsymbol{y}_0) = \det J\boldsymbol{f}(\boldsymbol{x}_0) \ne 0 \]
从而由函数导数十的定理3可知,存在\(\boldsymbol{x}_0\)的邻域\(H\)\(\boldsymbol{y}_0\)的邻域\(V\),其中\(H \subset D\),使得对每一点\(\boldsymbol{y} \in V\),方程\(\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}) = \boldsymbol{0}\)\(\boldsymbol{f}(\boldsymbol{x}) = \boldsymbol{y}\)\(H\)中有唯一解,记作\(\boldsymbol{g}(\boldsymbol{y})\),其中\(\boldsymbol{g} \in C^1(V)\),并且当\(\boldsymbol{y} \in V\)时,
\[ \begin{aligned} J\boldsymbol{g}(\boldsymbol{y}) &= -(J_{\boldsymbol{x}}\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}))^{-1}J_{\boldsymbol{y}}\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}) \\ &= -(J\boldsymbol{f}(\boldsymbol{x}))^{-1}(-\boldsymbol{I}_n) \\ &= (J\boldsymbol{f}(\boldsymbol{x}))^{-1} \end{aligned} \]
其中\(\boldsymbol{x} = \boldsymbol{g}(\boldsymbol{y})\)。令\(U = \boldsymbol{g}(V)\),可见\(V = \boldsymbol{f}(U)\)\(\boldsymbol{f}\)\(\boldsymbol{U}\)互为逆映射,最后证明\(U\)是开集,事实上有\(U = H \cap \boldsymbol{f}^{-1}(V)\),由于\(V\)是开集且\(\boldsymbol{f}\)是连续映射,从而根据函数极限七的定理3\(\boldsymbol{f}^{-1}(V)\)是开集,又英文\(H\)是开集,所以\(U\)也是开集。

定理2:逆映射定理

设开集\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^n\),满足:
(a)\(\boldsymbol{f} \in C^1(D)\)
(b)对每一个\(\boldsymbol{x} \in D\),有\(\det J\boldsymbol{f}(\boldsymbol{x}) \ne 0\)
那么\(G = \boldsymbol{f}(D)\)是一开集,又如果:
(c)\(\boldsymbol{f}\)\(D\)上的单射,
那么:
(1)存在从\(G\)\(D\)上的映射\(\boldsymbol{f}^{-1}\),满足:对一切\(\boldsymbol{y} \in G\),有
\[ \boldsymbol{f} \circ \boldsymbol{f}^{-1}(\boldsymbol{y}) = \boldsymbol{y} \]
(2)\(\boldsymbol{f}^{-1} \in C^1(G)\)
(3)
\[ J\boldsymbol{f}^{-1}(\boldsymbol{y}) = (J\boldsymbol{f}(\boldsymbol{x}))^{-1} \quad (\boldsymbol{x} = J\boldsymbol{f}^{-1}(\boldsymbol{y})) \]

证:由于\(\boldsymbol{f}\)是单射,所以逆映射\(\boldsymbol{f}^{-1}\)必然存在,所以(1)成立;再由定理1可知,(2)(3)自然也成立。接下来只需证明\(\boldsymbol{f}(D)\)是开集,任取\(\boldsymbol{y} \in G\),由定理1可知,存在\(\boldsymbol{x}\)的一个邻域\(U\)\(\boldsymbol{y}\)的一个邻域\(V\),使得\(V = \boldsymbol{f}(U) \subset \boldsymbol{f}(D) = G\),即\(\boldsymbol{y}\)\(G\)的内点,从而\(G\)是开集。

定义2:正则映射

设开集\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^n\),满足以下三个条件:
(1)\(\boldsymbol{f} \in C^1(D)\)
(2)\(\boldsymbol{f}\)\(D\)上的单射;
(3)\(\det J\boldsymbol{f}(\boldsymbol{x}) \ne 0\)对一切\(\boldsymbol{x} \in D\)成立。
则称\(\boldsymbol{f}\)\(D\)上的一个正则映射。